Maximum size of square such that all submatrices of that size have sum less than K
Last Updated :
12 Jul, 2025
Given an N x M matrix of integers and an integer K, the task is to find the size of the maximum square sub-matrix (S x S), such that all square sub-matrices of the given matrix of that size have a sum less than K.
Examples:
Input: K = 30
mat[N][M] = {{1, 2, 3, 4, 6},
{5, 3, 8, 1, 2},
{4, 6, 7, 5, 5},
{2, 4, 8, 9, 4} };
Output: 2
Explanation:
All Sub-matrices of size 2 x 2
have sum less than 30
Input : K = 100
mat[N][M] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
Output: 3
Explanation:
All Sub-matrices of size 3 x 3
have sum less than 100
Naive Approach The basic solution is to choose the size S of the submatrix and find all the submatrices of that size and check that the sum of all sub-matrices is less than the given sum whereas, this can be improved by computing the sum of the matrix using this approach. Therefore, the task will be to choose the maximum size possible and the starting and ending position of every possible sub-matrices. Due to which the overall time complexity will be O(N3).
Below is the implementation of the above approach:
C++
// C++ implementation to find the
// maximum size square submatrix
// such that their sum is less than K
#include <bits/stdc++.h>
using namespace std;
// Size of matrix
#define N 4
#define M 5
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
void preProcess(int mat[N][M],
int aux[N][M])
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
int sumQuery(int aux[N][M], int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1][tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
int maximumSquareSize(int mat[N][M], int K)
{
int aux[N][M];
preProcess(mat, aux);
// Loop to choose the size of matrix
for (int i = min(N, M); i >= 1; i--) {
bool satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
int main()
{
int K = 30;
int mat[N][M] = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
cout << maximumSquareSize(mat, K);
return 0;
}
// Java implementation to find the
// maximum size square submatrix
// such that their sum is less than K
class GFG{
// Size of matrix
static final int N = 4;
static final int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [][]mat,
int [][]aux)
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [][]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1][tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [][]mat, int K)
{
int [][]aux = new int[N][M];
preProcess(mat, aux);
// Loop to choose the size of matrix
for (int i = Math.min(N, M); i >= 1; i--) {
boolean satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
public static void main(String[] args)
{
int K = 30;
int mat[][] = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
System.out.print(maximumSquareSize(mat, K));
}
}
// This code is contributed by PrinciRaj1992
# Python3 implementation to find the
# maximum size square submatrix
# such that their sum is less than K
# Size of matrix
N = 4
M = 5
# Function to preprocess the matrix
# for computing the sum of every
# possible matrix of the given size
def preProcess(mat, aux):
# Loop to copy the first row
# of the matrix into the aux matrix
for i in range (M):
aux[0][i] = mat[0][i]
# Computing the sum column-wise
for i in range (1, N):
for j in range (M):
aux[i][j] = (mat[i][j] +
aux[i - 1][j])
# Computing row wise sum
for i in range (N):
for j in range (1, M):
aux[i][j] += aux[i][j - 1]
# Function to find the sum of a
# submatrix with the given indices
def sumQuery(aux, tli, tlj, rbi, rbj):
# Overall sum from the top to
# right corner of matrix
res = aux[rbi][rbj]
# Removing the sum from the top
# corer of the matrix
if (tli > 0):
res = res - aux[tli - 1][rbj]
# Remove the overlapping sum
if (tlj > 0):
res = res - aux[rbi][tlj - 1]
# Add the sum of top corner
# which is subtracted twice
if (tli > 0 and tlj > 0):
res = (res +
aux[tli - 1][tlj - 1])
return res
# Function to find the maximum
# square size possible with the
# such that every submatrix have
# sum less than the given sum
def maximumSquareSize(mat, K):
aux = [[0 for x in range (M)]
for y in range (N)]
preProcess(mat, aux)
# Loop to choose the size of matrix
for i in range (min(N, M), 0, -1):
satisfies = True
# Loop to find the sum of the
# matrix of every possible submatrix
for x in range (N):
for y in range (M) :
if (x + i - 1 <= N - 1 and
y + i - 1 <= M - 1):
if (sumQuery(aux, x, y,
x + i - 1,
y + i - 1) > K):
satisfies = False
if (satisfies == True):
return (i)
return 0
# Driver Code
if __name__ == "__main__":
K = 30
mat = [[1, 2, 3, 4, 6],
[5, 3, 8, 1, 2],
[4, 6, 7, 5, 5],
[2, 4, 8, 9, 4]]
print( maximumSquareSize(mat, K))
# This code is contributed by Chitranayal
// C# implementation to find the
// maximum size square submatrix
// such that their sum is less than K
using System;
public class GFG{
// Size of matrix
static readonly int N = 4;
static readonly int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [,]mat,
int [,]aux)
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0,i] = mat[0,i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i,j] = mat[i,j] +
aux[i - 1,j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i,j] += aux[i,j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [,]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi,rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1,rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi,tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1,tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [,]mat, int K)
{
int [,]aux = new int[N,M];
preProcess(mat, aux);
// Loop to choose the size of matrix
for (int i = Math.Min(N, M); i >= 1; i--) {
bool satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
public static void Main(String[] args)
{
int K = 30;
int [,]mat = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
Console.Write(maximumSquareSize(mat, K));
}
}
// This code contributed by PrinciRaj1992
<script>
// JavaScript implementation to find the
// maximum size square submatrix
// such that their sum is less than K
// Size of matrix
let N = 4;
let M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
function preProcess(mat,aux)
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (let i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (let i = 1; i < N; i++)
for (let j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (let i = 0; i < N; i++)
for (let j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
function sumQuery(aux,tli,tlj,rbi,rbj)
{
// Overall sum from the top to
// right corner of matrix
let res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1][tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
function maximumSquareSize(mat,k)
{
let aux = new Array(N);
for(let i=0;i<N;i++)
{
aux[i]=new Array(M);
}
preProcess(mat, aux);
// Loop to choose the size of matrix
for (let i = Math.min(N, M); i >= 1; i--) {
let satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (let x = 0; x < N; x++) {
for (let y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
let mat = [[1, 2, 3, 4, 6],
[5, 3, 8, 1, 2],
[4, 6, 7, 5, 5],
[2, 4, 8, 9, 4]];
let K = 30;
document.write(maximumSquareSize(mat, K));
// This code is contributed by avanitrachhadiya2155
</script>
- Time complexity: O(N3)
- Auxiliary Space: O(N2)
Efficient Approach: The key observation is, if a square of side s is the maximum size satisfying the condition, then all sizes smaller than it will satisfy the condition. Using this we can reduce our search space at each step by half which is precisely the idea of Binary Search. Below is the illustration of the steps of the approach:
- Search Space: The search space for this problem will be from [1, min(N, M)]. That is the search space for binary search is defined as -
low = 1
high = min(N, M)
- Next Search Space: In each iteration find the mid of the search space and then Finally, check that all subarrays of that size have the sum less than K. If all subarrays of that size have sum less than K. Then the next search space possible will be in the right of the middle. Otherwise, the next search space possible will be in the left of the middle. That is less than the middle.
- Case 1: Condition when the all the subarrays of size mid have sum less than K. Then -
if checkSubmatrix(mat, mid, K):
low = mid + 1- Case 2: Condition when the all the subarrays of size mid have sum greater than K. Then -
if not checkSubmatrix(mat, mid, K):
high = mid - 1
Below is the implementation of the above approach:
C++
// C++ implementation to find the
// maximum size square submatrix
// such that their sum is less than K
#include <bits/stdc++.h>
using namespace std;
// Size of matrix
#define N 4
#define M 5
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
void preProcess(int mat[N][M],
int aux[N][M])
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
int sumQuery(int aux[N][M], int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1][tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
bool check(int mid, int aux[N][M],
int K)
{
bool satisfies = true;
// Iterating through all possible
// submatrices of given size
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + mid - 1, y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
int maximumSquareSize(int mat[N][M],
int K)
{
int aux[N][M];
preProcess(mat, aux);
// Search space
int low = 1, high = min(N, M);
int mid;
// Binary search for size
while (high - low > 1) {
mid = (low + high) / 2;
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K)) {
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
int main()
{
int K = 30;
int mat[N][M] = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
cout << maximumSquareSize(mat, K);
return 0;
}
// Java implementation to find the
// maximum size square submatrix
// such that their sum is less than K
class GFG{
// Size of matrix
static final int N = 4;
static final int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [][]mat,
int [][]aux)
{
// Loop to copy the first row of
// the matrix into the aux matrix
for(int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for(int i = 1; i < N; i++)
for(int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for(int i = 0; i < N; i++)
for(int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [][]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1][tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
static boolean check(int mid, int [][]aux,
int K)
{
boolean satisfies = true;
// Iterating through all possible
// submatrices of given size
for(int x = 0; x < N; x++)
{
for(int y = 0; y < M; y++)
{
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1)
{
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [][]mat,
int K)
{
int [][]aux = new int[N][M];
preProcess(mat, aux);
// Search space
int low = 1, high = Math.min(N, M);
int mid;
// Binary search for size
while (high - low > 1)
{
mid = (low + high) / 2;
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K))
{
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
public static void main(String[] args)
{
int K = 30;
int [][]mat = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
System.out.print(maximumSquareSize(mat, K));
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation to find the
# maximum size square submatrix
# such that their sum is less than K
# Function to preprocess the matrix
# for computing the sum of every
# possible matrix of the given size
def preProcess(mat, aux):
# Loop to copy the first row
# of the matrix into the aux matrix
for i in range(5):
aux[0][i] = mat[0][i]
# Computing the sum column-wise
for i in range(1, 4):
for j in range(5):
aux[i][j] = (mat[i][j] +
aux[i - 1][j])
# Computing row wise sum
for i in range(4):
for j in range(1, 5):
aux[i][j] += aux[i][j - 1]
return aux
# Function to find the sum of a
# submatrix with the given indices
def sumQuery(aux, tli, tlj, rbi, rbj):
# Overall sum from the top to
# right corner of matrix
res = aux[rbi][rbj]
# Removing the sum from the top
# corer of the matrix
if (tli > 0):
res = res - aux[tli - 1][rbj]
# Remove the overlapping sum
if (tlj > 0):
res = res - aux[rbi][tlj - 1]
# Add the sum of top corner
# which is subtracted twice
if (tli > 0 and tlj > 0):
res = res + aux[tli - 1][tlj - 1]
return res
# Function to check whether square
# sub matrices of size mid satisfy
# the condition or not
def check(mid, aux, K):
satisfies = True
# Iterating through all possible
# submatrices of given size
for x in range(4):
for y in range(5):
if (x + mid - 1 <= 4 - 1 and
y + mid - 1 <= 5 - 1):
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K):
satisfies = False
return True if satisfies == True else False
# Function to find the maximum
# square size possible with the
# such that every submatrix have
# sum less than the given sum
def maximumSquareSize(mat, K):
aux = [[0 for i in range(5)]
for i in range(4)]
aux = preProcess(mat, aux)
# Search space
low , high = 1, min(4, 5)
mid = 0
# Binary search for size
while (high - low > 1):
mid = (low + high) // 2
# Check if the mid satisfies
# the given condition
if (check(mid, aux, K)):
low = mid
else:
high = mid
if (check(high, aux, K)):
return high
return low
# Driver Code
if __name__ == '__main__':
K = 30
mat = [ [ 1, 2, 3, 4, 6 ],
[ 5, 3, 8, 1, 2 ],
[ 4, 6, 7, 5, 5 ],
[ 2, 4, 8, 9, 4 ] ]
print(maximumSquareSize(mat, K))
# This code is contributed by mohit kumar 29
// C# implementation to find the
// maximum size square submatrix
// such that their sum is less than K
using System;
class GFG{
// Size of matrix
static readonly int N = 4;
static readonly int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [,]mat,
int [,]aux)
{
// Loop to copy the first row of
// the matrix into the aux matrix
for(int i = 0; i < M; i++)
aux[0, i] = mat[0, i];
// Computing the sum column-wise
for(int i = 1; i < N; i++)
for(int j = 0; j < M; j++)
aux[i, j] = mat[i, j] +
aux[i - 1, j];
// Computing row wise sum
for(int i = 0; i < N; i++)
for(int j = 1; j < M; j++)
aux[i, j] += aux[i, j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [,]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi, rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1, rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi, tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1, tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
static bool check(int mid, int [,]aux,
int K)
{
bool satisfies = true;
// Iterating through all possible
// submatrices of given size
for(int x = 0; x < N; x++)
{
for(int y = 0; y < M; y++)
{
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1)
{
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [,]mat,
int K)
{
int [,]aux = new int[N, M];
preProcess(mat, aux);
// Search space
int low = 1, high = Math.Min(N, M);
int mid;
// Binary search for size
while (high - low > 1)
{
mid = (low + high) / 2;
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K))
{
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
public static void Main(String[] args)
{
int K = 30;
int [,]mat = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
Console.Write(maximumSquareSize(mat, K));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript implementation to find the
// maximum size square submatrix
// such that their sum is less than K
// Size of matrix
let N = 4;
let M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
function preProcess(mat,aux)
{
// Loop to copy the first row of
// the matrix into the aux matrix
for(let i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for(let i = 1; i < N; i++)
for(let j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for(let i = 0; i < N; i++)
for(let j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
function sumQuery(aux,tli,tlj,rbi,rbj)
{
// Overall sum from the top to
// right corner of matrix
let res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1][tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
function check(mid,aux,K)
{
let satisfies = true;
// Iterating through all possible
// submatrices of given size
for(let x = 0; x < N; x++)
{
for(let y = 0; y < M; y++)
{
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1)
{
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
function maximumSquareSize(mat,K)
{
let aux = new Array(N);
for(let i=0;i<N;i++)
{
aux[i]=new Array(M);
}
preProcess(mat, aux);
// Search space
let low = 1, high = Math.min(N, M);
let mid;
// Binary search for size
while (high - low > 1)
{
mid = Math.floor((low + high) / 2);
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K))
{
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
let K = 30;
let mat = [[1, 2, 3, 4, 6 ],
[ 5, 3, 8, 1, 2 ],
[ 4, 6, 7, 5, 5 ],
[ 2, 4, 8, 9, 4 ] ];
document.write(maximumSquareSize(mat, K));
// This code is contributed by rag2127
</script>
Performance Analysis:
- Time Complexity: O(N2 * log(N) )
- Auxiliary Space: O(N2)
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