Merge branch 'main' of https://github.com/pranjal030404/leetcode.doocs

Author: pranjal030404

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README.md

@@ -303,29 +303,28 @@ class Solution {
 #### C
 
 ```c
-int lengthOfLongestSubstring(char *s) {
-  int freq[256] = {0};
-  int l = 0, r = 0;
-  int ans = 0;
-  int len = strlen(s);
-
-  for (r = 0; r < len; r++) {
-    char c = s[r];
-    freq[(unsigned char)c]++;
-
-    while (freq[(unsigned char)c] > 1) {
-      freq[(unsigned char)s[l]]--;
-      l++;
-    }
+int lengthOfLongestSubstring(char* s) {
+    int freq[256] = {0};
+    int l = 0, r = 0;
+    int ans = 0;
+    int len = strlen(s);
+
+    for (r = 0; r < len; r++) {
+        char c = s[r];
+        freq[(unsigned char) c]++;
+
+        while (freq[(unsigned char) c] > 1) {
+            freq[(unsigned char) s[l]]--;
+            l++;
+        }
 
-    if (ans < r - l + 1) {
-      ans = r - l + 1;
+        if (ans < r - l + 1) {
+            ans = r - l + 1;
+        }
     }
-  }
 
-  return ans;
+    return ans;
 }
-
 ```
 
 <!-- tabs:end -->

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README_EN.md

@@ -301,30 +301,28 @@ class Solution {
 #### C
 
 ```c
+int lengthOfLongestSubstring(char* s) {
+    int freq[256] = {0};
+    int l = 0, r = 0;
+    int ans = 0;
+    int len = strlen(s);
+
+    for (r = 0; r < len; r++) {
+        char c = s[r];
+        freq[(unsigned char) c]++;
+
+        while (freq[(unsigned char) c] > 1) {
+            freq[(unsigned char) s[l]]--;
+            l++;
+        }
 
-int lengthOfLongestSubstring(char *s) {
-  int freq[256] = {0};
-  int l = 0, r = 0;
-  int ans = 0;
-  int len = strlen(s);
-
-  for (r = 0; r < len; r++) {
-    char c = s[r];
-    freq[(unsigned char)c]++;
-
-    while (freq[(unsigned char)c] > 1) {
-      freq[(unsigned char)s[l]]--;
-      l++;
-    }
-
-    if (ans < r - l + 1) {
-      ans = r - l + 1;
+        if (ans < r - l + 1) {
+            ans = r - l + 1;
+        }
     }
-  }
 
-  return ans;
+    return ans;
 }
-
 ```
 
 <!-- tabs:end -->

solution/0000-0099/0003.Longest Substring Without Repeating Characters/Solution.c

@@ -1,22 +1,22 @@
-int lengthOfLongestSubstring(char *s) {
-  int freq[256] = {0};
-  int l = 0, r = 0;
-  int ans = 0;
-  int len = strlen(s);
+int lengthOfLongestSubstring(char* s) {
+    int freq[256] = {0};
+    int l = 0, r = 0;
+    int ans = 0;
+    int len = strlen(s);
 
-  for (r = 0; r < len; r++) {
-    char c = s[r];
-    freq[(unsigned char)c]++;
+    for (r = 0; r < len; r++) {
+        char c = s[r];
+        freq[(unsigned char) c]++;
 
-    while (freq[(unsigned char)c] > 1) {
-      freq[(unsigned char)s[l]]--;
-      l++;
-    }
+        while (freq[(unsigned char) c] > 1) {
+            freq[(unsigned char) s[l]]--;
+            l++;
+        }
 
-    if (ans < r - l + 1) {
-      ans = r - l + 1;
+        if (ans < r - l + 1) {
+            ans = r - l + 1;
+        }
     }
-  }
 
-  return ans;
+    return ans;
 }

solution/0000-0099/0004.Median of Two Sorted Arrays/README.md

@@ -353,30 +353,30 @@ proc medianOfTwoSortedArrays(nums1: seq[int], nums2: seq[int]): float =
 #### C
 
 ```c
-int findKth(int *nums1, int m, int i, int *nums2, int n, int j, int k) {
-  if (i >= m)
-    return nums2[j + k - 1];
-  if (j >= n)
-    return nums1[i + k - 1];
-  if (k == 1)
-    return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
-
-  int p = k / 2;
-
-  int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
-  int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
-
-  if (x < y)
-    return findKth(nums1, m, i + p, nums2, n, j, k - p);
-  else
-    return findKth(nums1, m, i, nums2, n, j + p, k - p);
+int findKth(int* nums1, int m, int i, int* nums2, int n, int j, int k) {
+    if (i >= m)
+        return nums2[j + k - 1];
+    if (j >= n)
+        return nums1[i + k - 1];
+    if (k == 1)
+        return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
+
+    int p = k / 2;
+
+    int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
+    int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
+
+    if (x < y)
+        return findKth(nums1, m, i + p, nums2, n, j, k - p);
+    else
+        return findKth(nums1, m, i, nums2, n, j + p, k - p);
 }
 
-double findMedianSortedArrays(int *nums1, int m, int *nums2, int n) {
-  int total = m + n;
-  int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
-  int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
-  return (a + b) / 2.0;
+double findMedianSortedArrays(int* nums1, int m, int* nums2, int n) {
+    int total = m + n;
+    int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
+    int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
+    return (a + b) / 2.0;
 }
 ```
 

solution/0000-0099/0004.Median of Two Sorted Arrays/README_EN.md

@@ -349,30 +349,30 @@ proc medianOfTwoSortedArrays(nums1: seq[int], nums2: seq[int]): float =
 #### C
 
 ```c
-int findKth(int *nums1, int m, int i, int *nums2, int n, int j, int k) {
-  if (i >= m)
-    return nums2[j + k - 1];
-  if (j >= n)
-    return nums1[i + k - 1];
-  if (k == 1)
-    return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
-
-  int p = k / 2;
-
-  int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
-  int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
-
-  if (x < y)
-    return findKth(nums1, m, i + p, nums2, n, j, k - p);
-  else
-    return findKth(nums1, m, i, nums2, n, j + p, k - p);
+int findKth(int* nums1, int m, int i, int* nums2, int n, int j, int k) {
+    if (i >= m)
+        return nums2[j + k - 1];
+    if (j >= n)
+        return nums1[i + k - 1];
+    if (k == 1)
+        return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
+
+    int p = k / 2;
+
+    int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
+    int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
+
+    if (x < y)
+        return findKth(nums1, m, i + p, nums2, n, j, k - p);
+    else
+        return findKth(nums1, m, i, nums2, n, j + p, k - p);
 }
 
-double findMedianSortedArrays(int *nums1, int m, int *nums2, int n) {
-  int total = m + n;
-  int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
-  int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
-  return (a + b) / 2.0;
+double findMedianSortedArrays(int* nums1, int m, int* nums2, int n) {
+    int total = m + n;
+    int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
+    int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
+    return (a + b) / 2.0;
 }
 ```
 

solution/0000-0099/0004.Median of Two Sorted Arrays/Solution.c

@@ -1,25 +1,25 @@
-int findKth(int *nums1, int m, int i, int *nums2, int n, int j, int k) {
-  if (i >= m)
-    return nums2[j + k - 1];
-  if (j >= n)
-    return nums1[i + k - 1];
-  if (k == 1)
-    return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
+int findKth(int* nums1, int m, int i, int* nums2, int n, int j, int k) {
+    if (i >= m)
+        return nums2[j + k - 1];
+    if (j >= n)
+        return nums1[i + k - 1];
+    if (k == 1)
+        return nums1[i] < nums2[j] ? nums1[i] : nums2[j];
 
-  int p = k / 2;
+    int p = k / 2;
 
-  int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
-  int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
+    int x = (i + p - 1 < m) ? nums1[i + p - 1] : INT_MAX;
+    int y = (j + p - 1 < n) ? nums2[j + p - 1] : INT_MAX;
 
-  if (x < y)
-    return findKth(nums1, m, i + p, nums2, n, j, k - p);
-  else
-    return findKth(nums1, m, i, nums2, n, j + p, k - p);
+    if (x < y)
+        return findKth(nums1, m, i + p, nums2, n, j, k - p);
+    else
+        return findKth(nums1, m, i, nums2, n, j + p, k - p);
 }
 
-double findMedianSortedArrays(int *nums1, int m, int *nums2, int n) {
-  int total = m + n;
-  int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
-  int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
-  return (a + b) / 2.0;
-}
+double findMedianSortedArrays(int* nums1, int m, int* nums2, int n) {
+    int total = m + n;
+    int a = findKth(nums1, m, 0, nums2, n, 0, (total + 1) / 2);
+    int b = findKth(nums1, m, 0, nums2, n, 0, (total + 2) / 2);
+    return (a + b) / 2.0;
+}