Find a subarray of size K whose sum is a perfect square

Given an array arr[] and an integer K, the task is to find a subarray of length K having a sum which is a perfect square. If no such subarray exists, then print -1. Otherwise, print the subarray.

Note: There can be more than one possible subarray. Print any one of them.
Examples:

Input: arr[] = {20, 34, 51, 10, 99, 87, 23, 45}, K = 3 
Output: {10, 99, 87} 
Explanation: The sum of the elements of the subarray {10, 99, 87} is 196, which is a perfect square (16² = 196).

Input: arr[] = {20, 34, 51, 10, 99, 87, 23, 45}, K = 4 
Output: -1 
Explanation: None of the subarrays of size K has a sum which is a perfect square.

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays of size K and check whether the sum of any subarray generated is a perfect square or not. If found to be true, print that subarray. If no subarray satisfies the condition, print “-1”. 
Time Complexity: O(N*K) 
Auxiliary Space: O(K)

Efficient Approach: An efficient approach is to use a sliding window technique to find the sum of a contiguous subarray of size K and then check if the sum is a perfect square or not using Binary Search. Below are the steps to solve this problem:

1. Compute the sum of the first K array elements and store it in a variable, say curr_sum.
2. Then iterate over the remaining array elements one by one, and update curr_sum by adding i^th element and removing (i – K)^th element from the array.
3. For every value of curr_sum obtained, check if it is a perfect square number or not.
4. If found to be true for any value of curr_sum, then print the corresponding subarray.
5. Otherwise, print -1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if a given number
// is a perfect square or not
bool isPerfectSquare(int n)
{
    // Find square root of n
    double sr = sqrt(n);

    // Check if the square root
    // is an integer or not
    return ((sr - floor(sr)) == 0);
}

// Function to print the subarray
// whose sum is a perfect square
void SubarrayHavingPerfectSquare(
    vector<int> arr, int k)
{
    pair<int, int> ans;
    int sum = 0, i;

    // Sum of first K elements
    for (i = 0; i < k; i++) {
        sum += arr[i];
    }

    bool found = false;

    // If the first k elements have
    // a sum as perfect square
    if (isPerfectSquare(sum)) {
        ans.first = 0;
        ans.second = i - 1;
    }
    else {

        // Iterate through the array
        for (int j = i;
             j < arr.size(); j++) {

            sum = sum + arr[j] - arr[j - k];

            // If sum is perfect square
            if (isPerfectSquare(sum)) {
                found = true;
                ans.first = j - k + 1;
                ans.second = j;
            }
        }

        for (int k = ans.first;
             k <= ans.second; k++) {
            cout << arr[k] << " ";
        }
    }

    // If subarray not found
    if (found == false) {
        cout << "-1";
    }
}

// Driver Code
int main()
{
    // Given array
    vector<int> arr;

    arr = { 20, 34, 51, 10,
            99, 87, 23, 45 };

    // Given subarray size K
    int K = 3;

    // Function Call
    SubarrayHavingPerfectSquare(arr, K);

    return 0;
}

Java

// Java program for
// the above approach
class GFG{

static class pair
{
  int first, second;
}
// Function to check if a given number
// is a perfect square or not
static boolean isPerfectSquare(int n)
{
  // Find square root of n
  double sr = Math.sqrt(n);

  // Check if the square root
  // is an integer or not
  return ((sr - Math.floor(sr)) == 0);
}

// Function to print the subarray
// whose sum is a perfect square
static void SubarrayHavingPerfectSquare(int[] arr,
                                        int k)
{
  pair ans = new pair();
  int sum = 0, i;

  // Sum of first K elements
  for (i = 0; i < k; i++)
  {
    sum += arr[i];
  }

  boolean found = false;

  // If the first k elements have
  // a sum as perfect square
  if (isPerfectSquare(sum))
  {
    ans.first = 0;
    ans.second = i - 1;
  }
  else
  {
    // Iterate through the array
    for (int j = i; j < arr.length; j++)
    {
      sum = sum + arr[j] - arr[j - k];

      // If sum is perfect square
      if (isPerfectSquare(sum))
      {
        found = true;
        ans.first = j - k + 1;
        ans.second = j;
      }
    }

    for (int k1 = ans.first;
             k1 <= ans.second; k1++)
    {
      System.out.print(arr[k1] + " ");
    }
  }

  // If subarray not found
  if (found == false)
  {
    System.out.print("-1");
  }
}

// Driver Code
public static void main(String[] args)
{
  // Given array
  int []arr = {20, 34, 51, 10,
               99, 87, 23, 45};

  // Given subarray size K
  int K = 3;

  // Function Call
  SubarrayHavingPerfectSquare(arr, K);

}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
from math import sqrt, ceil, floor

# Function to check if a given number
# is a perfect square or not
def isPerfectSquare(n):

    # Find square root of n
    sr = sqrt(n)

    # Check if the square root
    # is an integer or not
    return ((sr - floor(sr)) == 0)

# Function to print the subarray
# whose sum is a perfect square
def SubarrayHavingPerfectSquare(arr, k):

    ans = [ 0, 0 ]
    sum = 0

    # Sum of first K elements
    i = 0
    while i < k:
        sum += arr[i]
        i += 1

    found = False

    # If the first k elements have
    # a sum as perfect square
    if (isPerfectSquare(sum)):
        ans[0] = 0
        ans[1] = i - 1

    else:

        # Iterate through the array
        for j in range(i, len(arr)):
            sum = sum + arr[j] - arr[j - k]

            # If sum is perfect square
            if (isPerfectSquare(sum)):
                found = True
                ans[0] = j - k + 1
                ans[1] = j

        for k in range(ans[0], ans[1] + 1):
            print(arr[k], end = " ")

    # If subarray not found
    if (found == False):
        print("-1")

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [ 20, 34, 51, 10,
            99, 87, 23, 45 ]

    # Given subarray size K
    K = 3

    # Function call
    SubarrayHavingPerfectSquare(arr, K)

# This code is contributed by mohit kumar 29

C#

// C# program for
// the above approach
using System;
class GFG{

class pair
{
  public int first, second;
}

// Function to check if a given number
// is a perfect square or not
static bool isPerfectSquare(int n)
{
  // Find square root of n
  double sr = Math.Sqrt(n);

  // Check if the square root
  // is an integer or not
  return ((sr - Math.Floor(sr)) == 0);
}

// Function to print the subarray
// whose sum is a perfect square
static void SubarrayHavingPerfectSquare(int[] arr,
                                        int k)
{
  pair ans = new pair();
  int sum = 0, i;

  // Sum of first K elements
  for (i = 0; i < k; i++)
  {
    sum += arr[i];
  }

  bool found = false;

  // If the first k elements have
  // a sum as perfect square
  if (isPerfectSquare(sum))
  {
    ans.first = 0;
    ans.second = i - 1;
  }
  else
  {
    // Iterate through the array
    for (int j = i; j < arr.Length; j++)
    {
      sum = sum + arr[j] - arr[j - k];

      // If sum is perfect square
      if (isPerfectSquare(sum))
      {
        found = true;
        ans.first = j - k + 1;
        ans.second = j;
      }
    }

    for (int k1 = ans.first;
             k1 <= ans.second; k1++)
    {
      Console.Write(arr[k1] + " ");
    }
  }

  // If subarray not found
  if (found == false)
  {
    Console.Write("-1");
  }
}

// Driver Code
public static void Main(String[] args)
{
  // Given array
  int []arr = {20, 34, 51, 10,
               99, 87, 23, 45};

  // Given subarray size K
  int K = 3;

  // Function Call
  SubarrayHavingPerfectSquare(arr, K);

}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript program for the above approach

// Function to check if a given number
// is a perfect square or not
function isPerfectSquare(n)
{
    // Find square root of n
    var sr = Math.sqrt(n);

    // Check if the square root
    // is an integer or not
    return ((sr - Math.floor(sr)) == 0);
}

// Function to print the subarray
// whose sum is a perfect square
function SubarrayHavingPerfectSquare( arr, k)
{
    var ans = [];
    var sum = 0, i;

    // Sum of first K elements
    for (i = 0; i < k; i++) {
        sum += arr[i];
    }

    var found = false;

    // If the first k elements have
    // a sum as perfect square
    if (isPerfectSquare(sum)) {
        ans[0] = 0;
        ans[1] = i - 1;
    }
    else {

        // Iterate through the array
        for (var j = i;
             j < arr.length; j++) {

            sum = sum + arr[j] - arr[j - k];

            // If sum is perfect square
            if (isPerfectSquare(sum)) {
                found = true;
                ans[0] = j - k + 1;
                ans[1] = j;
            }
        }

        for (var k = ans[0];
             k <= ans[1]; k++) {
            document.write( arr[k] + " ");
        }
    }

    // If subarray not found
    if (found == false) {
        document.write( "-1");
    }
}

// Driver Code
// Given array
var arr = [20, 34, 51, 10,
        99, 87, 23, 45 ];
// Given subarray size K
var K = 3;
// Function Call
SubarrayHavingPerfectSquare(arr, K);


</script>

Output:

10 99 87

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