Given a unsorted integer array arr[] and an integer K. The task is to count the number of subarray with exactly K Perfect Square Numbers.
Examples:
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
Simple Approach:
Generate all the subarrays and count the number of perfect numbers in the given subarray if the count is equal to K increment the count for ans variable.
Time Complexity: O(N2)
Efficient Approach:
- Traverse the given array arr[] and check whether the element is Perfect Square or not.
- If the current element is Perfect Square then change the value of array at that index to 1, Else change the value at that index to 0.
- Now the given array is converted into Binary Array.
- Now, Find the count of subarray with sum equals to K in the above Binary Array using the approach discussed in this article.
Below is the implementation of the above approach.
C++
// C++ program to Count of subarrays having // exactly K perfect square numbers. #include <bits/stdc++.h> 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<int, int> 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<Integer, Integer> 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<int, int> prevSum = new Dictionary<int, int>(); 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
<script> // 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 </script> |
Output:
4
Time Complexity: O(N)
Space Complexity: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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