Given a binary matrix of size N * M, the task is to find the largest area sub-matrix such that all elements in it are same i.e. either all are 0 or all are 1. Print the largest possible area of such matrix.
Examples:
Input: mat[][] = {
{1, 1, 0, 1, 0, 0, 0, 0},
{0, 1, 1, 1, 1, 0, 0, 1},
{1, 0, 0, 1, 1, 1, 0, 0},
{0, 1, 1, 0, 1, 1, 0, 0},
{1, 0, 1, 1, 1, 1, 1, 0},
{0, 0, 1, 1, 1, 1, 1, 1} }
Output: 10
Largest submatrix with all equal elements starts from
mat[4][2] (top-left corner) and ends mat[5][6] (borrom-right corner).Input: mat[][] = {
{1, 0},
{0, 1}}
Output: 1
Approach:
- Try to find the largest sub-matrix with all 1s and the same can be applied to find the largest submatrix with all 0s.
- Maintain a matrix dp[N][M], where dp[i][j] represents the number of consecutive 1s present in the jth column starting from the ith row till the last row. This matrix can be easily filled by traversing each column from bottom to top.
- Now utilize the fact that for each row i, dp[i][j] represents the largest consecutive 1s till last row in the jth column. This problem is now same as the problem of finding the largest area rectangle present in the histogram which has been discussed in this article.
- The approach mentioned in the previous approach has to be applied for every row of the matrix to find the maximum area sub-matrix.
The same can be applied for finding the maximum area sub-matrix with all 0s.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>#define row 6#define col 8using namespace std;// Function to find the maximum rectangular// area under given histogram with n barsint cal(int hist[], int n){ // Create an empty stack. The stack holds indexes // of hist[] array. The bars stored in stack are // always in increasing order of their heights. stack<int> s; // Initialize max area int max_area = 0; // To store top of the stack int tp; // To store area with top bar int area_with_top; // as the smallest bar // Run through all bars of given histogram int i = 0; while (i < n) { // If this bar is higher than the bar on top // stack, push it to stack if (s.empty() || hist[s.top()] <= hist[i]) s.push(i++); // If this bar is lower than top of stack, // then calculate area of rectangle with stack // top as the smallest (or minimum height) bar. // 'i' is 'right index' for the top and element // before top in stack is 'left index' else { // Store the top index tp = s.top(); // Pop the top s.pop(); // Calculate the area with hist[tp] stack // as smallest bar area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1); // Update max area, if needed if (max_area < area_with_top) max_area = area_with_top; } } // Now pop the remaining bars from stack and calculate // area with every popped bar as the smallest bar while (s.empty() == false) { tp = s.top(); s.pop(); area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1); if (max_area < area_with_top) max_area = area_with_top; } return max_area;}// Function to find largest sub matrix// with all equal elementsint largestMatrix(int a[][col]){ // To find largest sub matrix // with all elements 1 int dp[row][col]; // Fill dp[][] by traversing each // column from bottom to up for (int i = 0; i < col; i++) { int cnt = 0; for (int j = row - 1; j >= 0; j--) { dp[j][i] = 0; if (a[j][i] == 1) { cnt++; dp[j][i] = cnt; } else { cnt = 0; } } } int ans = -1; for (int i = 0; i < row; i++) { // Maintain the histogram array int hist[col]; for (int j = 0; j < col; j++) { hist[j] = dp[i][j]; } // Find maximum area rectangle in Histogram ans = max(ans, cal(hist, col)); } // To fill dp[][] for finding largest // sub matrix with all elements 0 for (int i = 0; i < col; i++) { int cnt = 0; for (int j = row - 1; j >= 0; j--) { dp[j][i] = 0; if (a[j][i] == 0) { cnt++; dp[j][i] = cnt; } else { cnt = 0; } } } for (int i = 0; i < row; i++) { // Maintain the histogram array int hist[col]; for (int j = 0; j < col; j++) { hist[j] = dp[i][j]; } // Find maximum area rectangle in Histogram ans = max(ans, cal(hist, col)); } return ans;}// Driver codeint main(){ int a[row][col] = { { 1, 1, 0, 1, 0, 0, 0, 0 }, { 0, 1, 1, 1, 1, 0, 0, 1 }, { 1, 0, 0, 1, 1, 1, 0, 0 }, { 0, 1, 1, 0, 1, 1, 0, 0 }, { 1, 0, 1, 1, 1, 1, 1, 0 }, { 0, 0, 1, 1, 1, 1, 1, 1 } }; cout << largestMatrix(a); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static int row = 6;static int col = 8;// Function to find the maximum rectangular// area under given histogram with n barsstatic int cal(int hist[], int n){ // Create an empty stack. The stack holds indexes // of hist[] array. The bars stored in stack are // always in increasing order of their heights. Stack<Integer> s = new Stack<>(); // Initialize max area int max_area = 0; // To store top of the stack int tp; // To store area with top bar int area_with_top; // as the smallest bar // Run through all bars of given histogram int i = 0; while (i < n) { // If this bar is higher than the bar on top // stack, push it to stack if (s.empty() || hist[s.peek()] <= hist[i]) s.push(i++); // If this bar is lower than top of stack, // then calculate area of rectangle with stack // top as the smallest (or minimum height) bar. // 'i' is 'right index' for the top and element // before top in stack is 'left index' else { // Store the top index tp = s.peek(); // Pop the top s.pop(); // Calculate the area with hist[tp] stack // as smallest bar area_with_top = hist[tp] * (s.empty() ? i : i - s.peek() - 1); // Update max area, if needed if (max_area < area_with_top) max_area = area_with_top; } } // Now pop the remaining bars from stack and calculate // area with every popped bar as the smallest bar while (s.empty() == false) { tp = s.peek(); s.pop(); area_with_top = hist[tp] * (s.empty() ? i : i - s.peek() - 1); if (max_area < area_with_top) max_area = area_with_top; } return max_area;}// Function to find largest sub matrix// with all equal elementsstatic int largestMatrix(int a[][]){ // To find largest sub matrix // with all elements 1 int [][]dp = new int[row][col]; // Fill dp[][] by traversing each // column from bottom to up for (int i = 0; i < col; i++) { int cnt = 0; for (int j = row - 1; j >= 0; j--) { dp[j][i] = 0; if (a[j][i] == 1) { cnt++; dp[j][i] = cnt; } else { cnt = 0; } } } int ans = -1; for (int i = 0; i < row; i++) { // Maintain the histogram array int []hist = new int[col]; for (int j = 0; j < col; j++) { hist[j] = dp[i][j]; } // Find maximum area rectangle in Histogram ans = Math.max(ans, cal(hist, col)); } // To fill dp[][] for finding largest // sub matrix with all elements 0 for (int i = 0; i < col; i++) { int cnt = 0; for (int j = row - 1; j >= 0; j--) { dp[j][i] = 0; if (a[j][i] == 0) { cnt++; dp[j][i] = cnt; } else { cnt = 0; } } } for (int i = 0; i < row; i++) { // Maintain the histogram array int []hist = new int[col]; for (int j = 0; j < col; j++) { hist[j] = dp[i][j]; } // Find maximum area rectangle in Histogram ans = Math.max(ans, cal(hist, col)); } return ans;}// Driver codepublic static void main(String[] args){ int a[][] = {{ 1, 1, 0, 1, 0, 0, 0, 0 }, { 0, 1, 1, 1, 1, 0, 0, 1 }, { 1, 0, 0, 1, 1, 1, 0, 0 }, { 0, 1, 1, 0, 1, 1, 0, 0 }, { 1, 0, 1, 1, 1, 1, 1, 0 }, { 0, 0, 1, 1, 1, 1, 1, 1 }}; System.out.println(largestMatrix(a));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python 3 implementation of the approachrow = 6col = 8# Function to find the maximum rectangular# area under given histogram with n barsdef cal(hist, n): # Create an empty stack. The stack holds indexes # of hist[] array. The bars stored in stack are # always in increasing order of their heights. s = []; # Initialize max area max_area = 0 # Run through all bars of given histogram i = 0 while (i < n) : # If this bar is higher than the bar on top # stack, push it to stack if (len(s) == 0 or( hist[s[-1]] <= hist[i])): s.append(i) i += 1 # If this bar is lower than top of stack, # then calculate area of rectangle with stack # top as the smallest (or minimum height) bar. # 'i' is 'right index' for the top and element # before top in stack is 'left index' else : # Store the top index tp = s[-1] # Pop the top s.pop() # Calculate the area with hist[tp] stack # as smallest bar if len(s) == 0: area_with_top = hist[tp]*i else: area_with_top = hist[tp]*(i -s[-1] - 1) # Update max area, if needed if (max_area < area_with_top): max_area = area_with_top # Now pop the remaining bars from stack and calculate # area with every popped bar as the smallest bar while (len(s)!=0): tp = s[-1] s.pop() if len(s)==0: area_with_top = hist[tp]*i else: area_with_top = hist[tp]*(i - s[-1] - 1) if (max_area < area_with_top): max_area = area_with_top return max_area# Function to find largest sub matrix# with all equal elementsdef largestMatrix(a): # To find largest sub matrix # with all elements 1 dp = [[ 0 for x in range(col)] for y in range(row)] # Fill dp[][] by traversing each # column from bottom to up for i in range(col): cnt = 0 for j in range( row - 1, -1 ,-1): dp[j][i] = 0 if (a[j][i] == 1) : cnt+=1 dp[j][i] = cnt else : cnt = 0 # print("cnt ",cnt) ans = -1 for i in range( row ): # Maintain the histogram array hist = [0]*col for j in range(col): hist[j] = dp[i][j] # Find maximum area rectangle in Histogram ans = max(ans, cal(hist, col)) # To fill dp[][] for finding largest # sub matrix with all elements 0 for i in range( col): cnt = 0 for j in range( row - 1, -1 ,-1): dp[j][i] = 0 if (a[j][i] == 0): cnt +=1 dp[j][i] = cnt else: cnt = 0 for i in range( row) : # Maintain the histogram array hist = [0]*col for j in range(col): hist[j] = dp[i][j] # Find maximum area rectangle in Histogram ans = max(ans, cal(hist, col)) return ans# Driver codeif __name__ == "__main__": a = [ [1, 1, 0, 1, 0, 0, 0, 0 ], [ 0, 1, 1, 1, 1, 0, 0, 1 ], [ 1, 0, 0, 1, 1, 1, 0, 0 ], [ 0, 1, 1, 0, 1, 1, 0, 0 ], [ 1, 0, 1, 1, 1, 1, 1, 0 ], [ 0, 0, 1, 1, 1, 1, 1, 1 ]] print(largestMatrix(a))# This code is contributed by chitranayal |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {static int row = 6;static int col = 8;// Function to find the maximum rectangular// area under given histogram with n barsstatic int cal(int []hist, int n){ // Create an empty stack. The stack holds indexes // of hist[] array. The bars stored in stack are // always in increasing order of their heights. Stack<int> s = new Stack<int>(); // Initialize max area int max_area = 0; // To store top of the stack int tp; // To store area with top bar int area_with_top; // as the smallest bar // Run through all bars of given histogram int i = 0; while (i < n) { // If this bar is higher than the bar on top // stack, push it to stack if (s.Count == 0 || hist[s.Peek()] <= hist[i]) s.Push(i++); // If this bar is lower than top of stack, // then calculate area of rectangle with stack // top as the smallest (or minimum height) bar. // 'i' is 'right index' for the top and element // before top in stack is 'left index' else { // Store the top index tp = s.Peek(); // Pop the top s.Pop(); // Calculate the area with hist[tp] stack // as smallest bar area_with_top = hist[tp] * (s.Count == 0 ? i : i - s.Peek() - 1); // Update max area, if needed if (max_area < area_with_top) max_area = area_with_top; } } // Now pop the remaining bars from stack and calculate // area with every popped bar as the smallest bar while (s.Count == 0 == false) { tp = s.Peek(); s.Pop(); area_with_top = hist[tp] * (s.Count == 0 ? i : i - s.Peek() - 1); if (max_area < area_with_top) max_area = area_with_top; } return max_area;}// Function to find largest sub matrix// with all equal elementsstatic int largestMatrix(int [,]a){ // To find largest sub matrix // with all elements 1 int [,]dp = new int[row, col]; // Fill dp[,] by traversing each // column from bottom to up for (int i = 0; i < col; i++) { int cnt = 0; for (int j = row - 1; j >= 0; j--) { dp[j, i] = 0; if (a[j, i] == 1) { cnt++; dp[j, i] = cnt; } else { cnt = 0; } } } int ans = -1; for (int i = 0; i < row; i++) { // Maintain the histogram array int []hist = new int[col]; for (int j = 0; j < col; j++) { hist[j] = dp[i, j]; } // Find maximum area rectangle in Histogram ans = Math.Max(ans, cal(hist, col)); } // To fill dp[,] for finding largest // sub matrix with all elements 0 for (int i = 0; i < col; i++) { int cnt = 0; for (int j = row - 1; j >= 0; j--) { dp[j, i] = 0; if (a[j, i] == 0) { cnt++; dp[j, i] = cnt; } else { cnt = 0; } } } for (int i = 0; i < row; i++) { // Maintain the histogram array int []hist = new int[col]; for (int j = 0; j < col; j++) { hist[j] = dp[i, j]; } // Find maximum area rectangle in Histogram ans = Math.Max(ans, cal(hist, col)); } return ans;}// Driver codepublic static void Main(String[] args){ int [,]a = {{ 1, 1, 0, 1, 0, 0, 0, 0 }, { 0, 1, 1, 1, 1, 0, 0, 1 }, { 1, 0, 0, 1, 1, 1, 0, 0 }, { 0, 1, 1, 0, 1, 1, 0, 0 }, { 1, 0, 1, 1, 1, 1, 1, 0 }, { 0, 0, 1, 1, 1, 1, 1, 1 }}; Console.WriteLine(largestMatrix(a));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approachlet row = 6;let col = 8;// Function to find the maximum rectangular// area under given histogram with n barsfunction cal(hist, n){ // Create an empty stack. The stack holds indexes // of hist[] array. The bars stored in stack are // always in increasing order of their heights. let s = []; // Initialize max area let max_area = 0; // To store top of the stack let tp; // To store area with top bar let area_with_top; // as the smallest bar // Run through all bars of given histogram let i = 0; while (i < n) { // If this bar is higher than the bar on top // stack, push it to stack if (s.length == 0 || hist[s[s.length - 1]] <= hist[i]) s.push(i++); // If this bar is lower than top of stack, // then calculate area of rectangle with stack // top as the smallest (or minimum height) bar. // 'i' is 'right index' for the top and element // before top in stack is 'left index' else { // Store the top index tp = s[s.length - 1]; // Pop the top s.pop(); // Calculate the area with hist[tp] stack // as smallest bar area_with_top = hist[tp] * (s.length == 0 ? i : i - s[s.length - 1] - 1); // Update max area, if needed if (max_area < area_with_top) max_area = area_with_top; } } // Now pop the remaining bars from // stack and calculate area with // every popped bar as the smallest bar while (s.length != 0) { tp = s[s.length - 1]; s.pop(); area_with_top = hist[tp] * (s.length == 0 ? i : i - s[s.length - 1] - 1); if (max_area < area_with_top) max_area = area_with_top; } return max_area;}// Function to find largest sub matrix// with all equal elementsfunction largestMatrix(a){ // To find largest sub matrix // with all elements 1 let dp = new Array(row); for(let i = 0; i < row; i++) { dp[i] = new Array(col); } // Fill dp[][] by traversing each // column from bottom to up for(let i = 0; i < col; i++) { let cnt = 0; for(let j = row - 1; j >= 0; j--) { dp[j][i] = 0; if (a[j][i] == 1) { cnt++; dp[j][i] = cnt; } else { cnt = 0; } } } let ans = -1; for(let i = 0; i < row; i++) { // Maintain the histogram array let hist = new Array(col); for(let j = 0; j < col; j++) { hist[j] = dp[i][j]; } // Find maximum area rectangle in Histogram ans = Math.max(ans, cal(hist, col)); } // To fill dp[][] for finding largest // sub matrix with all elements 0 for(let i = 0; i < col; i++) { let cnt = 0; for(let j = row - 1; j >= 0; j--) { dp[j][i] = 0; if (a[j][i] == 0) { cnt++; dp[j][i] = cnt; } else { cnt = 0; } } } for(let i = 0; i < row; i++) { // Maintain the histogram array let hist = new Array(col); for(let j = 0; j < col; j++) { hist[j] = dp[i][j]; } // Find maximum area rectangle in Histogram ans = Math.max(ans, cal(hist, col)); } return ans;}// Driver codelet a = [ [ 1, 1, 0, 1, 0, 0, 0, 0 ], [ 0, 1, 1, 1, 1, 0, 0, 1 ], [ 1, 0, 0, 1, 1, 1, 0, 0 ], [ 0, 1, 1, 0, 1, 1, 0, 0 ], [ 1, 0, 1, 1, 1, 1, 1, 0 ], [ 0, 0, 1, 1, 1, 1, 1, 1 ]]; document.write(largestMatrix(a));// This code is contributed by rag2127</script> |
Output:
10
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