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

• Difficulty Level : Hard
• Last Updated : 14 Aug, 2021

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:

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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 ``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 cummulative 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 >& matrix, ``int` `k)``{``    ``// Stores the number of rows``    ``// and columns``    ``int` `row = matrix.size();``    ``int` `col = matrix.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 > 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)``        ``Set sumSet = ``new` `HashSet();` `        ``// 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``            ``ArrayList list = ``new` `ArrayList();``            ``list.addAll(sumSet);` `            ``int` `it = Integer.MIN_VALUE;``            ``for` `(``int` `e : list)``                ``if` `(e >= (curSum - k)) {``                    ``it = e;``                    ``break``;``                ``}` `            ``// If upper_bound of (cSum - K)``            ``// exists, then update max sum``            ``if` `(it != Integer.MIN_VALUE) {``                ``curMax = Math.max(curMax,``                                  ``curSum - it);``            ``}` `            ``// Insert cummulative 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`

## 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 cummulative 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 cummulative 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

 ``
Output:
`2`

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

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