Given a Matrix arr[][] of size M x N having positive integers and a number K, the task is to find the size of the largest square sub-matrix whose sum of elements is less than or equals to K.
Example:
Input:
arr[][] = { { 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 } },
K = 4
Output:
2
Explanation:
Maximum size square Sub-Matrix
with sum less than or equals to 22
1 1
1 1
Size is 4.
Input:
arr[][] = { { 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 },
{ 1, 1, 3, 2, 4, 3, 2 } },
K = 22
Output:
3
Explanation:
Maximum size square Sub-Matrix
with sum less than or equals to 22
1 1 3
1 1 3
1 1 3
Size is 3.
Approach:
- For the given matrix arr[][] create a prefix sum matrix(say sum[][]) such that sum[i][j] stores the sum of all the elements of the matrix of size i x j.
- For each row in prefix sum matrix sum[][] using Binary Search do the following:
- Perform Binary search with the lower limit as 0 end the upper limit as to maximum size of square matrix.
- Find the middle index (say mid).
- If the sum of elements of all possible square matrix of size mid is less than or equals to K, then update the lower limit as mid + 1 to find the maximum sum with size greater than mid.
- Else Update the upper limit as mid – 1 to find the maximum sum with size less than mid.
- Keep updating the maximum size of square matrix in each iteration for the given valid condition above.
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 size// of matrix with sum <= Kvoid findMaxMatrixSize(vector<vector<int> > arr, int K){ int i, j; // N size of rows and M size of cols int n = arr.size(); int m = arr[0].size(); // To store the prefix sum of matrix int sum[n + 1][m + 1]; // Create Prefix Sum for (int i = 0; i <= n; i++) { // Traverse each rows for (int j = 0; j <= m; j++) { if (i == 0 || j == 0) { sum[i][j] = 0; continue; } // Update the prefix sum // till index i x j sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1]; } } // To store the maximum size of // matrix with sum <= K int ans = 0; // Traverse the sum matrix for (i = 1; i <= n; i++) { for (j = 1; j <= m; j++) { // Index out of bound if (i + ans - 1 > n || j + ans - 1 > m) break; int mid, lo = ans; // Maximum possible size // of matrix int hi = min(n - i + 1, m - j + 1); // Binary Search while (lo < hi) { // Find middle index mid = (hi + lo + 1) / 2; // Check whether sum <= K // or not // If Yes check for other // half of the search if (sum[i + mid - 1][j + mid - 1] + sum[i - 1][j - 1] - sum[i + mid - 1][j - 1] - sum[i - 1][j + mid - 1] <= K) { lo = mid; } // Else check it in first // half else { hi = mid - 1; } } // Update the maximum size matrix ans = max(ans, lo); } } // Print the final answer cout << ans << endl;}// Driver Codeint main(){ vector<vector<int> > arr; arr = { { 1, 1, 3, 2, 4, 3, 2 }, { 1, 1, 3, 2, 4, 3, 2 }, { 1, 1, 3, 2, 4, 3, 2 } }; // Given target sum int K = 4; // Function Call findMaxMatrixSize(arr, K); return 0;} |
chevron_right
filter_none
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to find the maximum size // of matrix with sum <= K static void findMaxMatrixSize(int[][] arr, int K) { int i, j; // N size of rows and M size of cols int n = arr.length; int m = arr[0].length; // To store the prefix sum of matrix int[][] sum = new int[n + 1][m + 1]; // Create prefix sum for (i = 0; i <= n; i++) { // Traverse each rows for (j = 0; j <= m; j++) { if (i == 0 || j == 0) { sum[i][j] = 0; continue; } // Update the prefix sum // till index i x j sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1]; } } // To store the maximum size of // matrix with sum <= K int ans = 0; // Traverse the sum matrix for (i = 1; i <= n; i++) { for (j = 1; j <= m; j++) { // Index out of bound if (i + ans - 1 > n || j + ans - 1 > m) break; int mid, lo = ans; // Maximum possible size // of matrix int hi = Math.min(n - i + 1, m - j + 1); // Binary Search while (lo < hi) { // Find middle index mid = (hi + lo + 1) / 2; // Check whether sum <= K // or not // If Yes check for other // half of the search if (sum[i + mid - 1][j + mid - 1] + sum[i - 1][j - 1] - sum[i + mid - 1][j - 1] - sum[i - 1][j + mid - 1] <= K) { lo = mid; } // Else check it in first // half else { hi = mid - 1; } } // Update the maximum size matrix ans = Math.max(ans, lo); } } // Print the final answer System.out.print(ans + "\n"); } // Driver Code public static void main(String[] args) { int[][] arr = { { 1, 1, 3, 2, 4, 3, 2 }, { 1, 1, 3, 2, 4, 3, 2 }, { 1, 1, 3, 2, 4, 3, 2 } }; // Given target sum int K = 4; // Function Call findMaxMatrixSize(arr, K); }}// This code is contributed by 29AjayKumar |
chevron_right
filter_none
Python3
# Python3 program for the above approach# Function to find the maximum size# of matrix with sum <= Kdef findMaxMatrixSize(arr, K): # N size of rows and M size of cols n = len(arr) m = len(arr[0]) # To store the prefix sum of matrix sum = [[0 for i in range(m + 1)] for j in range(n + 1)] # Create Prefix Sum for i in range(n + 1): # Traverse each rows for j in range(m+1): if (i == 0 or j == 0): sum[i][j] = 0 continue # Update the prefix sum # till index i x j sum[i][j] = arr[i - 1][j - 1] + sum[i - 1][j] + \ sum[i][j - 1]-sum[i - 1][j - 1] # To store the maximum size of # matrix with sum <= K ans = 0 # Traverse the sum matrix for i in range(1, n + 1): for j in range(1, m + 1): # Index out of bound if (i + ans - 1 > n or j + ans - 1 > m): break mid = ans lo = ans # Maximum possible size # of matrix hi = min(n - i + 1, m - j + 1) # Binary Search while (lo < hi): # Find middle index mid = (hi + lo + 1) // 2 # Check whether sum <= K # or not # If Yes check for other # half of the search if (sum[i + mid - 1][j + mid - 1] + sum[i - 1][j - 1] - sum[i + mid - 1][j - 1] - sum[i - 1][j + mid - 1] <= K): lo = mid # Else check it in first # half else: hi = mid - 1 # Update the maximum size matrix ans = max(ans, lo) # Print the final answer print(ans - 1)# Driver Codeif __name__ == '__main__': arr = [[1, 1, 3, 2, 4, 3, 2], [1, 1, 3, 2, 4, 3, 2], [1, 1, 3, 2, 4, 3, 2]] # Given target sum K = 4 # Function Call findMaxMatrixSize(arr, K)# This code is contributed by Surendra_Gangwar |
chevron_right
filter_none
C#
// C# program for the above approachusing System;class GFG { // Function to find the maximum size // of matrix with sum <= K static void findMaxMatrixSize(int[, ] arr, int K) { int i, j; // N size of rows and M size of cols int n = arr.GetLength(0); int m = arr.GetLength(1); // To store the prefix sum of matrix int[, ] sum = new int[n + 1, m + 1]; // Create prefix sum for (i = 0; i <= n; i++) { // Traverse each rows for (j = 0; j <= m; j++) { if (i == 0 || j == 0) { sum[i, j] = 0; continue; } // Update the prefix sum // till index i x j sum[i, j] = arr[i - 1, j - 1] + sum[i - 1, j] + sum[i, j - 1] - sum[i - 1, j - 1]; } } // To store the maximum size // of matrix with sum <= K int ans = 0; // Traverse the sum matrix for (i = 1; i <= n; i++) { for (j = 1; j <= m; j++) { // Index out of bound if (i + ans - 1 > n || j + ans - 1 > m) break; int mid, lo = ans; // Maximum possible size // of matrix int hi = Math.Min(n - i + 1, m - j + 1); // Binary Search while (lo < hi) { // Find middle index mid = (hi + lo + 1) / 2; // Check whether sum <= K // or not // If Yes check for other // half of the search if (sum[i + mid - 1, j + mid - 1] + sum[i - 1, j - 1] - sum[i + mid - 1, j - 1] - sum[i - 1, j + mid - 1] <= K) { lo = mid; } // Else check it in first // half else { hi = mid - 1; } } // Update the maximum size matrix ans = Math.Max(ans, lo); } } // Print the readonly answer Console.Write(ans + "\n"); } // Driver Code public static void Main(String[] args) { int[, ] arr = { { 1, 1, 3, 2, 4, 3, 2 }, { 1, 1, 3, 2, 4, 3, 2 }, { 1, 1, 3, 2, 4, 3, 2 } }; // Given target sum int K = 4; // Function Call findMaxMatrixSize(arr, K); }}// This code is contributed by Amit Katiyar |
chevron_right
filter_none
Output
2
Time Complexity: O(N*N*log(N))
Auxiliary Space: O(M*N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Maximum size of square such that all submatrices of that size have sum less than K
- Count numbers whose maximum sum of distinct digit-sum is less than or equals M
- Find Maximum Length Of A Square Submatrix Having Sum Of Elements At-Most K
- Maximum sum subarray of size K with sum less than X
- Largest possible square submatrix with maximum AND value
- Submatrix of given size with maximum 1's
- Check if minimum element in array is less than or equals half of every other element
- Check if a matrix contains a square submatrix with 0 as boundary element
- Maximum sum subarray having sum less than or equal to given sum using Set
- Maximum sum of any submatrix of a Matrix which is sorted row-wise and column-wise
- Sum of all array elements less than X and greater than Y for Q queries
- Maximum value in a matrix which contain intersecting concentric submatrix
- Smallest submatrix with Kth maximum XOR
- Queries for greater than and not less than
- Maximum length L such that the sum of all subarrays of length L is less than K
- Minimum sum submatrix in a given 2D array
- Count of submatrix with sum X in a given Matrix
- Submatrix Sum Queries
- Smallest submatrix required to be removed such that sum of the remaining matrix is divisible by K
- Smallest subarray of size greater than K with sum greater than a given value
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
