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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 .

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?

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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

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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

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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++

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    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

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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

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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

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        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?

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'**|*|**|**'.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

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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

Install LeetCode

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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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