恰好具有 K 个完全平方数的子数组的计数

给定一个未排序的整数数组arr[]和一个整数K 。任务是计算恰好具有 K 个完美平方数的子数组的数量。
例子：

Input: arr[] = {2, 4, 9, 3}, K = 2 
Output: 4 
Explanation: 
Since total number of perfect square number in the array are 2. 
So the 4 subarrays with 2 perfect square number are: 
1.{2, 4, 9} 
2.{2, 4, 9, 3} 
3.{4, 9} 
4.{4, 9, 3}
Input: arr[] = {4, 2, 5}, K = 3 
Output: 0 
 

简单的方法：
生成所有子数组并计算给定子数组中完全数的数量，如果计数等于 K，则增加 ans 变量的计数。
时间复杂度： O(N ² )
有效的方法：

1. 遍历给定的数组 arr[] 并检查元素是否为完美正方形。
2. 如果当前元素是 Perfect Square，则将该索引处的数组值更改为 1，否则将该索引处的值更改为 0。
3. 现在给定的数组被转换为二进制数组。
4. 现在，使用本文讨论的方法在上述二进制数组中找到总和等于 K 的子数组的数量。

下面是上述方法的实现。

C++

// C++ program to Count of subarrays having
// exactly K perfect square numbers.
 
#include 
using namespace std;
 
// A utility function to check if
// the number n is perfect square
// or not
bool isPerfectSquare(long double x)
{
    // Find floating point value of
    // square root of x.
    long double sr = sqrt(x);
 
    // If square root is an integer
    return ((sr - floor(sr)) == 0);
}
 
// Function to find number of subarrays
// with sum exactly equal to k
int findSubarraySum(int arr[], int n, int K)
{
    // STL map to store number of subarrays
    // starting from index zero having
    // particular value of sum.
    unordered_map prevSum;
 
    int res = 0;
 
    // To store the sum of element traverse
    // so far
    int currsum = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Add current element to currsum
        currsum += arr[i];
 
        // If currsum = K, then a new
        // subarray is found
        if (currsum == K) {
            res++;
        }
 
        // If currsum > K then find the
        // no. of subarrays with sum
        // currsum - K and exclude those
        // subarrays
        if (prevSum.find(currsum - K)
            != prevSum.end())
            res += (prevSum[currsum - K]);
 
        // Add currsum to count of
        // different values of sum
        prevSum[currsum]++;
    }
 
    // Return the final result
    return res;
}
 
// Function to count the subarray with K
// perfect square numbers
void countSubarray(int arr[], int n, int K)
{
    // Update the array element
    for (int i = 0; i < n; i++) {
 
        // If current element is perfect
        // square then update the
        // arr[i] to 1
        if (isPerfectSquare(arr[i])) {
            arr[i] = 1;
        }
 
        // Else change arr[i] to 0
        else {
            arr[i] = 0;
        }
    }
 
    // Function Call
    cout << findSubarraySum(arr, n, K);
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 4, 9, 2 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countSubarray(arr, N, K);
    return 0;
}

Java

// Java program to Count of subarrays having
// exactly K perfect square numbers.
import java.util.*;
 
class GFG {
 
// A utility function to check if
// the number n is perfect square
// or not
static boolean isPerfectSquare(double x)
{
         
    // Find floating point value of
    // square root of x.
    double sr = Math.sqrt(x);
 
    // If square root is an integer
    return ((sr - Math.floor(sr)) == 0);
}
 
// Function to find number of subarrays
// with sum exactly equal to k
static int findSubarraySum(int arr[],
                           int n, int K)
{
     
    // Map to store number of subarrays
    // starting from index zero having
    // particular value of sum.
    Map prevSum = new HashMap<>();
 
    int res = 0;
 
    // To store the sum of element
    // traverse so far
    int currsum = 0;
 
    for(int i = 0; i < n; i++)
    {
        
       // Add current element to currsum
       currsum += arr[i];
        
       // If currsum = K, then a new
       // subarray is found
       if (currsum == K)
       {
           res++;
       }
        
       // If currsum > K then find the
       // no. of subarrays with sum
       // currsum - K and exclude those
       // subarrays
       if (prevSum.containsKey(currsum - K))
           res += (prevSum.get(currsum - K));
        
       // Add currsum to count of
       // different values of sum
       prevSum.put(currsum,
                   prevSum.getOrDefault(currsum, 0) + 1);
    }
     
    // Return the final result
    return res;
}
 
// Function to count the subarray with K
// perfect square numbers
static void countSubarray(int arr[], int n, int K)
{
     
    // Update the array element
    for(int i = 0; i < n; i++)
    {
         
       // If current element is perfect
       // square then update the
       // arr[i] to 1
       if (isPerfectSquare(arr[i]))
       {
           arr[i] = 1;
       }
        
       // Else change arr[i] to 0
       else
       {
           arr[i] = 0;
       }
    }
 
    // Function Call
    System.out.println(findSubarraySum(arr, n, K));
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 9, 2 };
    int K = 2;
    int N = arr.length;
 
    // Function Call
    countSubarray(arr, N, K);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to count of subarrays
# having exactly K perfect square numbers.
from collections import defaultdict
import math
 
# A utility function to check if
# the number n is perfect square
# or not
def isPerfectSquare(x):
 
    # Find floating point value of
    # square root of x.
    sr = math.sqrt(x)
 
    # If square root is an integer
    return ((sr - math.floor(sr)) == 0)
 
# Function to find number of subarrays
# with sum exactly equal to k
def findSubarraySum(arr, n, K):
 
    # STL map to store number of subarrays
    # starting from index zero having
    # particular value of sum.
    prevSum = defaultdict(int)
 
    res = 0
 
    # To store the sum of element traverse
    # so far
    currsum = 0
 
    for i in range(n):
 
        # Add current element to currsum
        currsum += arr[i]
 
        # If currsum = K, then a new
        # subarray is found
        if (currsum == K):
            res += 1
 
        # If currsum > K then find the
        # no. of subarrays with sum
        # currsum - K and exclude those
        # subarrays
        if ((currsum - K) in prevSum):
            res += (prevSum[currsum - K])
 
        # Add currsum to count of
        # different values of sum
        prevSum[currsum] += 1
 
    # Return the final result
    return res
 
# Function to count the subarray with K
# perfect square numbers
def countSubarray(arr, n, K):
 
    # Update the array element
    for i in range(n):
 
        # If current element is perfect
        # square then update the
        # arr[i] to 1
        if (isPerfectSquare(arr[i])):
            arr[i] = 1
 
        # Else change arr[i] to 0
        else:
            arr[i] = 0
 
    # Function Call
    print(findSubarraySum(arr, n, K))
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 2, 4, 9, 2 ]
    K = 2
    N = len(arr)
 
    # Function Call
    countSubarray(arr, N, K)
 
# This code is contributed by chitranayal

C#

// C# program to count of subarrays having
// exactly K perfect square numbers.
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
     
// A utility function to check if
// the number n is perfect square
// or not
static bool isPerfectSquare(double x)
{
         
    // Find floating point value of
    // square root of x.
    double sr = Math.Sqrt(x);
 
    // If square root is an integer
    return ((sr - Math.Floor(sr)) == 0);
}
 
// Function to find number of subarrays
// with sum exactly equal to k
static int findSubarraySum(int []arr,
                           int n, int K)
{
     
    // Map to store number of subarrays
    // starting from index zero having
    // particular value of sum.
    Dictionary prevSum = new Dictionary();
                                              
    int res = 0;
 
    // To store the sum of element
    // traverse so far
    int currsum = 0;
 
    for(int i = 0; i < n; i++)
    {
         
        // Add current element to currsum
        currsum += arr[i];
             
        // If currsum = K, then a new
        // subarray is found
        if (currsum == K)
        {
            res++;
        }
             
        // If currsum > K then find the
        // no. of subarrays with sum
        // currsum - K and exclude those
        // subarrays
        if (prevSum.ContainsKey(currsum - K))
            res += (prevSum[currsum - K]);
             
        // Add currsum to count of
        // different values of sum
        if(prevSum.ContainsKey(currsum))
        {
            prevSum[currsum]++;
        }
        else
        {
            prevSum[currsum] = 1;
        }
    }
     
    // Return the final result
    return res;
}
 
// Function to count the subarray with K
// perfect square numbers
static void countSubarray(int []arr, int n,
                                     int K)
{
     
    // Update the array element
    for(int i = 0; i < n; i++)
    {
         
        // If current element is perfect
        // square then update the
        // arr[i] to 1
        if (isPerfectSquare(arr[i]))
        {
            arr[i] = 1;
        }
             
        // Else change arr[i] to 0
        else
        {
            arr[i] = 0;
        }
    }
 
    // Function call
    Console.Write(findSubarraySum(arr, n, K));
}
 
// Driver Code
public static void Main(string[] args)
{
    int []arr = { 2, 4, 9, 2 };
    int K = 2;
    int N = arr.Length;
 
    // Function call
    countSubarray(arr, N, K);
}
}
 
// This code is contributed by rutvik_56

Javascript

// Javascript program to Count of subarrays having
// exactly K perfect square numbers.
     
// A utility function to check if
// the number n is perfect square
// or not   
function isPerfectSquare(x)
{
 
    // Find floating point value of
    // square root of x.
    let sr = Math.sqrt(x);
   
    // If square root is an integer
    return ((sr - Math.floor(sr)) == 0);
}
 
// Function to find number of subarrays
// with sum exactly equal to k
function findSubarraySum(arr, n, k)
{
 
    // Map to store number of subarrays
    // starting from index zero having
    // particular value of sum.
    let prevSum = new Map(); 
    let res = 0;
   
    // To store the sum of element
    // traverse so far
    let currsum = 0; 
    for(let i = 0; i < n; i++)
    {
          
       // Add current element to currsum
       currsum += arr[i];
          
       // If currsum = K, then a new
       // subarray is found
       if (currsum == K)
       {
           res++;
       }
          
       // If currsum > K then find the
       // no. of subarrays with sum
       // currsum - K and exclude those
       // subarrays
       if (prevSum.has(currsum - K))
           res += (prevSum.get(currsum - K));
          
       // Add currsum to count of
       // different values of sum
       prevSum.set(currsum,
                   prevSum.get(currsum)==null?1:prevSum.get(currsum) + 1);
    }
       
    // Return the final result
    return res;
}
 
// Function to count the subarray with K
// perfect square numbers
function countSubarray(arr, n, k)
{
 
    // Update the array element
    for(let i = 0; i < n; i++)
    {
           
       // If current element is perfect
       // square then update the
       // arr[i] to 1
       if (isPerfectSquare(arr[i]))
       {
           arr[i] = 1;
       }
          
       // Else change arr[i] to 0
       else
       {
           arr[i] = 0;
       }
    }
   
    // Function Call
    document.write(findSubarraySum(arr, n, K));
}
 
// Driver code
let arr=[2, 4, 9, 2];
let K = 2;
let N = arr.length;
 
// Function Call
countSubarray(arr, N, K);
 
// This code is contributed by avanitrachhadiya2155

输出：

4

时间复杂度： O(N)
空间复杂度： O(N)

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