Merge pull request neetcode-gh#2840 from Abe0770/main

Create 0221-maximal-square.cpp

Author: felivalencia3

cpp/0221-maximal-square.cpp

@@ -0,0 +1,33 @@
+/*
+  Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
+
+  Ex. Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
+      Output: 4
+
+  Time  : O(m*n)
+  Space : O(m*n)
+*/
+
+class Solution {
+public:
+    int maximalSquare(vector<vector<char>>& matrix) {
+        int rows = matrix.size(), cols = matrix[0].size();
+
+        vector<vector<int>> dp (rows+1, vector<int>(cols+1, 0));
+        int maxi = 0;
+        for(int i = rows-1 ; i >= 0; --i) {
+            for(int j = cols-1 ; j >=0 ; --j) {
+                if(matrix[i][j] == '1') {
+                    int right = dp[i][j+1], dia = dp[i+1][j+1], bottom = dp[i+1][j];
+                    
+                    dp[i][j] = 1 + min(right, min(dia, bottom));
+                    maxi = max(maxi, dp[i][j]);
+                }
+                else {
+                    dp[i][j] = 0;
+                }
+            }
+        }
+        return maxi*maxi;
+    }
+};

cpp/1024-video-stitching.cpp

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+/*
+  You are given a series of video clips from a sporting event that lasted time seconds. These video clips can be overlapping with each other and have varying lengths.
+  Each video clip is described by an array clips where clips[i] = [starti, endi] indicates that the ith clip started at starti and ended at endi.
+  We can cut these clips into segments freely.
+  For example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
+  Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event [0, time]. If the task is impossible, return -1.
+
+  Ex. Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 10
+      Output: 3
+      Explanation: We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
+      Then, we can reconstruct the sporting event as follows:
+      We cut [1,9] into segments [1,2] + [2,8] + [8,9].
+      Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
+
+  Time  : O(N log N)
+  Space : O(1)
+*/
+
+class Solution {
+public:
+    int videoStitching(vector<vector<int>>& clips, int time) {
+        int maxi = 0;
+        sort(clips.begin() , clips.end());
+        for(int i = 0 ; i < clips.size() ; i++) 
+            maxi = max(maxi, clips[i][1]);
+
+        if(maxi < time)
+            return -1;
+
+        int count = 0, endTime = 0, covered = INT_MIN;
+        for(int i = 0; endTime < time ; ) {
+            ++count;
+            while(i < clips.size() && clips[i][0] <= endTime)
+                covered = max(covered, clips[i++][1]);
+
+            if(endTime == covered)
+                return -1;
+            endTime = covered;
+        }
+        return count;
+    }
+};