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Check if the sum of a subarray within a given range is a perfect square or not
• Last Updated : 09 Apr, 2021

Given an array arr[] of size N and an array range[], the task is to check if the sum of the subarray {range, .. , range} is a perfect square or not. If the sum is a perfect square, then print the square root of the sum. Otherwise, print -1.

Example :

Input: arr[] = {2, 19, 33, 48, 90, 100}, range = [1, 3]
Output: 10
Explanation:
The sum of element from position 1 to position 3 is 19 + 33 + 48 = 100, which is a perfect square of 10.

Input: arr[] = {13, 15, 30, 55, 87}, range = [0, 1]
Output: -1

Naive Approach: The simplest approach is to iterate over the subarray and check if the sum of the subarray is a perfect square or not.

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Binary Search to calculate the square root of the sum of the subarray. Follow the steps below:

1. Calculate the sum of the subarray in the given range[].
2. Now, use binary search to find the square root of the sum in the range 0 to sum.
3. Find the mid element from the start and last value and compare the value of mid2 with the sum.
4. If the value of mid2 is equal to the sum, a perfect square is found, then return mid.
5. If the value of mid2 is greater than the sum, then the answer can lie only on the right side of the range after the mid element. So recur for right and reduce the search space to 0 to mid – 1.
6. If the value of mid2 is less than the sum, reduce the search space to mid + 1 to sum.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of``// the above approach``#include ``using` `namespace` `std;` `// Function to calculate the square``// root of the sum of a subarray in``// a given range``int` `checkForPerfectSquare(vector<``int``> arr,``                          ``int` `i, ``int` `j)``{``    ``int` `mid, sum = 0;` `    ``// Calculate the sum of array``    ``// elements within a given range``    ``for` `(``int` `m = i; m <= j; m++) {``        ``sum += arr[m];``    ``}` `    ``// Finding the square root``    ``int` `low = 0, high = sum / 2;``    ``while` `(low <= high) {``        ``mid = low + (high - low) / 2;` `        ``// If a perfect square is found``        ``if` `(mid * mid == sum) {``            ``return` `mid;``        ``}` `        ``// Reduce the search space if``        ``// the value is greater than sum``        ``else` `if` `(mid * mid > sum) {``            ``high = mid - 1;``        ``}` `        ``// Reduce the search space if``        ``// the value if smaller than sum``        ``else` `{``            ``low = mid + 1;``        ``}``    ``}``    ``return` `-1;``}` `// Driver Code``int` `main()``{``    ``// Given Array``    ``vector<``int``> arr;``    ``arr = { 2, 19, 33, 48, 90, 100 };` `    ``// Given range``    ``int` `L = 1, R = 3;` `    ``// Function Call``    ``cout << checkForPerfectSquare(arr, L, R);``    ``return` `0;``}`

## Java

 `// Java implementation of``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to calculate the square``// root of the sum of a subarray in``// a given range``static` `int` `checkForPerfectSquare(``int` `[]arr,``                                 ``int` `i, ``int` `j)``{``  ``int` `mid, sum = ``0``;` `  ``// Calculate the sum of array``  ``// elements within a given range``  ``for` `(``int` `m = i; m <= j; m++)``  ``{``    ``sum += arr[m];``  ``}` `  ``// Finding the square root``  ``int` `low = ``0``, high = sum / ``2``;``  ``while` `(low <= high)``  ``{``    ``mid = low + (high - low) / ``2``;` `    ``// If a perfect square``    ``// is found``    ``if` `(mid * mid == sum)``    ``{``      ``return` `mid;``    ``}` `    ``// Reduce the search space``    ``// if the value is greater``    ``// than sum``    ``else` `if` `(mid * mid > sum)``    ``{``      ``high = mid - ``1``;``    ``}` `    ``// Reduce the search space``    ``// if the value if smaller``    ``// than sum``    ``else``    ``{``      ``low = mid + ``1``;``    ``}``  ``}``  ``return` `-``1``;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``// Given Array``  ``int` `[]arr = {``2``, ``19``, ``33``,``               ``48``, ``90``, ``100``};` `  ``// Given range``  ``int` `L = ``1``, R = ``3``;` `  ``// Function Call``  ``System.out.print(``         ``checkForPerfectSquare(arr, L, R));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of``# the above approach` `# Function to calculate the square``# root of the sum of a subarray in``# a given range``def` `checkForPerfectSquare(arr, i, j):` `    ``mid, ``sum` `=` `0``, ``0` `    ``# Calculate the sum of array``    ``# elements within a given range``    ``for` `m ``in` `range``(i, j ``+` `1``):``        ``sum` `+``=` `arr[m]` `    ``# Finding the square root``    ``low ``=` `0``    ``high ``=` `sum` `/``/` `2``    ` `    ``while` `(low <``=` `high):``        ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2` `        ``# If a perfect square is found``        ``if` `(mid ``*` `mid ``=``=` `sum``):``            ``return` `mid` `        ``# Reduce the search space if``        ``# the value is greater than sum``        ``elif` `(mid ``*` `mid > ``sum``):``            ``high ``=` `mid ``-` `1` `        ``# Reduce the search space if``        ``# the value if smaller than sum``        ``else``:``            ``low ``=` `mid ``+` `1` `    ``return` `-``1` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Given Array``    ``arr ``=` `[ ``2``, ``19``, ``33``, ``48``, ``90``, ``100` `]` `    ``# Given range``    ``L ``=` `1``    ``R ``=` `3` `    ``# Function call``    ``print``(checkForPerfectSquare(arr, L, R))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of``// the above approach``using` `System;``class` `GFG{` `// Function to calculate the square``// root of the sum of a subarray in``// a given range``static` `int` `checkForPerfectSquare(``int` `[]arr,``                                 ``int` `i, ``int` `j)``{``  ``int` `mid, sum = 0;` `  ``// Calculate the sum of array``  ``// elements within a given range``  ``for` `(``int` `m = i; m <= j; m++)``  ``{``    ``sum += arr[m];``  ``}` `  ``// Finding the square root``  ``int` `low = 0, high = sum / 2;``  ``while` `(low <= high)``  ``{``    ``mid = low + (high - low) / 2;` `    ``// If a perfect square``    ``// is found``    ``if` `(mid * mid == sum)``    ``{``      ``return` `mid;``    ``}` `    ``// Reduce the search space``    ``// if the value is greater``    ``// than sum``    ``else` `if` `(mid * mid > sum)``    ``{``      ``high = mid - 1;``    ``}` `    ``// Reduce the search space``    ``// if the value if smaller``    ``// than sum``    ``else``    ``{``      ``low = mid + 1;``    ``}``  ``}``  ``return` `-1;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``// Given Array``  ``int` `[]arr = {2, 19, 33,``               ``48, 90, 100};` `  ``// Given range``  ``int` `L = 1, R = 3;` `  ``// Function Call``  ``Console.Write(checkForPerfectSquare(arr,``                                      ``L, R));}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`10`

Time Complexity: O(max(log(sum), N))
Auxiliary Space: O(1)

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