平均至少为 K 的平方子矩阵的计数
给定一个大小为NxM的矩阵arr[][]和一个整数K ，任务是找到给定矩阵中元素的平均值大于或等于K的平方子矩阵的计数。
例子：
Input: K = 4, arr[][] = {{2, 2, 3}, {3, 4, 5}, {4, 5, 5}}
Output: 7
Explanation:
The following square submatrices have an average greater than or equal to K:
- Square submatrices of dimension (1×1), formed by taking the elements at positions {(2, 2)}. The average of the submatrix is equal to 4.
- Square submatrices of dimension (1×1), formed by taking the elements at positions {(2, 3)}. The average of the submatrix is equal to 5.
- Square submatrices of dimension (1×1), formed by taking the elements at positions {(3, 1)}. The average of the submatrix is equal to 4.
- Square submatrices of dimension (1×1), formed by taking the elements at positions {(3, 2)}. The average of the submatrix is equal to 5.
- Square submatrices of dimension (1×1), formed by taking the elements at positions {(3, 3)}. The average of the submatrix is equal to 5.
- Square submatrices of dimension (2×2), formed by taking the elements at positions {(2, 1), (2, 2), (3, 1), (3, 2)}. The average of the submatrix is equal to (3+4+4+5 = 16)/4 = 4.
- Square submatrices of dimension (2×2), formed by taking the elements at positions {(2, 2), (2, 3), (3, 2), (3, 3)}. The average of the submatrix is equal to (4+5+5+5 = 19)/4 = 4.75.
Therefore, there are totals of 7 square submatrices with an average greater than or equals to K.
Input: K = 3, arr[][] = {{1, 1, 1}, {1, 1, 1}}
Output: 0
朴素方法：最简单的方法是生成所有可能的平方子矩阵，并检查子平方的所有元素的总和是否大于或等于K乘以子矩阵的大小。
时间复杂度： O(N 3 * M 3 )
辅助空间： O(1)
高效方法：上述方法可以使用前缀和矩阵进行优化，这导致子矩阵之和的恒定时间计算。请按照以下步骤解决问题：
- 初始化一个变量，比如count为0以存储平均值大于或等于K的子矩阵的计数。
- 计算矩阵arr[][]的前缀和并将其存储在一个向量中，比如pre[][] 。
- 使用变量i和j遍历矩阵的每个元素并执行以下步骤：
- 初始化两个变量，比如l为i和r为j 。
- 迭代直到l和r大于0 ，并且在每次迭代中执行以下步骤：
- 计算右下顶点为(i, j)和左上顶点为(l, r)的平方子矩阵之和，并将其存储在变量中，例如sum ，即sum = pre[i][j] – pre[l-1][r] – pre[l][r-1] + pre[l-1][r-1] 。
- 现在，如果K*(i-l+1)*(j-r+1)的值等于sum ，则将count增加1 。
- 将l和r减1 。
- 计算右下顶点为(i, j)和左上顶点为(l, r)的平方子矩阵之和，并将其存储在变量中，例如sum ，即sum = pre[i][j] – pre[l-1][r] – pre[l][r-1] + pre[l-1][r-1] 。
- 最后，完成以上步骤后，打印count的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
#define MAX 1000
// Function to count submatrixes with
// average greater than or equals to K
int cntMatrices(vector > arr, int N, int M,
int K)
{
// Stores count of submatrices
int cnt = 0;
// Stores the prefix sum of matrix
vector > pre(N + 1, vector(M + 1, 0));
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// Iterate over the range
// [1, M]
for (int j = 1; j <= M; j++) {
// Update the prefix sum
pre[i][j] = arr[i - 1][j - 1] + pre[i - 1][j]
+ pre[i][j - 1] - pre[i - 1][j - 1];
}
}
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// Iterate over the range
// [1, M]
for (int j = 1; j <= M; j++) {
// Iterate until l and r
// are greater than 0
for (int l = i, r = j; l > 0 && r > 0;
l--, r--) {
// Update count
int sum1 = (K * (i - l + 1) * (i - r + 1));
// Stores sum of submatrix
// with bottom right corner
// as (i, j) and top left
// corner as (l, r)
int sum2 = pre[i][j] - pre[l - 1][r]
- pre[l][r - 1]
+ pre[l - 1][r - 1];
// If sum1 is less than or
// equal to sum2
if (sum1 <= sum2)
// Increment cnt by 1
cnt++;
}
}
}
// Return cnt as the answer
return cnt;
}
// Driver Code
int main()
{
// Given Input
vector > arr
= { { 2, 2, 3 }, { 3, 4, 5 }, { 4, 5, 5 } };
int K = 4;
int N = arr.size();
int M = arr[0].size();
// Function Call
cout << cntMatrices(arr, N, M, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
static int MAX = 1000;
// Function to count submatrixes with
// average greater than or equals to K
static int cntMatrices(int[][] arr, int N,
int M, int K)
{
// Stores count of submatrices
int cnt = 0;
// Stores the prefix sum of matrix
int[][] pre = new int[N + 1][M + 1];
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// Iterate over the range
// [1, M]
for(int j = 1; j <= M; j++)
{
// Update the prefix sum
pre[i][j] = arr[i - 1][j - 1] + pre[i - 1][j] +
pre[i][j - 1] - pre[i - 1][j - 1];
}
}
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// Iterate over the range
// [1, M]
for(int j = 1; j <= M; j++)
{
// Iterate until l and r
// are greater than 0
for(int l = i, r = j;
l > 0 && r > 0; l--, r--)
{
// Update count
int sum1 = (K * (i - l + 1) *
(i - r + 1));
// Stores sum of submatrix
// with bottom right corner
// as (i, j) and top left
// corner as (l, r)
int sum2 = pre[i][j] - pre[l - 1][r] -
pre[l][r - 1] + pre[l - 1][r - 1];
// If sum1 is less than or
// equal to sum2
if (sum1 <= sum2)
// Increment cnt by 1
cnt++;
}
}
}
// Return cnt as the answer
return cnt;
}
// Driver Code
public static void main(String args[])
{
// Given Input
int[][] arr = { { 2, 2, 3 },
{ 3, 4, 5 },
{ 4, 5, 5 } };
int K = 4;
int N = arr.length;
int M = arr[0].length;
// Function Call
System.out.println( cntMatrices(arr, N, M, K));
}
}
// This code is contributed by avijitmondal1998Python3
# Python3 program for the above approach
# define MAX 1000
# Function to count submatrixes with
# average greater than or equals to K
def cntMatrices(arr, N, M, K):
# Stores count of submatrices
cnt = 0
# Stores the prefix sum of matrix
pre = [[0 for i in range(M + 1)]
for i in range(N + 1)]
# Iterate over the range [1, N]
for i in range(1, N + 1):
# Iterate over the range
# [1, M]
for j in range(1, M + 1):
# Update the prefix sum
pre[i][j] = (arr[i - 1][j - 1] +
pre[i - 1][j] +
pre[i][j - 1] -
pre[i - 1][j - 1])
# Iterate over the range [1, N]
for i in range(1, N + 1):
# Iterate over the range
# [1, M]
for j in range(1, M + 1):
# Iterate until l and r
# are greater than 0
l, r = i, j
while l > 0 and r > 0:
# Update count
sum1 = (K * (i - l + 1) * (i - r + 1))
# Stores sum of submatrix
# with bottom right corner
# as (i, j) and top left
# corner as (l, r)
sum2 = (pre[i][j] -
pre[l - 1][r] - pre[l][r - 1] +
pre[l - 1][r - 1])
# If sum1 is less than or
# equal to sum2
if (sum1 <= sum2):
# Increment cnt by 1
cnt += 1
l -= 1
r -= 1
# Return cnt as the answer
return cnt
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [ [ 2, 2, 3 ],
[ 3, 4, 5 ],
[ 4, 5, 5 ] ]
K = 4
N = len(arr)
M = len(arr[0])
# Function Call
print(cntMatrices(arr, N, M, K))
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
class GFG
{
static int MAX = 1000;
// Function to count submatrixes with
// average greater than or equals to K
static int cntMatrices(int[,] arr, int N,
int M, int K)
{
// Stores count of submatrices
int cnt = 0;
// Stores the prefix sum of matrix
int[,] pre = new int[N + 1, M + 1];
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// Iterate over the range
// [1, M]
for(int j = 1; j <= M; j++)
{
// Update the prefix sum
pre[i, j] = arr[i - 1, j - 1] + pre[i - 1, j] +
pre[i, j - 1] - pre[i - 1, j - 1];
}
}
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// Iterate over the range
// [1, M]
for(int j = 1; j <= M; j++)
{
// Iterate until l and r
// are greater than 0
for(int l = i, r = j;
l > 0 && r > 0; l--, r--)
{
// Update count
int sum1 = (K * (i - l + 1) *
(i - r + 1));
// Stores sum of submatrix
// with bottom right corner
// as (i, j) and top left
// corner as (l, r)
int sum2 = pre[i, j] - pre[l - 1, r] -
pre[l, r - 1] + pre[l - 1, r - 1];
// If sum1 is less than or
// equal to sum2
if (sum1 <= sum2)
// Increment cnt by 1
cnt++;
}
}
}
// Return cnt as the answer
return cnt;
}
// Driver code
public static void Main(string[] args)
{
// Given Input
int[,] arr = { { 2, 2, 3 },
{ 3, 4, 5 },
{ 4, 5, 5 } };
int K = 4;
int N = arr.GetLength(0);
int M = arr.GetLength(0);
// Function Call
Console.WriteLine( cntMatrices(arr, N, M, K));
}
}
// This code is contributed by sanjoy_62.Javascript
输出
7时间复杂度： O(M * N * (min(N, M))
辅助空间： O(M * N)