Solution of 0010 and 0011 in c language is done

Author: pranjal030404

solution/0000-0099/0010.Regular Expression Matching/README.md

@@ -579,6 +579,36 @@ class Solution {
 }
 ```
 
+#### C
+
+``` C
+bool isMatch(char *s, char *p) {
+  int m = strlen(s), n = strlen(p);
+  bool **dp = malloc((m + 1) * sizeof(bool *));
+  for (int i = 0; i <= m; i++)
+    dp[i] = calloc(n + 1, sizeof(bool));
+  dp[0][0] = 1;
+  for (int j = 2; j <= n; j++)
+    if (p[j - 1] == '*')
+      dp[0][j] = dp[0][j - 2];
+
+  for (int i = 1; i <= m; i++)
+    for (int j = 1; j <= n; j++)
+      if (p[j - 1] == '*')
+        dp[i][j] = dp[i][j - 2] ||
+                   ((p[j - 2] == '.' || p[j - 2] == s[i - 1]) && dp[i - 1][j]);
+      else
+        dp[i][j] =
+            (p[j - 1] == '.' || p[j - 1] == s[i - 1]) && dp[i - 1][j - 1];
+
+  bool res = dp[m][n];
+  for (int i = 0; i <= m; i++)
+    free(dp[i]);
+  free(dp);
+  return res;
+}
+```
+
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solution/0000-0099/0010.Regular Expression Matching/README_EN.md

@@ -578,6 +578,35 @@ class Solution {
 }
 ```
 
+#### C
+
+``` C
+bool isMatch(char *s, char *p) {
+  int m = strlen(s), n = strlen(p);
+  bool **dp = malloc((m + 1) * sizeof(bool *));
+  for (int i = 0; i <= m; i++)
+    dp[i] = calloc(n + 1, sizeof(bool));
+  dp[0][0] = 1;
+  for (int j = 2; j <= n; j++)
+    if (p[j - 1] == '*')
+      dp[0][j] = dp[0][j - 2];
+
+  for (int i = 1; i <= m; i++)
+    for (int j = 1; j <= n; j++)
+      if (p[j - 1] == '*')
+        dp[i][j] = dp[i][j - 2] ||
+                   ((p[j - 2] == '.' || p[j - 2] == s[i - 1]) && dp[i - 1][j]);
+      else
+        dp[i][j] =
+            (p[j - 1] == '.' || p[j - 1] == s[i - 1]) && dp[i - 1][j - 1];
+
+  bool res = dp[m][n];
+  for (int i = 0; i <= m; i++)
+    free(dp[i]);
+  free(dp);
+  return res;
+}
+```
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solution/0000-0099/0010.Regular Expression Matching/Solution.c

@@ -0,0 +1,23 @@
+bool isMatch(char* s, char* p) {
+    int m = strlen(s), n = strlen(p);
+    bool** dp = malloc((m + 1) * sizeof(bool*));
+    for (int i = 0; i <= m; i++)
+        dp[i] = calloc(n + 1, sizeof(bool));
+    dp[0][0] = 1;
+    for (int j = 2; j <= n; j++)
+        if (p[j - 1] == '*')
+            dp[0][j] = dp[0][j - 2];
+
+    for (int i = 1; i <= m; i++)
+        for (int j = 1; j <= n; j++)
+            if (p[j - 1] == '*')
+                dp[i][j] = dp[i][j - 2] || ((p[j - 2] == '.' || p[j - 2] == s[i - 1]) && dp[i - 1][j]);
+            else
+                dp[i][j] = (p[j - 1] == '.' || p[j - 1] == s[i - 1]) && dp[i - 1][j - 1];
+
+    bool res = dp[m][n];
+    for (int i = 0; i <= m; i++)
+        free(dp[i]);
+    free(dp);
+    return res;
+}

solution/0000-0099/0011.Container With Most Water/README.md

@@ -262,6 +262,22 @@ class Solution {
 }
 ```
 
+#### C
+
+ ``` C
+
+ int maxArea(int *h, int n) {
+  int l = 0, r = n - 1, max = 0;
+  while (l < r) {
+    int area = (r - l) * (h[l] < h[r] ? h[l++] : h[r--]);
+    if (area > max)
+      max = area;
+  }
+  return max;
+}
+
+ ```
+
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solution/0000-0099/0011.Container With Most Water/README_EN.md

@@ -259,6 +259,21 @@ class Solution {
 }
 ```
 
+#### C
+
+ ``` C
+
+ int maxArea(int *h, int n) {
+  int l = 0, r = n - 1, max = 0;
+  while (l < r) {
+    int area = (r - l) * (h[l] < h[r] ? h[l++] : h[r--]);
+    if (area > max)
+      max = area;
+  }
+  return max;
+}
+
+ ```
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solution/0000-0099/0011.Container With Most Water/Solution.c

@@ -0,0 +1,9 @@
+int maxArea(int* h, int n) {
+    int l = 0, r = n - 1, max = 0;
+    while (l < r) {
+        int area = (r - l) * (h[l] < h[r] ? h[l++] : h[r--]);
+        if (area > max)
+            max = area;
+    }
+    return max;
+}