Maximum sub-matrix area having count of 1’s one more than count of 0’s

Given a N x N binary matrix. The problem is finding the maximum area sub-matrix having a count of 1’s one more than count of 0’s.

Examples:

Input : mat[][] = { {1, 0, 0, 1},
                    {0, 1, 1, 1},
                    {1, 0, 0, 0},
                    {0, 1, 0, 1} }
Output : 9
The sub-matrix defined by the boundary values (1, 1) and (3, 3).
{ {1, 0, 0, 1},
  {0, 1, 1, 1},
  {1, 0, 0, 0},
  {0, 1, 0, 1} }


Naive Approach: Check every possible rectangle in given 2D matrix. This solution requires 4 nested loops and the time complexity of this solution would be O(n^4).

Efficient Approach: An efficient approach will be to use Longest subarray having count of 1s one more than count of 0s which reduces the time complexity to O(n^3). The idea is to fix the left and right columns one by one and find the maximum length contiguous rows having the count of 1’s one more than the count of 0’s for every left and right column pair. Find top and bottom row numbers (which have a maximum length) for every fixed left and right column pair. To find the top and bottom row numbers, calculate the sum of elements in every row from left to right and store these sums in an array say temp[] (consider 0 as -1 while adding it). So temp[i] indicates the sum of elements from left to right in row i. Using the approach in Longest subarray having count of 1s one more than count of 0s , temp[] array is used to get the maximum length subarray of temp[] having count of 1’s one more than a count of 0’s by obtaining the start and end row numbers, then these values can be used to find maximum possible area with left and right as boundary columns. To get the overall maximum area, compare this area with the maximum area so far.

Below is the implementation of the above approach:

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// C++ implementation to find
// the maximum area sub-matrix
// having count of 1's
// one more than count of 0's
#include <bits/stdc++.h>
  
using namespace std;
  
#define SIZE 10
  
// function to find the length of longest
// subarray having count of 1's one more
// than count of 0's
int lenOfLongSubarr(int arr[], int n,
                    int& start, int& finish)
{
    // unordered_map 'um' implemented as
    // hash table
    unordered_map<int, int> um;
    int sum = 0, maxLen = 0;
  
    // traverse the given array
    for (int i = 0; i < n; i++) {
  
        // accumulating sum
        sum += arr[i];
  
        // when subarray starts form index '0'
        if (sum == 1) {
            start = 0;
            finish = i;
            maxLen = i + 1;
        }
  
        // make an entry for 'sum' if it is
        // not present in 'um'
        else if (um.find(sum) == um.end())
            um[sum] = i;
  
        // check if 'sum-1' is present in 'um'
        // or not
        if (um.find(sum - 1) != um.end()) {
  
            // update 'start', 'finish'
            // and maxLength
            if (maxLen < (i - um[sum - 1]))
                start = um[sum - 1] + 1;
            finish = i;
            maxLen = i - um[sum - 1];
        }
    }
  
    // required maximum length
    return maxLen;
}
  
// function to find the maximum
// area sub-matrix having
// count of 1's one more than count of 0's
void largestSubmatrix(int mat[SIZE][SIZE], int n)
{
    // variables to store final
    // and intermediate results
    int finalLeft, finalRight, finalTop, finalBottom;
    int temp[n], maxArea = 0, len, start, finish;
  
    // set the left column
    for (int left = 0; left < n; 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 < n; right++) {
  
            // Calculate sum between current left and right
            // for every row 'i', consider '0' as '-1'
            for (int i = 0; i < n; ++i)
                temp[i] += mat[i][right] == 0 ? -1 : 1;
  
            // function to set the 'start' and 'finish'
            // variables having index values of
            // temp[] which contains the longest
            // subarray of temp[] having count of 1's
            // one more than count of 0's
            len = lenOfLongSubarr(temp, n, start, finish);
  
            // Compare with maximum area
            // so far and accordingly update the
            // final variables
            if ((len != 0) && (maxArea < (finish - start + 1)
                                             * (right - left + 1))) {
                finalLeft = left;
                finalRight = right;
                finalTop = start;
                finalBottom = finish;
                maxArea = (finish - start + 1) * (right - left + 1);
            }
        }
    }
  
    // Print final values
    cout << "(Top, Left): (" << finalTop << ", "
         << finalLeft << ")\n";
  
    cout << "(Bottom, Right): (" << finalBottom << ", "
         << finalRight << ")\n";
  
    cout << "Maximum area: " << maxArea;
}
  
// Driver Code
int main()
{
    int mat[SIZE][SIZE] = { { 1, 0, 0, 1 },
                            { 0, 1, 1, 1 },
                            { 1, 0, 0, 0 },
                            { 0, 1, 0, 1 } };
    int n = 4;
    largestSubmatrix(mat, n);
    return 0;
}

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Output:

(Top, Left): (1, 1)
(Bottom, Right): (3, 3)
Maximum area: 9

Time Complexity: O(N3).
Auxiliary Space: O(N).



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