C++ Solution of 48.Rotate_image

Leetcode-Easy/32.cpp_Longest Valid Parentheses.cpp

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+/*Explanation -:
+
+Create an empty stack and push -1 to it.
+The first element of the stack is used
+to provide a base for the next valid string.
+Initialize result as 0.
+
+If the character is '(' i.e. str[i] == '('),
+push index'i' to the stack.
+
+Else (if the character is ')')
+a) Pop an item from the stack (Most of the
+time an opening bracket)
+
+b) If the stack is not empty, then find the
+length of current valid substring by taking
+the difference between the current index and
+top of the stack. If current length is more
+than the result, then update the result.
+
+c) If the stack is empty, push the current index
+as a base for the next valid substring.
+
+Return result.
+
+If still not able to visualize, Try to do a dry run with pen/paper.
+
+C++ Code -:
+*/
+class Solution {
+public:
+	int longestValidParentheses(string s) 
+	{
+		int n=s.length();
+		int result=0;
+		stack<int> st;
+		st.push(-1);
+		for(int i=0;i<n;i++){
+			if(s[i]=='('){
+				st.push(i);
+			}
+			else{
+				if(!st.empty()){
+					st.pop();
+				}
+				if(!st.empty()){
+					result=max(result,i-st.top());
+				}
+				else{
+					st.push(i);
+				}
+			}
+		}
+		return result;
+
+	}
+};

Leetcode-Medium/48-Rotate_IMAGE_C++.cpp

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+class Solution {
+public:
+    void rotate(vector<vector<int>>& matrix) {
+          int n= matrix.size();
+        for(int i=0; i<n; i++)
+        {
+            for(int j=0; j<i; j++)
+            {
+                swap(matrix[i][j], matrix[j][i]);
+            }
+        }
+
+        for(int i=0; i<n; i++)
+            reverse(matrix[i].begin(), matrix[i].end());
+    }
+
+};