给定大小为M x N且具有正整数和数字K的Matrix arr [] [] ，任务是找到元素之和小于或等于K的最大平方子矩阵的大小。
例子：
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 4
1 1
1 1
Size is 2.
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. 方法：
- 对于给定的矩阵arr [] []创建一个前缀和矩阵(例如sum [] [] )，使得sum [i] [j]存储大小为ixj的矩阵所有元素的和。
- 对于使用二进制搜索的前缀总和矩阵sum [] []中的每一行，请执行以下操作：
- 进行二进制搜索，下限为0 ，上限为方矩阵的最大大小。
- 找到中间索引(例如mid )。
- 如果大小为mid的所有可能方阵的元素之和小于或等于K ，则将下限更新为mid +1，以找到大小大于mid的最大和。
- 否则将上限更新为中– 1，以找到大小小于中的最大和。
- 对于上述给定的有效条件，请在每次迭代中不断更新方阵的最大大小。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum size
// of matrix with sum <= K
void findMaxMatrixSize(vector > 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 Code
int main()
{
vector > 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;
} Java
// Java program for the above approach
import 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 29AjayKumarPython3
# Python3 program for the above approach
# Function to find the maximum size
# of matrix with sum <= K
def 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 Code
if __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_GangwarC#
// C# program for the above approach
using 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输出
2时间复杂度： O(N * N * log(N))
辅助空间： O(M * N)