Given an array, arr[] of size N, the task is to print the start and end indices of all subarrays whose sum is a perfect square.
Examples:
Input: arr[] = {65, 79, 81}
Output: (0, 1) (0, 2) (2, 2)
Explanation:
Subarray sum whose start and end index is (0, 1) = 65 + 79 = 144 = 122
Subarray sum whose start and end index is (0, 2} = 65 + 79 + 81 = 225 = 152
Subarray sum whose start and end index is {2, 2} = 81 = 92Input: arr[] = {1, 2, 3, 4, 5}
Output: (0, 0) (1, 3) (3, 3) (3, 4)
Approach: The problem can be solved using the Prefix Sum Array technique. The idea is to find the sum of all subarrays using the Prefix Sum Array. For each subarray, check if the sum of the subarray is a perfect square or not. If found to be true for any subarray, then print the start and end indices of that subarray. Follow the steps below to solve the problem.
- Initialize a variable, say currSubSum to store the current subarray sum.
- Iterate over the array to generate all possible subarrays of the given array.
- Calculate the sum of each subarray and for each subarray sum, check if it is a perfect square or not.
- If found to be true for any subarray, then print the start and end indices of the subarray.
Below is the implementation of the above approach:
C++14
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to print the start and end// indices of all subarrays whose sum// is a perfect squarevoid PrintIndexes(int arr[], int N){ for (int i = 0; i < N; i++) { // Stores the current // subarray sum int currSubSum = 0; for (int j = i; j < N; j++) { // Update current subarray sum currSubSum += arr[j]; // Stores the square root // of currSubSum int sq = sqrt(currSubSum); // Check if currSubSum is // a perfect square or not if (sq * sq == currSubSum) { cout << "(" << i << ", " << j << ") "; } } }}// Driver Codeint main(){ int arr[] = { 65, 79, 81 }; int N = sizeof(arr) / sizeof(arr[0]); PrintIndexes(arr, N);} |
Java
// Java program to implement // the above approach import java.io.*;class GFG{ // Function to print the start and end // indices of all subarrays whose sum // is a perfect square static void PrintIndexes(int arr[], int N) { for(int i = 0; i < N; i++) { // Stores the current // subarray sum int currSubSum = 0; for(int j = i; j < N; j++) { // Update current subarray sum currSubSum += arr[j]; // Stores the square root // of currSubSum int sq = (int)Math.sqrt(currSubSum); // Check if currSubSum is // a perfect square or not if (sq * sq == currSubSum) { System.out.print("(" + i + "," + j + ")" + " "); } } } } // Driver codepublic static void main (String[] args)throws java.lang.Exception{ int arr[] = { 65, 79, 81 }; PrintIndexes(arr, arr.length);}}// This code is contributed by bikram2001jha |
Python3
# Python3 program to implement# the above approachimport math# Function to print the start and end# indices of all subarrays whose sum# is a perfect squaredef PrintIndexes(arr, N): for i in range(N): # Stores the current # subarray sum currSubSum = 0 for j in range(i, N, 1): # Update current subarray sum currSubSum += arr[j] # Stores the square root # of currSubSum sq = int(math.sqrt(currSubSum)) # Check if currSubSum is # a perfect square or not if (sq * sq == currSubSum): print("(", i, ",", j, ")", end = " ") # Driver Codearr = [ 65, 79, 81 ]N = len(arr) PrintIndexes(arr, N)# This code is contributed by sanjoy_62 |
C#
// C# program to implement // the above approach using System;class GFG{ // Function to print the start // and end indices of all // subarrays whose sum // is a perfect square static void PrintIndexes(int []arr, int N) { for(int i = 0; i < N; i++) { // Stores the current // subarray sum int currSubSum = 0; for(int j = i; j < N; j++) { // Update current subarray sum currSubSum += arr[j]; // Stores the square root // of currSubSum int sq = (int)Math.Sqrt(currSubSum); // Check if currSubSum is // a perfect square or not if (sq * sq == currSubSum) { Console.Write("(" + i + "," + j + ")" + " "); } } } } // Driver codepublic static void Main(String[] args){ int []arr = {65, 79, 81}; PrintIndexes(arr, arr.Length);}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript program to implement// the above approach// Function to print the start and end // indices of all subarrays whose sum // is a perfect square function PrintIndexes(arr, N) { for(let i = 0; i < N; i++) { // Stores the current // subarray sum let currSubSum = 0; for(let j = i; j < N; j++) { // Update current subarray sum currSubSum += arr[j]; // Stores the square root // of currSubSum let sq = Math.floor(Math.sqrt(currSubSum)); // Check if currSubSum is // a perfect square or not if (sq * sq == currSubSum) { document.write("(" + i + "," + j + ")" + " "); } } } }// Driver Code let arr = [ 65, 79, 81 ]; PrintIndexes(arr, arr.length);</script> |
Output:
(0, 1) (0, 2) (2, 2)
Time Complexity: O(N2)
Auxiliary Space: O(1)
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