Maximum sum not exceeding K possible for any rectangle of a Matrix

Given a matrix mat[][] of dimensions N * M, and an integer K, the task is to find the maximum sum of any rectangle possible from the given matrix, whose sum of elements is at most K.

Examples:

Input: mat[][] ={{1, 0, 1}, {0, -2, 3}}, K = 2
Output: 2
Explanation: The maximum sum possible in any rectangle from the matrix is 2 (<= K), obtained from the matrix {{0, 1}, {-2, 3}}.

Input: mat[][] = {{2, 2, -1}}, K = 3
Output: 3
Explanation: The maximum sum rectangle is {{2, 2, -1}} is 3 ( <- K).

Naive Approach: The simplest approach is to check for all possible submatrices from the given matrix whether the sum of its elements is at most K or not. If found to be true, store the sum of that submatrix. Finally, print the maximum sum of such submatrices obtained. 

Time Complexity: O(N⁶)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using an approach similar to finding the maximum sum rectangle in a 2D matrix. The only difference is that the sum of the rectangle must not exceed K. The idea is to fix the left and right columns one by one and in each iteration, store the sum of each row in the current rectangle and find the maximum subarray sum less than K in this array. Follow the steps below to solve the problem:

• Initialize a variable, say res, that stores the maximum sum of elements of the submatrix having sum at most K.
• Iterate over the range [0, M – 1] using the variable i for the left column and perform the following steps:
  • Initialize an array V[] of size N, to store the sum of elements of each row in between the left and right column pair.
  • Iterate over the range [0, M – 1] using a variable j for the right column and perform the following steps:
    • Find the sum between current left and right column of every row and update the sum in the array V[].
    • Find the maximum sum subarray with sum less than K in V[] and store the result in ans.
    • If the value of ans is greater than the value of res, update res to ans.
• After completing the above steps, print the value of res as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum possible
// sum of  arectangle which is less than K
int maxSubarraySum(vector<int>& sum,
                   int k, int row)
{
    int curSum = 0, curMax = INT_MIN;

    // Stores the values (cum_sum - K)
    set<int> sumSet;

    // Insert 0 into the set sumSet
    sumSet.insert(0);

    // Traverse over the rows
    for (int r = 0; r < row; ++r) {

        // Get cumulative sum from [0 to i]
        curSum += sum[r];

        // Search for upperbound of
        // (cSum - K) in the hashmap
        auto it = sumSet.lower_bound(curSum - k);

        // If upper_bound of (cSum - K)
        // exists, then update max sum
        if (it != sumSet.end()) {
            curMax = max(curMax,
                         curSum - *it);
        }

        // Insert cumulative value
        // in the hashmap
        sumSet.insert(curSum);
    }

    // Return the maximum sum
    // which is less than K
    return curMax;
}

// Function to find the maximum sum of
// rectangle such that its sum is no
// larger than K
void maxSumSubmatrix(
    vector<vector<int> >& matrix, int k)
{
    // Stores the number of rows
    // and columns
    int row = matrix.size();
    int col = matrix[0].size();

    // Store the required result
    int ret = INT_MIN;

    // Set the left column
    for (int i = 0; i < col; ++i) {
        vector<int> sum(row, 0);

        // Set the right column for the
        // left column set by outer loop
        for (int j = i; j < col; ++j) {

            // Calculate sum between the
            // current left and right
            // for every row
            for (int r = 0; r < row; ++r) {
                sum[r] += matrix[r][j];
            }

            // Stores the sum of rectangle
            int curMax = maxSubarraySum(
                sum, k, row);

            // Update the overall maximum sum
            ret = max(ret, curMax);
        }
    }

    // Print the result
    cout << ret;
}

// Driver Code
int main()
{
    vector<vector<int> > matrix
        = { { 1, 0, 1 }, { 0, -2, 3 } };
    int K = 2;

    // Function Call
    maxSumSubmatrix(matrix, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG {

    // Function to find the maximum possible
    // sum of arectangle which is less than K
    static int maxSubarraySum(int[] sum, int k, int row)
    {
        int curSum = 0, curMax = Integer.MIN_VALUE;

        // Stores the values (cum_sum - K)
        TreeSet<Integer> sumSet = new TreeSet<Integer>();

        // Insert 0 into the set sumSet
        sumSet.add(0);

        // Traverse over the rows
        for (int r = 0; r < row; ++r) {

            // Get cumulative sum from [0 to i]
            curSum += sum[r];

            // Search for upperbound of
            // (cSum - K) in the hashmap
            Integer target = sumSet.ceiling(curSum - k);
            // If upper_bound of (cSum - K)
            // exists, then update max sum
            if (target != null) {
                curMax = Math.max(curMax, curSum - target);
            }

            // Insert cumulative value
            // in the hashmap
            sumSet.add(curSum);
        }

        // Return the maximum sum
        // which is less than K
        return curMax;
    }

    // Function to find the maximum sum of
    // rectangle such that its sum is no
    // larger than K
    static void maxSumSubmatrix(int[][] matrix, int k)
    {

        // Stores the number of rows
        // and columns
        int row = matrix.length;
        int col = matrix[0].length;

        // Store the required result
        int ret = Integer.MIN_VALUE;

        // Set the left column
        for (int i = 0; i < col; ++i) {
            int[] sum = new int[row];

            // Set the right column for the
            // left column set by outer loop
            for (int j = i; j < col; ++j) {

                // Calculate sum between the
                // current left and right
                // for every row
                for (int r = 0; r < row; ++r) {
                    sum[r] += matrix[r][j];
                }

                // Stores the sum of rectangle
                int curMax = maxSubarraySum(sum, k, row);

                // Update the overall maximum sum
                ret = Math.max(ret, curMax);
            }
        }

        // Print the result
        System.out.print(ret);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[][] matrix = { { 5, -4, -3, 4 },
                           { -3, -4, 4, 5 },
                           { 5, 1, 5, -4  } };
        int K = 8;

        // Function Call
        maxSumSubmatrix(matrix, K);
    }
}

// This code is contributed by avanitrachhadiya2155
// This code is improved by Snigdha Patil

Python3

# Python3 program for the above approach
from bisect import bisect_left, bisect_right
import sys

# Function to find the maximum possible
# sum of  arectangle which is less than K
def maxSubarraySum(sum, k, row):

    curSum, curMax = 0, -sys.maxsize - 1

    # Stores the values (cum_sum - K)
    sumSet = {}

    # Insert 0 into the set sumSet
    sumSet[0] = 1

    # Traverse over the rows
    for r in range(row):

        # Get cumulative sum from [0 to i]
        curSum += sum[r]

        # Search for upperbound of
        # (cSum - K) in the hashmap
        arr = list(sumSet.keys())

        it = bisect_left(arr, curSum - k)

        # If upper_bound of (cSum - K)
        # exists, then update max sum
        if (it != len(arr)):
            curMax = max(curMax, curSum - it)

        # Insert cumulative value
        # in the hashmap
        sumSet[curSum] = 1

    # Return the maximum sum
    # which is less than K
    return curMax

# Function to find the maximum sum of
# rectangle such that its sum is no
# larger than K
def maxSumSubmatrix(matrix, k):

    # Stores the number of rows
    # and columns
    row = len(matrix)
    col = len(matrix[0])

    # Store the required result
    ret = -sys.maxsize - 1

    # Set the left column
    for i in range(col):
        sum = [0] * (row)

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

            # Calculate sum between the
            # current left and right
            # for every row
            for r in range(row):
                sum[r] += matrix[r][j]

            # Stores the sum of rectangle
            curMax = maxSubarraySum(sum, k, row)

            # Update the overall maximum sum
            ret = max(ret, curMax)

    # Print the result
    print(ret)

# Driver Code
if __name__ == '__main__':

    matrix = [ [ 1, 0, 1 ], [ 0, -2, 3 ] ]
    K = 2

    # Function Call
    maxSumSubmatrix(matrix, K)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the maximum possible
// sum of  arectangle which is less than K
static int maxSubarraySum(int[] sum,int k, int row)
{
    int curSum = 0, curMax = Int32.MinValue;

    // Stores the values (cum_sum - K)
    HashSet<int> sumSet = new HashSet<int>();

    // Insert 0 into the set sumSet
    sumSet.Add(0);

    // Traverse over the rows
    for(int r = 0; r < row; ++r)
    {

        // Get cumulative sum from [0 to i]
        curSum += sum[r];

        // Search for upperbound of
        // (cSum - K) in the hashmap
        List<int> list = new List<int>();
        list.AddRange(sumSet);
        int it = list.LastIndexOf(curSum - k);

        // If upper_bound of (cSum - K)
        // exists, then update max sum
        if (it > -1)
        {
            curMax = Math.Max(curMax,
                              curSum - it);
        }

        // Insert cumulative value
        // in the hashmap
        sumSet.Add(curSum);
    }

    // Return the maximum sum
    // which is less than K
    return curMax;
}

// Function to find the maximum sum of
// rectangle such that its sum is no
// larger than K
static void maxSumSubmatrix(int[,] matrix, int k)
{

    // Stores the number of rows
    // and columns
    int row = matrix.GetLength(0);
    int col = matrix.GetLength(1);

    // Store the required result
    int ret = Int32.MinValue;

    // Set the left column
    for(int i = 0; i < col; ++i)
    {
        int[] sum = new int[row];

        // Set the right column for the
        // left column set by outer loop
        for(int j = i; j < col; ++j)
        {

            // Calculate sum between the
            // current left and right
            // for every row
            for(int r = 0; r < row; ++r)
            {
                sum[r] += matrix[r, j];
            }

            // Stores the sum of rectangle
            int curMax = maxSubarraySum(
                sum, k, row);

            // Update the overall maximum sum
            ret = Math.Max(ret, curMax);
        }
    }

    // Print the result
    Console.Write(ret);
}

// Driver Code
static public void Main()
{
    int[,] matrix = { { 1, 0, 1 },
                      { 0, -2, 3 } };
    int K = 2;

    // Function Call
    maxSumSubmatrix(matrix, K);
}
}

// This code is contributed by rag2127

Javascript

<script>

// JavaScript program for the above approach

// Function to find the maximum possible
// sum of  arectangle which is less than K
function maxSubarraySum(sum,k,row)
{
    let curSum = 0, curMax = Number.MIN_VALUE;

    // Stores the values (cum_sum - K)
    let sumSet = new Set();

    // Insert 0 into the set sumSet
    sumSet.add(0);

    // Traverse over the rows
    for(let r = 0; r < row; ++r)
    {

        // Get cumulative sum from [0 to i]
        curSum += sum[r];

        // Search for upperbound of
        // (cSum - K) in the hashmap
        let list = [];
        list=Array.from(sumSet);

        let it = list.lastIndexOf(curSum - k);

        // If upper_bound of (cSum - K)
        // exists, then update max sum
        if (it >-1)
        {
            curMax = Math.max(curMax,
                              curSum - it);
        }

        // Insert cumulative value
        // in the hashmap
        sumSet.add(curSum);
    }

    // Return the maximum sum
    // which is less than K
    return curMax;
}

// Function to find the maximum sum of
// rectangle such that its sum is no
// larger than K
function maxSumSubmatrix(matrix,  k)
{

    // Stores the number of rows
    // and columns
    let row = matrix.length;
    let col = matrix[0].length;

    // Store the required result
    let ret = Number.MIN_VALUE;

    // Set the left column
    for(let i = 0; i < col; ++i)
    {
        let sum = new Array(row);
         for(let j=0;j<row;j++)
            sum[j]=0;
        // Set the right column for the
        // left column set by outer loop
        for(let j = i; j < col; ++j)
        {

            // Calculate sum between the
            // current left and right
            // for every row
            for(let r = 0; r < row; ++r)
            {
                sum[r] += matrix[r][j];
            }

            // Stores the sum of rectangle
            let curMax = maxSubarraySum(
                sum, k, row);

            // Update the overall maximum sum
            ret = Math.max(ret, curMax);
        }
    }

    // Print the result
    document.write(ret);
}

// Driver Code
let matrix=[[ 1, 0, 1 ],
            [ 0, -2, 3 ]];
let K = 2;

maxSumSubmatrix(matrix, K)

// This code is contributed by patel2127

</script>

Output:

2

Time Complexity: O(N³*log(N))
Auxiliary Space: O(N)