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

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

- 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

Security

- 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

- 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

Reuse

- 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?

CopyDownload

```
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

CopyDownload

```
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

CopyDownload

```
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++

CopyDownload

```
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

CopyDownload

```
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 "Process Tasks Using Servers" asked in leetcode weekly contest 243

CopyDownload

```
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

CopyDownload

```
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 "stars and bars" implementation?

CopyDownload

```
'**|*|**|**'.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?

CopyDownload

```
>>> 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

CopyDownload

```
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
```

QUESTION

LeetCode TwoSum question returning buggy answer (C Implementation)

Asked 2021-Jun-15 at 16:06Here 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 16:06You 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.

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

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You can use LeetCode like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the LeetCode component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

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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