From b58f570df8f3ddc3efa681de00553e15a57aa6c9 Mon Sep 17 00:00:00 2001
From: Sai pradyumna Goud Chiragoni
 <143107589+Saipradyumnagoud@users.noreply.github.com>
Date: Mon, 24 Jun 2024 13:40:16 +0530
Subject: [PATCH] Added 861-score-after-flipping-matrix.md

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+---
+id: score-after-flipping-matrix
+title: Score After Flipping Matrix
+level: medium
+sidebar_label: Score After Flipping Matrix
+tags:
+  - Array
+  - Greedy
+  - Matrix
+  - Java
+description: "This document provides solutions for the Score After Flipping Matrix problem."
+---
+
+## Problem Statement
+
+You are given an `m x n` binary matrix `grid`.
+
+A move consists of choosing any row or column and toggling each value in that row or column (i.e., changing all 0's to 1's, and all 1's to 0's).
+
+Every row of the matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.
+
+Return the highest possible score after making any number of moves (including zero moves).
+
+**Example 1:**
+
+Input: `grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]`
+
+Output: `39`
+
+Explanation: `0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39`
+
+**Example 2:**
+
+Input: `grid = [[0]]`
+
+Output: `1`
+
+**Constraints:**
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m, n <= 20`
+- `grid[i][j]` is `0` or `1`.
+
+## Solutions
+
+### Approach
+
+To maximize the score, follow these steps:
+
+1. **Row Flipping:**
+   - Ensure that each row starts with a `1` by flipping rows where the first element is `0`.
+
+2. **Column Flipping:**
+   - For each column, if the number of `0`s exceeds the number of `1`s, flip the column to maximize the number of `1`s in that column.
+
+3. **Calculate Score:**
+   - Convert each row from binary to decimal and sum these values to get the final score.
+
+### Java 
+
+```java
+class Solution {
+    public int matrixScore(int[][] grid) {
+        int m = grid.length;
+        int n = grid[0].length;
+        int i = 0, j = 0;
+
+        for (i = 0; i < m; i++) {
+            if (grid[i][0] == 0) {
+                flipRow(grid, i);
+            }
+        }
+
+        for (j = 0; j < n; j++) {
+            if (searchColZeroes(grid, j) > searchColOnes(grid, j)) {
+                flipCol(grid, j);
+            }
+        }
+
+        int sum = 0;
+        for (i = 0; i < grid.length; i++) {
+            int count = 0;
+            int temp = grid[0].length - 1;
+            for (j = 0; j < grid[0].length; j++) {
+                count += (int) (grid[i][j] * Math.pow(2, temp--));
+            }
+            sum += count;
+        }
+        return sum;
+    }
+
+    public int searchColZeroes(int[][] grid, int j) {
+        int count = 0;
+        for (int temp = 0; temp < grid.length; temp++) {
+            if (grid[temp][j] == 0) {
+                count++;
+            }
+        }
+        return count;
+    }
+
+    public int searchColOnes(int[][] grid, int j) {
+        int count = 0;
+        for (int temp = 0; temp < grid.length; temp++) {
+            if (grid[temp][j] == 1) {
+                count++;
+            }
+        }
+        return count;
+    }
+
+    public void flipCol(int[][] grid, int j) {
+        for (int temp = 0; temp < grid.length; temp++) {
+            grid[temp][j] = (grid[temp][j] == 0) ? 1 : 0;
+        }
+    }
+
+    public void flipRow(int[][] grid, int i) {
+        for (int temp = 0; temp < grid[0].length; temp++) {
+            grid[i][temp] = (grid[i][temp] == 0) ? 1 : 0;
+        }
+    }
+}
+```
+
+### Python 
+```Python 
+class Solution:
+    def matrixScore(self, grid: List[List[int]]) -> int:
+        m = len(grid)
+        n = len(grid[0])
+        
+        # Step 1: Ensure all rows start with '1' by flipping rows where grid[i][0] == 0
+        for i in range(m):
+            if grid[i][0] == 0:
+                self.flipRow(grid, i)
+        
+        # Step 2: Ensure each column has more '1's than '0's by flipping columns if necessary
+        for j in range(n):
+            if self.searchColZeroes(grid, j) > self.searchColOnes(grid, j):
+                self.flipCol(grid, j)
+        
+        # Step 3: Calculate the matrix score
+        score = 0
+        for i in range(m):
+            row_sum = 0
+            for j in range(n):
+                row_sum += grid[i][j] * (1 << (n - 1 - j))  # Equivalent to grid[i][j] * 2^(n-1-j)
+            score += row_sum
+        
+        return score
+    
+    def searchColZeroes(self, grid: List[List[int]], j: int) -> int:
+        count = 0
+        for i in range(len(grid)):
+            if grid[i][j] == 0:
+                count += 1
+        return count
+    
+    def searchColOnes(self, grid: List[List[int]], j: int) -> int:
+        count = 0
+        for i in range(len(grid)):
+            if grid[i][j] == 1:
+                count += 1
+        return count
+    
+    def flipCol(self, grid: List[List[int]], j: int) -> None:
+        for i in range(len(grid)):
+            grid[i][j] = 1 - grid[i][j]
+    
+    def flipRow(self, grid: List[List[int]], i: int) -> None:
+        for j in range(len(grid[0])):
+            grid[i][j] = 1 - grid[i][j]
+
+```