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 Kint 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 Kvoid 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 Codeint 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 approachimport 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 approachfrom bisect import bisect_left, bisect_rightimport sys# Function to find the maximum possible# sum of arectangle which is less than Kdef 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 Kdef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{ // Function to find the maximum possible// sum of arectangle which is less than Kstatic 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 Kstatic 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 Codestatic 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 Kfunction 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 Kfunction 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 Codelet 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)
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