Create 02 - Top-Down | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/15 - Largest Square Formed in a Matrix/02 - Top-Down | DP | Approach.cpp

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+// User function Template for C++
+class Solution {
+  public:
+    // Recursive function with memoization to find the size of the largest square ending at (i, j)
+    int solve(vector<vector<int>>& mat, int i, int j, vector<vector<int>>& dp) {
+        // Base condition: If out of bounds, return 0
+        if (i >= mat.size() || j >= mat[0].size()) return 0;
+
+        // If the value is already computed, return it
+        if (dp[i][j] != -1) return dp[i][j];
+        
+        // If the current cell contains 0, it cannot form a square
+        if (mat[i][j] == 0) return dp[i][j] = 0;
+
+        // Recursive calls to calculate the size of squares in three directions:
+        int right = solve(mat, i, j - 1, dp);         // Square ending to the left
+        int diagnol = solve(mat, i - 1, j, dp);       // Square ending diagonally
+        int down = solve(mat, i - 1, j - 1, dp);      // Square ending above
+
+        // Store the size of the square ending at (i, j) in the dp table
+        return dp[i][j] = 1 + min({right, diagnol, down});
+    }
+
+    // Main function to find the largest square submatrix with all 1s
+    int maxSquare(vector<vector<int>>& mat) {
+        int maxi = 0; // Variable to store the maximum square size
+        int n = mat.size();
+        int m = mat[0].size();
+
+        // Create a memoization table and initialize it with -1
+        vector<vector<int>> dp(n, vector<int>(m, -1));
+
+        // Iterate through the matrix and calculate the largest square ending at each cell
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                maxi = max(maxi, solve(mat, i, j, dp)); // Update the maximum square size
+            }
+        }
+
+        return maxi; // Return the maximum square size
+    }
+};