Given a 2D matrix, find the largest rectangular sub-matrix whose sum is 0. for example consider the following N x M input matrix
Examples:
Input : 1, 2, 3
-3, -2, -1
1, 7, 5
Output : 1, 2, 3
-3, -2, -1
Input : 9, 7, 16, 5
1, -6, -7, 3
1, 8, 7, 9
7, -2, 0, 10
Output :-6, -7
8, 7
-2, 0
The naive solution for this problem is to check every possible rectangle in given 2D array. This solution requires 4 nested loops and time complexity of this solution would be O(n^4).
The solution is based on Maximum sum rectangle in a 2D matrix. The idea is to reduce the problem to 1 D array. We can use Hashing to find maximum length of sub-array in 1-D array in O(n) time. 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 is 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 on temp, and no. of elements is greater than previous no. of elements then update the values of final row_up, final row_down, final col_left, final col_right.
Implementation:
CPP
// A C++ program to find Largest rectangular// sub-matrix whose sum is 0#include <bits/stdc++.h>using namespace std;const int MAX = 100;// This function basically finds largest 0// sum subarray in temp[0..n-1]. If 0 sum// does't exist, then it returns false. Else// it returns true and sets starting and// ending indexes as starti and endj.bool sumZero(int temp[], int* starti, int* endj, int n){ // Map to store the previous sums map<int, int> presum; int sum = 0; // Initialize sum of elements // Initialize length of sub-array with sum 0 int max_length = 0; // Traverse through the given array for (int i = 0; i < n; i++) { // Add current element to sum sum += temp[i]; if (temp[i] == 0 && max_length == 0) { *starti = i; *endj = i; max_length = 1; } if (sum == 0) { if (max_length < i + 1) { *starti = 0; *endj = i; } max_length = i + 1; } // Look for this sum in Hash table if (presum.find(sum) != presum.end()) { // store previous max_length so // that we can check max_length // is updated or not int old = max_length; // If this sum is seen before, // then update max_len max_length = max(max_length, i - presum[sum]); if (old < max_length) { // If max_length is updated then // enter and update start and end // point of array *endj = i; *starti = presum[sum] + 1; } } else // Else insert this sum with // index in hash table presum[sum] = i; } // Return true if max_length is non-zero return (max_length != 0);}// The main function that finds Largest rectangle// sub-matrix in a[][] whose sum is 0.void sumZeroMatrix(int a[][MAX], int row, int col){ int temp[row]; // Variables to store the final output int fup = 0, fdown = 0, fleft = 0, fright = 0; int sum; int up, down; int maxl = INT_MIN; // 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' for (int i = 0; i < row; i++) temp[i] += a[i][right]; // Find largest subarray with 0 sum in // temp[]. The sumZero() function also // sets values of start and finish. So // 'sum' is sum of rectangle between (start, // left) and (finish, right) which is // boundary columns strictly as left and right. bool sum = sumZero(temp, &up, &down, row); int ele = (down - up + 1) * (right - left + 1); // Compare no. of elements with previous // no. of elements in sub-Matrix. // If new sub-matrix has more elements // then update maxl and final boundaries // like fup, fdown, fleft, fright if (sum && ele > maxl) { fup = up; fdown = down; fleft = left; fright = right; maxl = ele; } } } // If there is no change in boundaries // than check if a[0][0] is 0 // If it not zero then print // that no such zero-sum sub-matrix exists if (fup == 0 && fdown == 0 && fleft == 0 && fright == 0 && a[0][0] != 0) { cout << "No zero-sum sub-matrix exists"; return; } // Print final values for (int j = fup; j <= fdown; j++) { for (int i = fleft; i <= fright; i++) cout << a[j][i] << " "; cout << endl; }}// Driver program to test above functionsint main(){ int a[][MAX] = { { 9, 7, 16, 5 }, { 1, -6, -7, 3 }, { 1, 8, 7, 9 }, { 7, -2, 0, 10 } }; int row = 4, col = 4; sumZeroMatrix(a, row, col); return 0;} |
Java
// A Java program to find Largest rectangular// sub-matrix whose sum is 0import java.util.*;public class Main { static final int MAX = 100; // This function basically finds largest 0 // sum subarray in temp[0..n-1]. If 0 sum // does't exist, then it returns false. Else // it returns true and sets starting and // ending indexes as starti and endj. static boolean sumZero(int[] temp, int[] starti, int[] endj, int n) { // Map to store the previous sums Map<Integer, Integer> presum = new HashMap<>(); int sum = 0; // Initialize length of sub-array with sum 0 int max_length = 0; // Traverse through the given array for (int i = 0; i < n; i++) { sum += temp[i]; if (temp[i] == 0 && max_length == 0) { starti[0] = i; endj[0] = i; max_length = 1; } if (sum == 0) { if (max_length < i + 1) { starti[0] = 0; endj[0] = i; } max_length = i + 1; } // Look for this sum in Hash table if (presum.containsKey(sum)) { // store previous max_length so // that we can` check max_length // is updated or not int old = max_length; // If this sum is seen before, // then update max_len max_length = Math.max(max_length, i - presum.get(sum)); if (old < max_length) { // If max_length is updated then // enter and update start and end // point of array endj[0] = i; starti[0] = presum.get(sum) + 1; } } else { // Else insert this sum with // index in hash table presum.put(sum, i); } } // Return true if max_length is non-zero return (max_length != 0); } // The main function that finds Largest rectangle // sub-matrix in a[][] whose sum is 0. static void sumZeroMatrix(int[][] a, int row, int col) { int[] temp = new int[row]; // Variables to store the final output int fup = 0, fdown = 0, fleft = 0, fright = 0; int maxl = Integer.MIN_VALUE; // Set the left column for (int left = 0; left < col; left++) { // Initialize all elements of temp as 0 Arrays.fill(temp, 0); // 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' for (int i = 0; i < row; i++) { temp[i] += a[i][right]; } int[] up = new int[1]; int[] down = new int[1]; // Find largest subarray with 0 sum in // temp[]. The sumZero() function also // sets values of start and finish. So // 'sum' is sum of rectangle between (start, // left) and (finish, right) which is // boundary columns strictly as left and // right. boolean s = sumZero(temp, up, down, row); int ele = (down[0] - up[0] + 1) * (right - left + 1); // Compare no. of elements with previous // no. of elements in sub-Matrix. // If new sub-matrix has more elements // then update maxl and final boundaries // like fup, fdown, fleft, fright if (s && ele > maxl) { fup = up[0]; fdown = down[0]; fleft = left; fright = right; maxl = ele; } } } // If there is no change in boundaries // than check if a[0][0] is 0 // If it not zero then print // that no such zero-sum sub-matrix exists if (fup == 0 && fdown == 0 && fleft == 0 && fright == 0 && a[0][0] != 0) { System.out.println( "No zero-sum sub-matrix exists"); return; } // Print final values for (int j = fup; j <= fdown; j++) { for (int i = fleft; i <= fright; i++) { System.out.print(a[j][i] + " "); } System.out.println(); } } // Driver program to test above functions public static void main(String[] args) { int[][] a = { { 9, 7, 16, 5 }, { 1, -6, -7, 3 }, { 1, 8, 7, 9 }, { 7, -2, 0, 10 } }; int row = 4, col = 4; sumZeroMatrix(a, row, col); }}// This code is contributed by shiv1o43g |
Python3
# A Python program to find Largest rectangular# sub-matrix whose sum is 0from typing import List, Tuplefrom collections import defaultdict# This function basically finds largest 0# sum subarray in temp[0..n-1]. If 0 sum# does't exist, then it returns false. Else# it returns true and sets starting and# ending indexes as starti and endj.def sum_zero(temp: List[int], starti: List[int], endj: List[int], n: int) -> bool: presum = defaultdict(int) sum_ = 0 # Initialize length of sub-array with sum 0 max_length = 0 for i in range(n): sum_ += temp[i] if temp[i] == 0 and max_length == 0: starti[0] = i endj[0] = i max_length = 1 if sum_ == 0: if max_length < i + 1: starti[0] = 0 endj[0] = i max_length = i + 1 if sum_ in presum: old = max_length max_length = max(max_length, i - presum[sum_]) if old < max_length: endj[0] = i starti[0] = presum[sum_] + 1 else: presum[sum_] = i return max_length != 0# The main function that finds Largest rectangle# sub-matrix in a[][] whose sum is 0.def sum_zero_matrix(a: List[List[int]], row: int, col: int) -> None: temp = [0] * row # Variables to store the final output fup = fdown = fleft = fright = 0 maxl = float('-inf') # Set the left column for left in range(col): temp = [0] * row # Set the right column for the left column # set by outer loop for right in range(left, col): # Calculate sum between current left # and right for every row 'i' for i in range(row): temp[i] += a[i][right] up, down = [0], [0] # Find largest subarray with 0 sum in # temp[]. The sumZero() function also # sets values of start and finish. So # 'sum' is sum of rectangle between (start, # left) and (finish, right) which is # boundary columns strictly as left and right. s = sum_zero(temp, up, down, row) ele = (down[0] - up[0] + 1) * (right - left + 1) # Compare no. of elements with previous # no. of elements in sub-Matrix. # If new sub-matrix has more elements # then update maxl and final boundaries # like fup, fdown, fleft, fright if s and ele > maxl: fup = up[0] fdown = down[0] fleft = left fright = right maxl = ele # If there is no change in boundaries # than check if a[0][0] is 0 # If it not zero then print # that no such zero-sum sub-matrix exists if fup == fdown == fleft == fright == 0 and a[0][0] != 0: # Print final values print("No zero-sum sub-matrix exists") return for j in range(fup, fdown+1): for i in range(fleft, fright+1): print(a[j][i], end=" ") print()# Driver program to test above functionsif __name__ == '__main__': a = [[9, 7, 16, 5], [1, -6, -7, 3], [1, 8, 7, 9], [7, -2, 0, 10]] row, col = 4, 4 sum_zero_matrix(a, row, col)# This code is contributed by shivhack999 |
C#
// A C# program to find Largest rectangular// sub-matrix whose sum is 0using System;using System.Collections.Generic;class Program { const int MAX = 100; // This function basically finds largest 0 // sum subarray in temp[0..n-1]. If 0 sum // does't exist, then it returns false. Else // it returns true and sets starting and // ending indexes as starti and endj. static bool SumZero(int[] temp, out int starti, out int endj, int n) { var presum = new Dictionary<int, int>(); int sum = 0; int max_length = 0; starti = -1; endj = -1; for (int i = 0; i < n; i++) { sum += temp[i]; if (temp[i] == 0 && max_length == 0) { starti = i; endj = i; max_length = 1; } if (sum == 0) { if (max_length < i + 1) { starti = 0; endj = i; } max_length = i + 1; } if (presum.TryGetValue(sum, out int j)) { int old = max_length; max_length = Math.Max(max_length, i - j); if (old < max_length) { endj = i; starti = presum[sum] + 1; } } else { presum[sum] = i; } } return (max_length != 0); } // The main function that finds Largest rectangle // sub-matrix in a[][] whose sum is 0. static void SumZeroMatrix(int[, ] a, int row, int col) { int[] temp = new int[row]; int fup = 0, fdown = 0, fleft = 0, fright = 0; int sum; int up, down; int maxl = int.MinValue; for (int left = 0; left < col; left++) { Array.Clear(temp, 0, temp.Length); for (int right = left; right < col; right++) { for (int i = 0; i < row; i++) { temp[i] += a[i, right]; } bool sumZero = SumZero(temp, out up, out down, row); int ele = (down - up + 1) * (right - left + 1); if (sumZero && ele > maxl) { fup = up; fdown = down; fleft = left; fright = right; maxl = ele; } } } if (fup == 0 && fdown == 0 && fleft == 0 && fright == 0 && a[0, 0] != 0) { Console.WriteLine( "No zero-sum sub-matrix exists"); return; } for (int j = fup; j <= fdown; j++) { for (int i = fleft; i <= fright; i++) { Console.Write(a[j, i] + " "); } Console.WriteLine(); } } // Driver program to test above functions static void Main() { int[, ] a = { { 9, 7, 16, 5 }, { 1, -6, -7, 3 }, { 1, 8, 7, 9 }, { 7, -2, 0, 10 } }; int row = 4, col = 4; SumZeroMatrix(a, row, col); }}// This Code is contributed by Gaurav_Arora |
Output
-6 -7 8 7 -2 0
Time Complexity: O(n3)
Auxiliary Space: O(1)
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