Given a binary matrix. The problem is to find the largest area rectangular sub-matrix with equal number of 1’s and 0’s. Examples:
Input : mat[][] = { {0, 0, 1, 1},
{0, 1, 1, 0},
{1, 1, 1, 0},
{1, 0, 0, 1} }
Output : 8 sq. units
(Top, left): (0, 0)
(Bottom, right): (3, 1)
Input : mat[][] = { {0, 0, 1, 1},
{0, 1, 1, 1} }
Output : 6 sq. units
The naive solution for this problem is to check every possible rectangle in given 2D array by counting the total number of 1’s and 0’s in that rectangle. This solution requires 4 nested loops and time complexity of this solution would be O(n^4).
An efficient solution is based on Largest rectangular sub-matrix whose sum is 0 which reduces the time complexity to O(n^3). First of all consider every ‘0’ in the matrix as ‘-1’. Now, the idea is to reduce the problem to 1-D array. We fix the left and right columns one by one and find the largest sub-array with 0 sum contiguous rows for every left and right column pair. We basically find top and bottom row numbers (which have sum zero) for every fixed left and right column pair. To find the top and bottom row numbers, calculate sum of elements in every row from left to right and store these sums in an array say temp[].
So temp[i] indicates sum of elements from left to right in row i. If we find largest subarray with 0 sum in temp[], we can get the index positions of rectangular sub-matrix with sum equal to 0 (i.e. having equal number of 1’s and 0’s). With this process we can find the largest area rectangular sub-matrix with sum equal to 0 (i.e. having equal number of 1’s and 0’s). We can use Hashing technique to find maximum length sub-array with sum equal to 0 in 1-D array in O(n) time.
Implementation:
CPP
// C++ implementation to find largest area rectangular// submatrix with equal number of 1's and 0's#include <bits/stdc++.h>using namespace std;#define MAX_ROW 10#define MAX_COL 10// This function basically finds largest 0// sum subarray in arr[0..n-1]. If 0 sum// doesn't exist, then it returns false. Else// it returns true and sets starting and// ending indexes as start and end.bool subArrWithSumZero(int arr[], int& start, int& end, int n){ // to store cumulative sum int sum[n]; // Initialize all elements of sum[] to 0 memset(sum, 0, sizeof(sum)); // map to store the indexes of sum unordered_map<int, int> um; // build up the cumulative sum[] array sum[0] = arr[0]; for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + arr[i]; // to store the maximum length subarray // with sum equal to 0 int maxLen = 0; // traverse to the sum[] array for (int i = 0; i < n; i++) { // if true, then there is a subarray // with sum equal to 0 from the // beginning up to index 'i' if (sum[i] == 0) { // update the required variables start = 0; end = i; maxLen = (i + 1); } // else if true, then sum[i] has not // seen before in 'um' else if (um.find(sum[i]) == um.end()) um[sum[i]] = i; // sum[i] has been seen before in the // unordered_map 'um' else { // if previous subarray length is smaller // than the current subarray length, then // update the required variables if (maxLen < (i - um[sum[i]])) { maxLen = (i - um[sum[i]]); start = um[sum[i]] + 1; end = i; } } } // if true, then there is no // subarray with sum equal to 0 if (maxLen == 0) return false; // else return true return true;}// function to find largest area rectangular// submatrix with equal number of 1's and 0'svoid maxAreaRectWithSumZero(int mat[MAX_ROW][MAX_COL], int row, int col){ // to store intermediate values int temp[row], startRow, endRow; // to store the final outputs int finalLeft, finalRight, finalTop, finalBottom; finalLeft = finalRight = finalTop = finalBottom = -1; int maxArea = 0; // Set the left column for (int left = 0; left < col; left++) { // Initialize all elements of temp as 0 memset(temp, 0, sizeof(temp)); // Set the right column for the left column // set by outer loop for (int right = left; right < col; right++) { // Calculate sum between current left // and right for every row 'i' // consider value '1' as '1' and // value '0' as '-1' for (int i = 0; i < row; i++) temp[i] += mat[i][right] ? 1 : -1; // Find largest subarray with 0 sum in // temp[]. The subArrWithSumZero() function // also sets values of finalTop, finalBottom, // finalLeft and finalRight if there exists // a subarray with sum 0 in temp if (subArrWithSumZero(temp, startRow, endRow, row)) { int area = (right - left + 1) * (endRow - startRow + 1); // Compare current 'area' with previous area // and accordingly update final values if (maxArea < area) { finalTop = startRow; finalBottom = endRow; finalLeft = left; finalRight = right; maxArea = area; } } } } // if true then there is no rectangular submatrix // with equal number of 1's and 0's if (maxArea == 0) cout << "No such rectangular submatrix exists:"; // displaying the top left and bottom right boundaries // with the area of the rectangular submatrix else { cout << "(Top, Left): " << "(" << finalTop << ", " << finalLeft << ")" << endl; cout << "(Bottom, Right): " << "(" << finalBottom << ", " << finalRight << ")" << endl; cout << "Area: " << maxArea << " sq.units"; }}// Driver program to test aboveint main(){ int mat[MAX_ROW][MAX_COL] = { { 0, 0, 1, 1 }, { 0, 1, 1, 0 }, { 1, 1, 1, 0 }, { 1, 0, 0, 1 } }; int row = 4, col = 4; maxAreaRectWithSumZero(mat, row, col); return 0;} |
Java
// Java implementation to find largest area rectangular// submatrix with equal number of 1's and 0'simport java.io.*;import java.util.*;class GFG { // This function basically finds largest 0 // sum subarray in arr[0..n-1]. If 0 sum // doesn't exist, then it returns false. Else // it returns true and sets starting and // ending indexes as start and end. public static boolean subArrWithSumZero(int arr[], int start, int end, int n) { // to store cumulative sum int sum[] = new int[n]; // map to store the indexes of sum HashMap<Integer, Integer> um = new HashMap<Integer, Integer>(); // build up the cumulative sum[] array sum[0] = arr[0]; for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + arr[i]; // to store the maximum length subarray // with sum equal to 0 int maxLen = 0; // traverse to the sum[] array for (int i = 0; i < n; i++) { // if true, then there is a subarray // with sum equal to 0 from the // beginning up to index 'i' if (sum[i] == 0) { // update the required variables start = 0; end = i; maxLen = (i + 1); } // else if true, then sum[i] has not // seen before in 'um' else if (um.get(sum[i]) == null) um.put(sum[i], i); // sum[i] has been seen before in the // unordered_map 'um' else { // if previous subarray length is smaller // than the current subarray length, then // update the required variables if (maxLen < (i - um.get(sum[i]))) { maxLen = i - um.get(sum[i]); start = i - um.get(sum[i]) + 1; end = i; } } } // if true, then there is no // subarray with sum equal to 0 if (maxLen == 0) return false; // else return true return true; } // function to find largest area rectangular // submatrix with equal number of 1's and 0's public static void maxAreaRectWithSumZero(int mat[][], int row, int col) { // to store intermediate values int temp[] = new int[row]; int startRow = 0, endRow = 0; // to store the final outputs int finalLeft = -1, finalRight = -1, finalTop = -1, finalBottom = -1; int maxArea = 0; // Set the left column for (int left = 0; left < col; left++) { // Set the right column for the left column // set by outer loop for (int right = left; right < col; right++) { // Calculate sum between current left // and right for every row 'i' // consider value '1' as '1' and // value '0' as '-1' for (int i = 0; i < row; i++) temp[i] += (mat[i][right] != 0) ? 1 : -1; // Find largest subarray with 0 sum in // temp[]. The subArrWithSumZero() function // also sets values of finalTop, // finalBottom, finalLeft and finalRight if // there exists a subarray with sum 0 in // temp if (subArrWithSumZero(temp, startRow, endRow, row)) { int area = (right - left + 1) * (endRow - startRow + 1); // Compare current 'area' with previous // area and accordingly update final // values if (maxArea < area) { finalTop = startRow; finalBottom = endRow; finalLeft = left; finalRight = right; maxArea = area; } } } } // if true then there is no rectangular submatrix // with equal number of 1's and 0's if (maxArea == 0) System.out.print( "No such rectangular submatrix exists:"); // displaying the top left and bottom right // boundaries with the area of the rectangular // submatrix else { System.out.println("(Top, Left): " + "(" + finalTop + ", " + finalLeft + ")"); System.out.println("(Bottom, Right): " + "(" + finalBottom + ", " + finalRight + ")"); System.out.println("Area: " + maxArea + " sq.units"); } } // Driver Code public static void main(String[] args) { int mat[][] = { { 0, 0, 1, 1 }, { 0, 1, 1, 0 }, { 1, 1, 1, 0 }, { 1, 0, 0, 1 } }; int row = 4, col = 4; maxAreaRectWithSumZero(mat, row, col); }}// This code is contributed by Rohit Pradhan |
Python3
# code# this function first convert 0->-1. Then it creates an memory array# which stores sum of all element in row from column index i to j# then each array is passed into getmax functiondef solve(m,R,C): for i in range(R): for j in range(C): if m[i][j] == 0: m[i][j] = -1 maxval =0 temparr = [[0]*(C+1) for _ in range(R)] for i in range(C): for j in range(i, C): temp = [] for k in range(R): if i==0: temparr[k][j+1] = temparr[k][j]+m[k][j] temp.append(temparr[k][j+1]) else: temp.append(temparr[k][j+1]-m[k][i]) maxval = max(maxval, getmax(temp)*(j-i+1)) return maxval # This function basically finds largest 0# sum subarray in arr[0..n-1]. it returns # length of largest subarray in arr with sum of subarray is zero.def getmax(arr): dic = dict() currsum = 0 maxlen = 0 for i in range(len(arr)): currsum+= arr[i] if currsum ==0: maxlen = i+1 if currsum not in dic: dic[currsum] = i else: temp = i- dic[currsum] maxlen = max(temp, maxlen) return maxlenif __name__ == '__main__': # m = [[0, 0, 1, 1],[0, 1, 1, 0],[1, 1, 1, 0],[1, 0, 0, 1]] m = [[0, 0, 1, 1], [0, 1, 1, 1]] R = len(m) C = len(m[0]) ans = solve(m, R, C) print(ans) |
C#
// C# implementation to find largest area rectangular// submatrix with equal number of 1's and 0'susing System;using System.Collections.Generic;public class GFG { // This function basically finds largest 0 // sum subarray in arr[0..n-1]. If 0 sum // doesn't exist, then it returns false. Else // it returns true and sets starting and // ending indexes as start and end. public static bool subArrWithSumZero(int[] arr, int start, int end, int n) { // to store cumulative sum int[] sum = new int[n]; // map to store the indexes of sum Dictionary<int, int> um = new Dictionary<int, int>(); // build up the cumulative sum[] array sum[0] = arr[0]; for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + arr[i]; // to store the maximum length subarray // with sum equal to 0 int maxLen = 0; // traverse to the sum[] array for (int i = 0; i < n; i++) { // if true, then there is a subarray // with sum equal to 0 from the // beginning up to index 'i' if (sum[i] == 0) { // update the required variables start = 0; end = i; maxLen = (i + 1); } // else if true, then sum[i] has not // seen before in 'um' else if (!um.ContainsKey(sum[i])) um[sum[i]] = i; // sum[i] has been seen before in the // unordered_map 'um' else { // if previous subarray length is smaller // than the current subarray length, then // update the required variables if (maxLen < (i - um[sum[i]])) { maxLen = i - um[sum[i]]; start = i - um[sum[i]] + 1; end = i; } } } // if true, then there is no // subarray with sum equal to 0 if (maxLen == 0) return false; // else return true return true; } // function to find largest area rectangular // submatrix with equal number of 1's and 0's public static void maxAreaRectWithSumZero(int[, ] mat, int row, int col) { // to store intermediate values int[] temp = new int[row]; int startRow = 0, endRow = 0; // to store the final outputs int finalLeft = -1, finalRight = -1, finalTop = -1, finalBottom = -1; int maxArea = 0; // Set the left column for (int left = 0; left < col; left++) { // Set the right column for the left column // set by outer loop for (int right = left; right < col; right++) { // Calculate sum between current left // and right for every row 'i' // consider value '1' as '1' and // value '0' as '-1' for (int i = 0; i < row; i++) temp[i] += (mat[i, right] != 0) ? 1 : -1; // Find largest subarray with 0 sum in // temp[]. The subArrWithSumZero() function // also sets values of finalTop, // finalBottom, finalLeft and finalRight if // there exists a subarray with sum 0 in // temp if (subArrWithSumZero(temp, startRow, endRow, row)) { int area = (right - left + 1) * (endRow - startRow + 1); // Compare current 'area' with previous // area and accordingly update final // values if (maxArea < area) { finalTop = startRow; finalBottom = endRow; finalLeft = left; finalRight = right; maxArea = area; } } } } // if true then there is no rectangular submatrix // with equal number of 1's and 0's if (maxArea == 0) Console.Write( "No such rectangular submatrix exists:"); // displaying the top left and bottom right // boundaries with the area of the rectangular // submatrix else { Console.WriteLine("(Top, Left): " + "(" + finalTop + ", " + finalLeft + ")"); Console.WriteLine("(Bottom, Right): " + "(" + finalBottom + ", " + finalRight + ")"); Console.WriteLine("Area: " + maxArea + " sq.units"); } } static public void Main() { // Code int[, ] mat = { { 0, 0, 1, 1 }, { 0, 1, 1, 0 }, { 1, 1, 1, 0 }, { 1, 0, 0, 1 } }; int row = 4, col = 4; maxAreaRectWithSumZero(mat, row, col); }}// This code is contributed by lokesh |
Javascript
// JavaScript implementation to find largest area // rectangular submatrix with equal number of 1's and 0's// This function basically finds largest 0// sum subarray in arr[0..n-1]. If 0 sum// doesn't exist, then it returns false. Else// it returns true and sets starting and// ending indexes as start and end.function subArrWithSumZero(arr, start, end, n){ // to store cumulative sum var sum = new Array(n); // map to store the indexes of sum var um = new Map(); // build up the cumulative sum[] array sum[0] = arr[0]; for (let i = 1; i < n; i++) sum[i] = sum[i - 1] + arr[i]; // to store the maximum length subarray // with sum equal to 0 var maxLen = 0; for(let i=0;i<n;i++) { // if true, then there is a subarray // with sum equal to 0 from the // beginning up to index 'i' if (sum[i] == 0) { // update the required variables start = 0; end = i; maxLen = (i + 1); } // else if true, then sum[i] has not // seen before in 'um' else if(um.get(sum[i])==null){ um.set(sum[i], i); } // sum[i] has been seen before in the // unordered_map 'um' else { // if previous subarray length is smaller // than the current subarray length, then // update the required variables if (maxLen < (i - um.get(sum[i]))) { maxLen = i - um.get(sum[i]); start = i - um.get(sum[i]) + 1; end = i; } } } // if true, then there is no // subarray with sum equal to 0 if (maxLen == 0) return false; // else return true return true;}// function to find largest area rectangular// submatrix with equal number of 1's and 0'sfunction maxAreaRectWithSumZero(mat, row, col){ // to store intermediate values var temp = new Array(row); var startRow = 0, endRow = 0; // to store the final outputs var finalLeft = -1, finalRight = -1, finalTop = -1, finalBottom = -1; var maxArea = 0; // Set the left column for(let left = 0; left < col; left++) { // Set the right column for the left column // set by outer loop for(let right = left; right < col; right++) { // Calculate sum between current left // and right for every row 'i' // consider value '1' as '1' and // value '0' as '-1' for(let i = 0;i < row; i++){ temp[i] += (mat[i][right] != 0) ? 1 : -1; } // Find largest subarray with 0 sum in // temp[]. The subArrWithSumZero() function // also sets values of finalTop, // finalBottom, finalLeft and finalRight if // there exists a subarray with sum 0 in // temp if(subArrWithSumZero(temp, startRow, endRow, row)){ let area = (right - left + 1) * (endRow - startRow + 1); // Compare current 'area' with previous // area and accordingly update final // values if (maxArea < area) { finalTop = startRow; finalBottom = endRow; finalLeft = left; finalRight = right; maxArea = area; } } } } // if true then there is no rectangular submatrix // with equal number of 1's and 0's if (maxArea == 0){ console.log("No such rectangular submatrix exists:"); } // displaying the top left and bottom right // boundaries with the area of the rectangular // submatrix else{ console.log("(Top, Left): " + "(" + finalTop + ", " + finalLeft + ")" + "<br>"); console.log("(Bottom, Right): " + "(" + finalBottom + ", " + finalRight + ")" + "<br>"); console.log("Area: " + maxArea + " sq.units" + "<br>"); }}var mat = [ [ 0, 0, 1, 1 ], [ 0, 1, 1, 0 ], [ 1, 1, 1, 0 ], [ 1, 0, 0, 1 ] ] var row = 4, col = 4;maxAreaRectWithSumZero(mat, row, col);// This code is contributed by lokeshmvs21. |
Output
(Top, Left): (0, 0) (Bottom, Right): (3, 1) Area: 8 sq.units
Time Complexity: O(n3) Auxiliary Space: O(n) This article is contributed by Ayush Jauhari. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org.
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