给定一个N x M矩阵，其中N是给定矩阵中的行数， M是列数和整数K。任务是找到元素总和小于或等于K的正方形子矩阵的最大长度，如果没有此类正方形，则打印0 。
例子：
Input: r = 4, c = 4 , k = 6
matrix[][] = { {1, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 0},
{1, 0, 0, 0},
}
Output: 3
Explanation:
Square from (0,0) to (2,2) with
sum 5 is one of the valid answer.
Input: r = 4, c = 4 , k = 1
matrix[][] = { {2, 2, 2},
{2, 2, 2},
{2, 2, 2},
{2, 2, 2},
}
Output: 0
Explanation:
There is no valid answer.
方法：
这个想法是计算前缀和矩阵。一旦计算了前缀和矩阵，就可以以O(1)时间复杂度来计算所需的子矩阵和。使用滑动窗口技术来计算最大长度的正方形子矩阵。对于每个长度为cur_max + 1的正方形，其中cur_max是当前找到的正方形子矩阵的最大长度，我们检查当前子矩阵中具有cur_max + 1长度的元素之和是否小于或等于K 。如果是，则将结果加1，否则我们将继续进行检查。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to return maximum
// length of square submatrix
// having sum of elements at-most K
int maxLengthSquare(int row, int column,
int arr[][4], int k)
{
// Matrix to store prefix sum
int sum[row + 1][column + 1] ;
for(int i = 1; i <= row; i++)
for(int j = 0; j <= column; j++)
sum[i][j] = 0;
// Current maximum length
int cur_max = 1;
// Variable for storing
// maximum length of square
int max = 0;
for (int i = 1; i <= row; i++)
{
for (int j = 1; j <= column; j++)
{
// Calculating prefix sum
sum[i][j] = sum[i - 1][j] + sum[i][j - 1] +
arr[i - 1][j - 1] - sum[i - 1][j - 1];
// Checking whether there
// exits square with length
// cur_max+1 or not
if(i >= cur_max && j >= cur_max &&
sum[i][j] - sum[i - cur_max][j]
- sum[i][j - cur_max] +
sum[i - cur_max][j - cur_max] <= k)
{
max = cur_max++;
}
}
}
// Returning the
// maximum length
return max;
}
// Driver code
int main()
{
int row = 4, column = 4;
int matrix[4][4] = { {1, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 0},
{1, 0, 0, 0}
};
int k = 6;
int ans = maxLengthSquare(row, column, matrix, k);
cout << ans;
return 0;
}
// This code is contributed by AnkitRai01 Java
// Java implementation of
// the above approach
import java.util.*;
class GFG
{
// Function to return maximum
// length of square submatrix
// having sum of elements at-most K
public static int maxLengthSquare(int row,int column,
int[][] arr, int k)
{
// Matrix to store prefix sum
int sum[][] = new int[row + 1][column + 1];
// Current maximum length
int cur_max = 1;
// Variable for storing
// maximum length of square
int max = 0;
for (int i = 1; i <= row; i++)
{
for (int j = 1; j <= column; j++)
{
// Calculating prefix sum
sum[i][j] = sum[i - 1][j] + sum[i][j - 1] +
arr[i - 1][j - 1] - sum[i - 1][j - 1];
// Checking whether there
// exits square with length
// cur_max+1 or not
if(i >=cur_max&&j>=cur_max&&sum[i][j]-sum[i - cur_max][j]
- sum[i][j - cur_max]
+ sum[i - cur_max][j - cur_max] <= k){
max = cur_max++;
}
}
}
// Returning the
// maximum length
return max;
}
// Driver code
public static void main(String args[])
{
int row = 4 , column = 4;
int matrix[][] = { {1, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 0},
{1, 0, 0, 0}
};
int k = 6;
int ans = maxLengthSquare(row,column,matrix, k);
System.out.println(ans);
}
}Python3
# Python3 implementation of the above approach
import numpy as np
# Function to return maximum
# length of square submatrix
# having sum of elements at-most K
def maxLengthSquare(row, column, arr, k) :
# Matrix to store prefix sum
sum = np.zeros((row + 1, column + 1));
# Current maximum length
cur_max = 1;
# Variable for storing
# maximum length of square
max = 0;
for i in range(1, row + 1) :
for j in range(1, column + 1) :
# Calculating prefix sum
sum[i][j] = sum[i - 1][j] + sum[i][j - 1] + \
arr[i - 1][j - 1] - \
sum[i - 1][j - 1];
# Checking whether there
# exits square with length
# cur_max+1 or not
if(i >= cur_max and j >= cur_max and
sum[i][j] - sum[i - cur_max][j] - sum[i][j -
cur_max] + sum[i -
cur_max][j - cur_max] <= k) :
max = cur_max;
cur_max += 1;
# Returning the maximum length
return max;
# Driver code
if __name__ == "__main__" :
row = 4 ;
column = 4;
matrix = [[1, 1, 1, 1],
[1, 0, 0, 0],
[1, 0, 0, 0],
[1, 0, 0, 0]];
k = 6;
ans = maxLengthSquare(row, column, matrix, k);
print(ans);
# This code is contributed by AnkitRai01C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return maximum
// length of square submatrix
// having sum of elements at-most K
public static int maxLengthSquare(int row,int column,
int[,] arr, int k)
{
// Matrix to store prefix sum
int [,]sum = new int[row + 1,column + 1];
// Current maximum length
int cur_max = 1;
// Variable for storing
// maximum length of square
int max = 0;
for (int i = 1; i <= row; i++)
{
for (int j = 1; j <= column; j++)
{
// Calculating prefix sum
sum[i, j] = sum[i - 1, j] + sum[i, j - 1] +
arr[i - 1, j - 1] - sum[i - 1, j - 1];
// Checking whether there
// exits square with length
// cur_max+1 or not
if(i >=cur_max && j>=cur_max && sum[i, j]-sum[i - cur_max, j]
- sum[i, j - cur_max]
+ sum[i - cur_max, j - cur_max] <= k)
{
max = cur_max++;
}
}
}
// Returning the
// maximum length
return max;
}
// Driver code
public static void Main()
{
int row = 4 , column = 4;
int [,]matrix = { {1, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 0},
{1, 0, 0, 0}
};
int k = 6;
int ans = maxLengthSquare(row, column, matrix, k);
Console.WriteLine(ans);
}
}
// This code is contributed by AnkitRai01输出：
3
时间复杂度： O(N x M)