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LeetCode | LeetCode刷题记录与面试整理 | Learning library

 by   yuanguangxin Java Version: Current License: No License

 by   yuanguangxin Java Version: Current License: No License

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kandi X-RAY | LeetCode Summary

LeetCode is a Java library typically used in Tutorial, Learning, Example Codes, LeetCode applications. LeetCode has no vulnerabilities and it has medium support. However LeetCode has 3 bugs and it build file is not available. You can download it from GitHub.
LeetCode刷题记录与面试整理
Support
Support
Quality
Quality
Security
Security
License
License
Reuse
Reuse

kandi-support Support

  • LeetCode has a medium active ecosystem.
  • It has 5890 star(s) with 1626 fork(s). There are 191 watchers for this library.
  • It had no major release in the last 12 months.
  • There are 0 open issues and 5 have been closed. On average issues are closed in 0 days. There are no pull requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of LeetCode is current.
LeetCode Support
Best in #Learning
Average in #Learning
LeetCode Support
Best in #Learning
Average in #Learning

quality kandi Quality

  • LeetCode has 3 bugs (0 blocker, 0 critical, 3 major, 0 minor) and 267 code smells.
LeetCode Quality
Best in #Learning
Average in #Learning
LeetCode Quality
Best in #Learning
Average in #Learning

securitySecurity

  • LeetCode has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
  • LeetCode code analysis shows 0 unresolved vulnerabilities.
  • There are 6 security hotspots that need review.
LeetCode Security
Best in #Learning
Average in #Learning
LeetCode Security
Best in #Learning
Average in #Learning

license License

  • LeetCode does not have a standard license declared.
  • Check the repository for any license declaration and review the terms closely.
  • Without a license, all rights are reserved, and you cannot use the library in your applications.
LeetCode License
Best in #Learning
Average in #Learning
LeetCode License
Best in #Learning
Average in #Learning

buildReuse

  • LeetCode releases are not available. You will need to build from source code and install.
  • LeetCode has no build file. You will be need to create the build yourself to build the component from source.
  • LeetCode saves you 1466 person hours of effort in developing the same functionality from scratch.
  • It has 3273 lines of code, 224 functions and 133 files.
  • It has low code complexity. Code complexity directly impacts maintainability of the code.
LeetCode Reuse
Best in #Learning
Average in #Learning
LeetCode Reuse
Best in #Learning
Average in #Learning
Top functions reviewed by kandi - BETA

kandi has reviewed LeetCode and discovered the below as its top functions. This is intended to give you an instant insight into LeetCode implemented functionality, and help decide if they suit your requirements.

  • Returns the spiral order
    • set zeros into matrix
      • Resolves a mapping between start and end indices .
        • Write lrd from a tree .
          • The main loop .
            • Pop the current element .
              • push a value back to max
                • Adds an integer .
                  • Removes the next element from the queue .
                    • Gets the value at the top of the stack .

                      Get all kandi verified functions for this library.

                      Get all kandi verified functions for this library.

                      LeetCode Key Features

                      LeetCode刷题记录与面试整理

                      Why is the output an empty list, when I call a function that skips nodes in a linked-list?

                      copy iconCopydownload iconDownload
                      class Solution:
                          def removeElements(self, head: ListNode, val: int) -> ListNode:
                              curr = head
                              dummy = ListNode(0, head)  # Link to head!
                              prev = dummy
                      
                              while curr: 
                                  if curr.val == val:    
                                      prev.next = curr.next  # This is all that needs to happen
                                  else: 
                                      prev = curr
                                  curr = curr.next  # This should happen in both cases
                                  # The loop invariant is: prev.next == curr
                      
                              return dummy.next        
                      
                      class Solution:
                          def removeElements(self, head: ListNode, val: int) -> ListNode:
                              curr = head
                              dummy = ListNode(0, head)
                              prev = dummy
                      
                              def skipper(curr):
                                  while curr is not None and curr.val == val:
                                      curr = curr.next
                                  return curr
                              
                              while curr: 
                                  if curr.val == val:    
                                      prev.next = skipper(curr)
                                  else: 
                                      prev = curr
                                  curr = prev.next
                      
                              return dummy.next
                      
                      class Solution:
                          def removeElements(self, head: ListNode, val: int) -> ListNode:
                              curr = head
                              dummy = ListNode(0, head)  # Link to head!
                              prev = dummy
                      
                              while curr: 
                                  if curr.val == val:    
                                      prev.next = curr.next  # This is all that needs to happen
                                  else: 
                                      prev = curr
                                  curr = curr.next  # This should happen in both cases
                                  # The loop invariant is: prev.next == curr
                      
                              return dummy.next        
                      
                      class Solution:
                          def removeElements(self, head: ListNode, val: int) -> ListNode:
                              curr = head
                              dummy = ListNode(0, head)
                              prev = dummy
                      
                              def skipper(curr):
                                  while curr is not None and curr.val == val:
                                      curr = curr.next
                                  return curr
                              
                              while curr: 
                                  if curr.val == val:    
                                      prev.next = skipper(curr)
                                  else: 
                                      prev = curr
                                  curr = prev.next
                      
                              return dummy.next
                      

                      how can i check if a string contains another string throughout it

                      copy iconCopydownload iconDownload
                      import re
                      l1 = ["heollo", "leetcode"]
                      l2 = ["e", "oo"]
                      patterns = ['.*'.join(list(item)) for item in l2]
                      # patterns is ['e', 'o.*o']
                      result = [item for item in l1 if all(
                          [re.search(pattern, item) is not None for pattern in patterns]
                      )]
                      # ['heollo']
                      

                      Time and Space complexity of Palindrome Partitioning II

                      copy iconCopydownload iconDownload
                      T(0) = 1
                      T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                      # and for simplicity, I will just use
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      
                      T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                      T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                      T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                      T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                      
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                      T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                      
                      # with
                      T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                      T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                      
                      # we can substitute:
                      T(n) = 1 + T(n-1) + 2 * T(n-1)
                      T(n) = 1 + 3 * T(n-1)
                      
                      S(0) = 1
                      S(n) = 1 + S(n-1)
                      
                      T(0) = 1
                      T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                      # and for simplicity, I will just use
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      
                      T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                      T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                      T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                      T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                      
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                      T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                      
                      # with
                      T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                      T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                      
                      # we can substitute:
                      T(n) = 1 + T(n-1) + 2 * T(n-1)
                      T(n) = 1 + 3 * T(n-1)
                      
                      S(0) = 1
                      S(n) = 1 + S(n-1)
                      
                      T(0) = 1
                      T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                      # and for simplicity, I will just use
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      
                      T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                      T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                      T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                      T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                      
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                      T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                      
                      # with
                      T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                      T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                      
                      # we can substitute:
                      T(n) = 1 + T(n-1) + 2 * T(n-1)
                      T(n) = 1 + 3 * T(n-1)
                      
                      S(0) = 1
                      S(n) = 1 + S(n-1)
                      
                      T(0) = 1
                      T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                      # and for simplicity, I will just use
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      
                      T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                      T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                      T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                      T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                      
                      T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                      T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                      T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                      
                      # with
                      T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                      T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                      
                      # we can substitute:
                      T(n) = 1 + T(n-1) + 2 * T(n-1)
                      T(n) = 1 + 3 * T(n-1)
                      
                      S(0) = 1
                      S(n) = 1 + S(n-1)
                      

                      Unable to resolve memory read and segmentation fault error in C++

                      copy iconCopydownload iconDownload
                          void find_island(vector<vector<int>> & arg_grid, int* arg_coordinates) {
                              if (arg_grid[*arg_coordinates][*(arg_coordinates + 1)] == INT_MAX) {
                                  *last = *arg_coordinates;
                                  *(last + 1) = *(arg_coordinates + 1);
                                  map_island(arg_grid, arg_coordinates, 1);
                              } else {
                                  if (*(arg_coordinates + 1) < arg_grid[*arg_coordinates].size()) {
                                      *(arg_coordinates + 1) = *(arg_coordinates + 1) + 1;
                                      find_island(arg_grid, arg_coordinates);
                                  }
                                  if (*(arg_coordinates + 1) == arg_grid[*arg_coordinates].size()) {
                                      if (*arg_coordinates == arg_grid.size()) return;
                                      *arg_coordinates = *arg_coordinates + 1;
                                      *(arg_coordinates + 1) = 0;
                                      find_island(arg_grid, arg_coordinates);
                                  }
                              }
                          }
                      
                      *arg_coordinates = *arg_coordinates + 1;
                      if (*arg_coordinates == arg_grid.size()) return;
                      
                          void find_island(vector<vector<int>> & arg_grid, int* arg_coordinates) {
                              if (arg_grid[*arg_coordinates][*(arg_coordinates + 1)] == INT_MAX) {
                                  *last = *arg_coordinates;
                                  *(last + 1) = *(arg_coordinates + 1);
                                  map_island(arg_grid, arg_coordinates, 1);
                              } else {
                                  if (*(arg_coordinates + 1) < arg_grid[*arg_coordinates].size()) {
                                      *(arg_coordinates + 1) = *(arg_coordinates + 1) + 1;
                                      find_island(arg_grid, arg_coordinates);
                                  }
                                  if (*(arg_coordinates + 1) == arg_grid[*arg_coordinates].size()) {
                                      if (*arg_coordinates == arg_grid.size()) return;
                                      *arg_coordinates = *arg_coordinates + 1;
                                      *(arg_coordinates + 1) = 0;
                                      find_island(arg_grid, arg_coordinates);
                                  }
                              }
                          }
                      
                      *arg_coordinates = *arg_coordinates + 1;
                      if (*arg_coordinates == arg_grid.size()) return;
                      

                      Detecting signed integer multiplication overflow in C

                      copy iconCopydownload iconDownload
                      reversed += (x % 10) * tenPower;
                      
                      ((x % 10) * tenPower) > INT_MAX
                      
                      (reversed + (x % 10) * tenPower) > INT_MAX
                      
                      if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                            (((x % 10) * tenPower) > INT_MAX - reversed))
                          return 0;
                      
                      reversed += (x % 10) * tenPower;
                      
                      ((x % 10) * tenPower) > INT_MAX
                      
                      (reversed + (x % 10) * tenPower) > INT_MAX
                      
                      if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                            (((x % 10) * tenPower) > INT_MAX - reversed))
                          return 0;
                      
                      reversed += (x % 10) * tenPower;
                      
                      ((x % 10) * tenPower) > INT_MAX
                      
                      (reversed + (x % 10) * tenPower) > INT_MAX
                      
                      if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                            (((x % 10) * tenPower) > INT_MAX - reversed))
                          return 0;
                      
                      reversed += (x % 10) * tenPower;
                      
                      ((x % 10) * tenPower) > INT_MAX
                      
                      (reversed + (x % 10) * tenPower) > INT_MAX
                      
                      if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                            (((x % 10) * tenPower) > INT_MAX - reversed))
                          return 0;
                      
                      int reverse(int n) {
                          // 1st overflow checking...
                          // Check if the absolute value is greater than INT32_MAX.
                          // I converted "int" to "long" to get correct overflow value.
                          if (n < 0 && (long) n * -1l > INT32_MAX) {
                              return 0;
                          }
                          
                          int res = 0;
                          // Convert to absolute value.
                          int tmp = n < 0 ? n * -1 : n;
                          
                          while (tmp > 0) {
                              // Get the right most digit and add it to the current result.
                              res += tmp % 10;
                              // Remove the right most digit.
                              tmp /= 10;
                              
                              // "tmp" still has remaining numbers.
                              if (tmp > 0) {
                                  // 2nd overflow checking...
                                  // Check if reversed value will be greater than INT32_MAX when appending 0 to right most.
                                  // I converted "int" to "long" to get correct overflow value.
                                  if ((long) res * 10l > INT32_MAX) {
                                      return 0;
                                  }
                                  
                                  // Append 0 to right most value of result.
                                  // If result is equal to 0, do not append 0.
                                  res *= res == 0 ? 1 : 10;
                              }
                          }
                          
                          // Return result.
                          // If original value is negative, return negative result value..
                          return n < 0 ? res * -1 : res;
                      }
                      
                        if (x < 0) { pos = -x; }
                        ...
                        (x % 10) * tenPower
                        ....
                        reversed += (x % 10) * tenPower;
                        ...
                        if (x < 0) return -reversed;
                      
                        if (x < 0) { 
                          if (x < -INT_MAX) { puts("Overflow"); return 0; }
                          pos = -x;
                        }
                      
                        // x is >= 0 here, tenPower >= 1
                        int digit = x % 10;
                        if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
                        int digit10 = digit * tenPower;
                        if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
                        reversed += digit10;
                      
                      // Return 1 on overflow
                      int is_undefined_mult1(int a, int b) {
                        if (a > 0) {
                          if (b > 0) {
                            return a > INT_MAX / b;       // a positive, b positive
                          }
                          return b < INT_MIN / a;         // a positive, b not positive
                        }
                        if (b > 0) {
                          return a < INT_MIN / b;         // a not positive, b positive
                        }
                        return a != 0 && b < INT_MAX / a; // a not positive, b not positive
                      }
                      
                        int digit = x % 10;
                        if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
                        int digit10 = digit * tenPower;
                        if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
                        reversed += digit10;
                      
                        if (x < 0) { pos = -x; }
                        ...
                        (x % 10) * tenPower
                        ....
                        reversed += (x % 10) * tenPower;
                        ...
                        if (x < 0) return -reversed;
                      
                        if (x < 0) { 
                          if (x < -INT_MAX) { puts("Overflow"); return 0; }
                          pos = -x;
                        }
                      
                        // x is >= 0 here, tenPower >= 1
                        int digit = x % 10;
                        if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
                        int digit10 = digit * tenPower;
                        if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
                        reversed += digit10;
                      
                      // Return 1 on overflow
                      int is_undefined_mult1(int a, int b) {
                        if (a > 0) {
                          if (b > 0) {
                            return a > INT_MAX / b;       // a positive, b positive
                          }
                          return b < INT_MIN / a;         // a positive, b not positive
                        }
                        if (b > 0) {
                          return a < INT_MIN / b;         // a not positive, b positive
                        }
                        return a != 0 && b < INT_MAX / a; // a not positive, b not positive
                      }
                      
                        int digit = x % 10;
                        if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
                        int digit10 = digit * tenPower;
                        if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
                        reversed += digit10;
                      
                        if (x < 0) { pos = -x; }
                        ...
                        (x % 10) * tenPower
                        ....
                        reversed += (x % 10) * tenPower;
                        ...
                        if (x < 0) return -reversed;
                      
                        if (x < 0) { 
                          if (x < -INT_MAX) { puts("Overflow"); return 0; }
                          pos = -x;
                        }
                      
                        // x is >= 0 here, tenPower >= 1
                        int digit = x % 10;
                        if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
                        int digit10 = digit * tenPower;
                        if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
                        reversed += digit10;
                      
                      // Return 1 on overflow
                      int is_undefined_mult1(int a, int b) {
                        if (a > 0) {
                          if (b > 0) {
                            return a > INT_MAX / b;       // a positive, b positive
                          }
                          return b < INT_MIN / a;         // a positive, b not positive
                        }
                        if (b > 0) {
                          return a < INT_MIN / b;         // a not positive, b positive
                        }
                        return a != 0 && b < INT_MAX / a; // a not positive, b not positive
                      }
                      
                        int digit = x % 10;
                        if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
                        int digit10 = digit * tenPower;
                        if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
                        reversed += digit10;
                      
                      if digit.bit_length() >= 32:
                              return 0
                          else:
                              return reversed
                      

                      Not being able to find an edge case in question &quot;Process Tasks Using Servers&quot; asked in leetcode weekly contest 243

                      copy iconCopydownload iconDownload
                      serverPriority = [(name, index) for index, name in enumerate(servers)]
                              
                              serverPriority.sort()
                      
                              serverId = [i for i in range(len(servers))]
                              heapq.heapify(serverId)
                              
                              res = [-1] * len(tasks)
                              
                              runningServers = []
                              heapq.heapify(runningServers)
                              
                              currTime = 0
                              j = 0
                              while j < len(tasks):
                                  currTime = max(currTime, j)
                                  
                                  if len(runningServers) == 0:
                                      ID = heapq.heappop(serverId)
                                      res[j] = serverPriority[ID][1]
                                      
                                      heapq.heappush(runningServers, (currTime + tasks[j], ID))
                                  else:
                                      # Here, we are first freeing up the servers
                                      while len(runningServers) > 0 and currTime >= runningServers[0][0]:
                                          time, ID = heapq.heappop(runningServers)
                                          heapq.heappush(serverId, ID)
                                  
                                      if len(serverId) > 0:
                                          ID = heapq.heappop(serverId)
                                          res[j] = serverPriority[ID][1]
                      
                                          heapq.heappush(runningServers, (currTime + tasks[j], ID))
                                      else:
                                          """
                                           here, we are keeping record of the server 
                                           with earliest finish time.
                                          """
                                          time = runningServers[0][0]
                                          currTime = time
                                          j -= 1
                                  
                                  j += 1
                                  
                              return res
                      

                      Why does my operation of mod (%) on a 2 digit number return zero in python

                      copy iconCopydownload iconDownload
                              if number >= 1000:
                                  thousands = number - (number % 1000)              # 3549 - (3549 % 1000) = 3000
                      
                                  hundreds = number - thousands;                  #// 3549 - 3000 = 549
                                  tens = hundreds % 100                      #// 549 % 100 = 49
                                  hundreds = hundreds - (hundreds % 100)     #// 549 - (549 % 100) = 500
                      
                                  unit = tens % 10 # you input "units" here
                                  tens = tens - (tens % 10)                  #// 49 - (49 % 10) = 49 - 9 = 9
                      
                                  thou = thousands // 1000                     # 3
                      
                      MMMDXLIX
                      
                      def intToRoman(num: int) -> str:
                          roman_dict_key = [1,4,5,9,10,40,50,90,100,400,500,900,1000][::-1]
                          roman_dict_value = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"][::-1]
                          roman_dict = dict(zip(roman_dict_key,roman_dict_value))
                          ans = ""
                          for digit,roman in roman_dict.items():
                              while num>=digit:
                                  num -= digit
                                  ans += roman
                          return ans
                      
                              if number >= 1000:
                                  thousands = number - (number % 1000)              # 3549 - (3549 % 1000) = 3000
                      
                                  hundreds = number - thousands;                  #// 3549 - 3000 = 549
                                  tens = hundreds % 100                      #// 549 % 100 = 49
                                  hundreds = hundreds - (hundreds % 100)     #// 549 - (549 % 100) = 500
                      
                                  unit = tens % 10 # you input "units" here
                                  tens = tens - (tens % 10)                  #// 49 - (49 % 10) = 49 - 9 = 9
                      
                                  thou = thousands // 1000                     # 3
                      
                      MMMDXLIX
                      
                      def intToRoman(num: int) -> str:
                          roman_dict_key = [1,4,5,9,10,40,50,90,100,400,500,900,1000][::-1]
                          roman_dict_value = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"][::-1]
                          roman_dict = dict(zip(roman_dict_key,roman_dict_value))
                          ans = ""
                          for digit,roman in roman_dict.items():
                              while num>=digit:
                                  num -= digit
                                  ans += roman
                          return ans
                      
                              if number >= 1000:
                                  thousands = number - (number % 1000)              # 3549 - (3549 % 1000) = 3000
                      
                                  hundreds = number - thousands;                  #// 3549 - 3000 = 549
                                  tens = hundreds % 100                      #// 549 % 100 = 49
                                  hundreds = hundreds - (hundreds % 100)     #// 549 - (549 % 100) = 500
                      
                                  unit = tens % 10 # you input "units" here
                                  tens = tens - (tens % 10)                  #// 49 - (49 % 10) = 49 - 9 = 9
                      
                                  thou = thousands // 1000                     # 3
                      
                      MMMDXLIX
                      
                      def intToRoman(num: int) -> str:
                          roman_dict_key = [1,4,5,9,10,40,50,90,100,400,500,900,1000][::-1]
                          roman_dict_value = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"][::-1]
                          roman_dict = dict(zip(roman_dict_key,roman_dict_value))
                          ans = ""
                          for digit,roman in roman_dict.items():
                              while num>=digit:
                                  num -= digit
                                  ans += roman
                          return ans
                      

                      What is wrong with my &quot;stars and bars&quot; implementation?

                      copy iconCopydownload iconDownload
                      '**|*|**|**'.scan(/(?<=\|)\*+(?=\|)/).sum(&:size) 
                        #=> 3 
                      
                      def find_em(str, start_idx, end_idx)
                         start_idx.zip(end_idx).map do |s,e|
                           next 0 unless str[s..e].include?('|')
                           str[str.index('|', s)..str.rindex('|', e)].count('*')
                         end
                      end
                      
                      find_em('**|*|**|**', [1, 1, 1, 0, 0], [6, 7, 8, 1, 3])
                        #=> [1, 3, 3, 0, 0]
                      
                      '**|*|**|**'.scan(/(?<=\|)\*+(?=\|)/).sum(&:size) 
                        #=> 3 
                      
                      def find_em(str, start_idx, end_idx)
                         start_idx.zip(end_idx).map do |s,e|
                           next 0 unless str[s..e].include?('|')
                           str[str.index('|', s)..str.rindex('|', e)].count('*')
                         end
                      end
                      
                      find_em('**|*|**|**', [1, 1, 1, 0, 0], [6, 7, 8, 1, 3])
                        #=> [1, 3, 3, 0, 0]
                      
                      '**|*|**|**'.scan(/(?<=\|)\*+(?=\|)/).sum(&:size) 
                        #=> 3 
                      
                      def find_em(str, start_idx, end_idx)
                         start_idx.zip(end_idx).map do |s,e|
                           next 0 unless str[s..e].include?('|')
                           str[str.index('|', s)..str.rindex('|', e)].count('*')
                         end
                      end
                      
                      find_em('**|*|**|**', [1, 1, 1, 0, 0], [6, 7, 8, 1, 3])
                        #=> [1, 3, 3, 0, 0]
                      

                      Does Counter from collections maintain order after frequency has been considered, if all other frequencies are the same?

                      copy iconCopydownload iconDownload
                      >>> print(cnt.keys())
                      dict_keys(['l', 'e', 't', 'c', 'o', 'd'])
                      
                      def __repr__(self):
                          if not self:
                              return f'{self.__class__.__name__}()'
                          try:
                              # dict() preserves the ordering returned by most_common()
                              d = dict(self.most_common())
                          except TypeError:
                              # handle case where values are not orderable
                              d = dict(self)
                          return f'{self.__class__.__name__}({d!r})'
                      
                      >>> print(cnt.keys())
                      dict_keys(['l', 'e', 't', 'c', 'o', 'd'])
                      
                      def __repr__(self):
                          if not self:
                              return f'{self.__class__.__name__}()'
                          try:
                              # dict() preserves the ordering returned by most_common()
                              d = dict(self.most_common())
                          except TypeError:
                              # handle case where values are not orderable
                              d = dict(self)
                          return f'{self.__class__.__name__}({d!r})'
                      

                      Remove Nth Node From End of List LeetCode

                      copy iconCopydownload iconDownload
                      if not pointer1.next:
                          return pointer1.next
                      
                      def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
                          pointer1 = head
                          pointer2 = head
                          count = 0
                          while pointer2 is not None:
                              if count > n:
                                  pointer1 = pointer1.next
                              pointer2 = pointer2.next
                              count += 1
                          if count > n:
                              pointer1.next = pointer1.next.next
                              return head  
                          else:
                              return head.next
                      
                      if not pointer1.next:
                          return pointer1.next
                      
                      def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
                          pointer1 = head
                          pointer2 = head
                          count = 0
                          while pointer2 is not None:
                              if count > n:
                                  pointer1 = pointer1.next
                              pointer2 = pointer2.next
                              count += 1
                          if count > n:
                              pointer1.next = pointer1.next.next
                              return head  
                          else:
                              return head.next
                      

                      Community Discussions

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                      Trending Discussions on LeetCode

                      QUESTION

                      LeetCode TwoSum question returning buggy answer (C Implementation)

                      Asked 2021-Jun-15 at 16:06

                      Here is the question:

                      Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
                      
                      You may assume that each input would have exactly one solution, and you may not use the same element twice.
                      
                      You can return the answer in any order
                      

                      Here is my solution for the same:

                      int* twoSum(int* nums, int numsSize, int target, int* returnSize){
                          
                          int* result= (int*)malloc(2*sizeof(int));
                          int* iterator= nums;
                          int sum=0;
                          
                          for (int i=0; i<numsSize; i++)
                          {
                              for (int j=0; j<numsSize; j++)
                              {
                                  if (i!=j)
                                  { 
                                      sum=iterator[i]+iterator[j];
                                      if (sum==target)
                                      {
                                          result[0]=i;
                                          result[1]=j;
                                          return result;
                                      }
                                  }
                              }
                          }
                      
                          return NULL;
                      
                      }
                      
                      

                      I know that it is not the most optimal solution, but I assumed it would work. LeetCode shows me that the answer returned is [] or ]. It seems to work fine on VSCode. I was wondering if I could get some help regarding what the error might be.

                      ANSWER

                      Answered 2021-Jun-15 at 16:06

                      You have to assign the number of elements of the array to return (2 in this case) to what the argument returnSize points at (*returnSize) to tell the judge system how large the array you returned is.

                      Source https://stackoverflow.com/questions/67989752

                      Community Discussions, Code Snippets contain sources that include Stack Exchange Network

                      Vulnerabilities

                      No vulnerabilities reported

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                      Support

                      For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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