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: 

C++

// 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;
}

Java

// Java implementation to find
// the maximum area sub-matrix
// having count of 1's
// one more than count of 0's
import java.util.*;

class GFG
{

static int start, finish;

// Function to find the length of longest
// subarray having count of 1's one more
// than count of 0's
static int lenOfLongSubarr(int []arr, int n)
{

    // unordered_map 'um' implemented as
    // hash table
    HashMap<Integer,Integer> um = new HashMap<Integer, Integer>();

    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.containsKey(sum))
            um.put(sum,i);

        // Check if 'sum-1' is present in 'um'
        // or not
        if (um.containsKey(sum - 1))
        {

            // Update 'start', 'finish'
            // and maxLength
            if (maxLen < (i - um.get(sum - 1)))
                start = um.get(sum - 1) + 1;

            finish = i;
            maxLen = i - um.get(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
static void largestSubmatrix(int [][]mat, int n)
{

    // Variables to store final
    // and intermediate results
    int finalLeft = 0, finalRight = 0,
        finalTop = 0, finalBottom = 0;
    int maxArea = 0, len;
    finish = 0;
    start=0;

    int []temp = new int[n];

    // Set the left column
    for(int left = 0; left < n; 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 < 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);

            // 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
    System.out.print("(Top, Left): (" + finalTop +
                ", " + finalLeft + ")\n");

    System.out.print("(Bottom, Right): (" + finalBottom +
                    ", " + finalRight + ")\n");

    System.out.print("Maximum area: " + maxArea);

}

// Driver code
public static void main(String[] args)
{
    int [][]mat = new int[][]{ { 1, 0, 0, 1 },
                            { 0, 1, 1, 1 },
                            { 1, 0, 0, 0 },
                            { 0, 1, 0, 1 } };
    int n = 4;

    largestSubmatrix(mat, n);
}
}

// This code is contributed by pratham76

Python3

# Python implementation to find
# the maximum area sub-matrix
# having count of 1's
# one more than count of 0's

# function to find the length of longest
# subarray having count of 1's one more
# than count of 0's
def lenOfLongSubarr(arr, n, start, finish):

  # unordered_map 'um' implemented as
  # hash table
  um = {}
  sum = 0
  maxLen = 0

  # traverse the given array
  for i in range(n):

    # 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'
    elif (sum not in um):
      um[sum] = i

    # check if 'sum-1' is present in 'um'
    # or not
    if (sum - 1 in um):

      # 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,start,finish]

# function to find the maximum
# area sub-matrix having
# count of 1's one more than count of 0's
def largestSubmatrix(mat, n):

  # variables to store final
  # and intermediate results
  temp = []
  maxArea = 0

  # set the left column
  for left in range(n):

    # Initialize all elements of temp as 0
    temp = [0 for i in range(n)]

    # Set the right column for the
    # left column set by outer loop
    for right in range(left, n):

      # Calculate sum between current left and right
      # for every row 'i', consider '0' as '-1'
      for i in range(n):
        if mat[i][right] == 0:
          temp[i] -= 1
        else:
          temp[i] += 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
      start = 0
      finish = 0
      fc = lenOfLongSubarr(temp, n, start, finish)
      len = fc[0]
      start = fc[1]
      finish = fc[2]

      # Compare with maximum area
      # so far and accordingly update the
      # final variables
      if ((len != 0) and (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
  print("(Top, Left): (",finalTop,", ",finalLeft,")")
  print("(Bottom, Right): (",finalBottom, ", ",finalRight,")")
  print("Maximum area: ", maxArea)

# Driver Code
mat = [[1, 0, 0, 1 ], [ 0, 1, 1, 1 ], [ 1, 0, 0, 0 ], [ 0, 1, 0, 1 ]]
n = 4
largestSubmatrix(mat, n)

# This code is contributed by rohitsingh07052

C#

// C# implementation to find
// the maximum area sub-matrix
// having count of 1's
// one more than count of 0's
using System;
using System.Collections.Generic;
using System.Collections;

class GFG{

// Function to find the length of longest
// subarray having count of 1's one more
// than count of 0's
static int lenOfLongSubarr(int []arr, int n,
                       ref int start, ref int finish)
{

    // unordered_map 'um' implemented as
    // hash table
    Dictionary<int,
               int> um = new Dictionary<int,
                                        int>();

    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.ContainsKey(sum))
            um[sum] = i;

        // Check if 'sum-1' is present in 'um'
        // or not
        if (um.ContainsKey(sum - 1))
        {

            // 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
static void largestSubmatrix(int [,]mat, int n)
{

    // Variables to store final
    // and intermediate results
    int finalLeft = 0, finalRight = 0,
         finalTop = 0, finalBottom = 0;
    int maxArea = 0, len, start = 0, finish = 0;

    int []temp = new int[n];

    // Set the left column
    for(int left = 0; left < n; left++)
    {

        // Initialize all elements of temp as 0
        Array.Fill(temp, 0);

        // 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, ref start,
                                           ref 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
    Console.Write("(Top, Left): (" + finalTop +
                  ", " + finalLeft + ")\n");

    Console.Write("(Bottom, Right): (" + finalBottom +
                     ", " + finalRight + ")\n");

    Console.Write("Maximum area: " + maxArea);

}

// Driver code
public static void Main(string[] args)
{
    int [,]mat = new int[,]{ { 1, 0, 0, 1 },
                             { 0, 1, 1, 1 },
                             { 1, 0, 0, 0 },
                             { 0, 1, 0, 1 } };
    int n = 4;

    largestSubmatrix(mat, n);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// JavaScript implementation to find
// the maximum area sub-matrix
// having count of 1's
// one more than count of 0's

// function to find the length of longest
// subarray having count of 1's one more
// than count of 0's
function lenOfLongSubarr(arr, n, start, finish){

  // unordered_map 'um' implemented as
  // hash table
  let um = new Map()
  let sum = 0
  let maxLen = 0

  // traverse the given array
  for(let 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.has(sum) == false)
      um.set(sum,i)

    // check if 'sum-1' is present in 'um'
    // or not
    if (um.has(sum - 1)){

      // update 'start', 'finish'
      // and maxLength
      if (maxLen < (i - um.get(sum - 1))){
        start = um.get(sum - 1) + 1
        finish = i
        maxLen = i - um.get(sum - 1)
      }
    }
  }

  // required maximum length
  return [maxLen,start,finish]
}

// function to find the maximum
// area sub-matrix having
// count of 1's one more than count of 0's
function largestSubmatrix(mat, n){

  // variables to store final
  // and intermediate results
  let temp = []
  let maxArea = 0

  let finalLeft,finalRight,finalTop,finalBottom;

  // set the left column
  for(let left=0;left<n;left++){

    // Initialize all elements of temp as 0
    temp = new Array(n).fill(0)

    // Set the right column for the
    // left column set by outer loop
    for(let right = left;right<n;right++){

      // Calculate sum between current left and right
      // for every row 'i', consider '0' as '-1'
      for(let i=0;i<n;i++){
        if(mat[i][right] == 0)
          temp[i] -= 1
        else
          temp[i] += 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
      let start = 0
      let finish = 0
      let fc = lenOfLongSubarr(temp, n, start, finish)
      let len = fc[0]
      start = fc[1]
      finish = fc[2]

      // 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
  document.write("(Top, Left): (" + finalTop + ", " + finalLeft + ")","</br>")
  document.write("(Bottom, Right): (" + finalBottom  + ", " + finalRight + ")","</br>")
  document.write("Maximum area: "+  maxArea,"</br>")
}

// Driver Code
let mat = [[1, 0, 0, 1 ], [ 0, 1, 1, 1 ], [ 1, 0, 0, 0 ], [ 0, 1, 0, 1 ]]
let n = 4
largestSubmatrix(mat, n)

// This code is contributed by shinjanpatra

</script>

Output

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

Complexity Analysis:

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