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:
- Compute the sum of the first K array elements and store it in a variable, say curr_sum.
- Then iterate over the remaining array elements one by one, and update curr_sum by adding ith element and removing (i – K)th element from the array.
- For every value of curr_sum obtained, check if it is a perfect square number or not.
- If found to be true for any value of curr_sum, then print the corresponding subarray.
- Otherwise, print -1.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(int n)
{
double sr = sqrt(n);
return ((sr - floor(sr)) == 0);
}
void SubarrayHavingPerfectSquare(
vector<int> arr, int k)
{
pair<int, int> ans;
int sum = 0, i;
for (i = 0; i < k; i++) {
sum += arr[i];
}
bool found = false;
if (isPerfectSquare(sum)) {
ans.first = 0;
ans.second = i - 1;
}
else {
for (int j = i;
j < arr.size(); j++) {
sum = sum + arr[j] - arr[j - k];
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 (found == false) {
cout << "-1";
}
}
int main()
{
vector<int> arr;
arr = { 20, 34, 51, 10,
99, 87, 23, 45 };
int K = 3;
SubarrayHavingPerfectSquare(arr, K);
return 0;
}
|
Java
import java.io.*;
public class GFG{
static class pair
{
int first, second;
}
static boolean isPerfectSquare(int n)
{
double sr = Math.sqrt(n);
return ((sr - Math.floor(sr)) == 0);
}
static void SubarrayHavingPerfectSquare(int[] arr,
int k)
{
pair ans = new pair();
int sum = 0, i;
for (i = 0; i < k; i++)
{
sum += arr[i];
}
boolean found = false;
if (isPerfectSquare(sum))
{
ans.first = 0;
ans.second = i - 1;
}
else
{
for (int j = i; j < arr.length; j++)
{
sum = sum + arr[j] - arr[j - k];
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 (found == false)
{
System.out.print("-1");
}
}
public static void main(String[] args)
{
int []arr = {20, 34, 51, 10,
99, 87, 23, 45};
int K = 3;
SubarrayHavingPerfectSquare(arr, K);
}
}
|
Python3
from math import sqrt, ceil, floor
def isPerfectSquare(n):
sr = sqrt(n)
return ((sr - floor(sr)) == 0)
def SubarrayHavingPerfectSquare(arr, k):
ans = [ 0, 0 ]
sum = 0
i = 0
while i < k:
sum += arr[i]
i += 1
found = False
if (isPerfectSquare(sum)):
ans[0] = 0
ans[1] = i - 1
else:
for j in range(i, len(arr)):
sum = sum + arr[j] - arr[j - k]
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 (found == False):
print("-1")
if __name__ == '__main__':
arr = [ 20, 34, 51, 10,
99, 87, 23, 45 ]
K = 3
SubarrayHavingPerfectSquare(arr, K)
|
C#
using System;
class GFG{
class pair
{
public int first, second;
}
static bool isPerfectSquare(int n)
{
double sr = Math.Sqrt(n);
return ((sr - Math.Floor(sr)) == 0);
}
static void SubarrayHavingPerfectSquare(int[] arr,
int k)
{
pair ans = new pair();
int sum = 0, i;
for (i = 0; i < k; i++)
{
sum += arr[i];
}
bool found = false;
if (isPerfectSquare(sum))
{
ans.first = 0;
ans.second = i - 1;
}
else
{
for (int j = i; j < arr.Length; j++)
{
sum = sum + arr[j] - arr[j - k];
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 (found == false)
{
Console.Write("-1");
}
}
public static void Main(String[] args)
{
int []arr = {20, 34, 51, 10,
99, 87, 23, 45};
int K = 3;
SubarrayHavingPerfectSquare(arr, K);
}
}
|
Javascript
<script>
function isPerfectSquare(n)
{
var sr = Math.sqrt(n);
return ((sr - Math.floor(sr)) == 0);
}
function SubarrayHavingPerfectSquare( arr, k)
{
var ans = [];
var sum = 0, i;
for (i = 0; i < k; i++) {
sum += arr[i];
}
var found = false;
if (isPerfectSquare(sum)) {
ans[0] = 0;
ans[1] = i - 1;
}
else {
for (var j = i;
j < arr.length; j++) {
sum = sum + arr[j] - arr[j - k];
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 (found == false) {
document.write( "-1");
}
}
var arr = [20, 34, 51, 10,
99, 87, 23, 45 ];
var K = 3;
SubarrayHavingPerfectSquare(arr, K);
</script>
|
Time Complexity: O(N*log(sum))
Auxiliary Space:O(1)
Approach 2: Dynamic Programming:
To solve this problem using dynamic programming, we can create a 2D array where dp[i][j] represents the sum of the subarray from index i to index j. We can calculate the values of dp[i][j] using the following recurrence relation:
- dp[i][j] = dp[i][j-1] + arr[j]
- Then, we can iterate over all subarrays of size K and check if the sum of the subarray is a perfect square. If we find such a subarray, we can print it and return.
Here’s the codes:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(int n)
{
int sr = sqrt(n);
return ((sr * sr) == n);
}
void SubarrayHavingPerfectSquare(
vector<int>& arr, int K)
{
int n = arr.size();
int dp[n][n];
memset(dp, 0, sizeof(dp));
for (int i = 0; i < n; i++) {
dp[i][i] = arr[i];
for (int j = i + 1; j < n; j++) {
dp[i][j] = dp[i][j - 1] + arr[j];
}
}
bool found = false;
for (int i = 0; i <= n - K; i++) {
int sum = dp[i][i + K - 1];
if (isPerfectSquare(sum)) {
found = true;
for (int j = i; j < i + K; j++) {
cout << arr[j] << " ";
}
break;
}
}
if (found == false) {
cout << "-1";
}
}
int main()
{
vector<int> arr = { 20, 34, 51, 10, 99, 87, 23, 45 };
int K = 3;
SubarrayHavingPerfectSquare(arr, K);
return 0;
}
|
Java
import java.util.*;
public class SubarrayHavingPerfectSquare {
static boolean isPerfectSquare(int n) {
int sr = (int)Math.sqrt(n);
return (sr * sr == n);
}
static void subarrayHavingPerfectSquare(List<Integer> arr, int K) {
int n = arr.size();
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++) {
dp[i][i] = arr.get(i);
for (int j = i + 1; j < n; j++) {
dp[i][j] = dp[i][j - 1] + arr.get(j);
}
}
boolean found = false;
for (int i = 0; i <= n - K; i++) {
int sum = dp[i][i + K - 1];
if (isPerfectSquare(sum)) {
found = true;
for (int j = i; j < i + K; j++) {
System.out.print(arr.get(j) + " ");
}
break;
}
}
if (found == false) {
System.out.print("-1");
}
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(20, 34, 51, 10, 99, 87, 23, 45);
int K = 3;
subarrayHavingPerfectSquare(arr, K);
}
}
|
Python3
import math
def isPerfectSquare(n):
sr = int(math.sqrt(n))
return sr * sr == n
def SubarrayHavingPerfectSquare(arr, K):
n = len(arr)
dp = [[0] * n for _ in range(n)]
for i in range(n):
dp[i][i] = arr[i]
for j in range(i + 1, n):
dp[i][j] = dp[i][j - 1] + arr[j]
found = False
for i in range(n - K + 1):
sum = dp[i][i + K - 1]
if isPerfectSquare(sum):
found = True
for j in range(i, i + K):
print(arr[j], end=" ")
break
if not found:
print("-1")
if __name__ == "__main__":
arr = [20, 34, 51, 10, 99, 87, 23, 45]
K = 3
SubarrayHavingPerfectSquare(arr, K)
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
static bool IsPerfectSquare(int n)
{
int sr = (int)Math.Sqrt(n);
return (sr * sr == n);
}
static void SubarrayHavingPerfectSquare(List<int> arr, int K)
{
int n = arr.Count;
int[,] dp = new int[n, n];
for (int i = 0; i < n; i++)
{
dp[i, i] = arr[i];
for (int j = i + 1; j < n; j++)
{
dp[i, j] = dp[i, j - 1] + arr[j];
}
}
bool found = false;
for (int i = 0; i <= n - K; i++)
{
int sum = dp[i, i + K - 1];
if (IsPerfectSquare(sum))
{
found = true;
for (int j = i; j < i + K; j++)
{
Console.Write(arr[j] + " ");
}
break;
}
}
if (!found)
{
Console.Write("-1");
}
}
public static void Main(string[] args)
{
List<int> arr = new List<int> { 20, 34, 51, 10, 99, 87, 23, 45 };
int K = 3;
SubarrayHavingPerfectSquare(arr, K);
}
}
|
Javascript
function isPerfectSquare(n) {
let sr = Math.floor(Math.sqrt(n));
return sr * sr === n;
}
function subarrayHavingPerfectSquare(arr, K) {
let n = arr.length;
let dp = new Array(n).fill(0).map(() => new Array(n).fill(0));
for (let i = 0; i < n; i++) {
dp[i][i] = arr[i];
for (let j = i + 1; j < n; j++) {
dp[i][j] = dp[i][j - 1] + arr[j];
}
}
let found = false;
for (let i = 0; i <= n - K; i++) {
let sum = dp[i][i + K - 1];
if (isPerfectSquare(sum)) {
found = true;
for (let j = i; j < i + K; j++) {
console.log(arr[j]);
}
break;
}
}
if (!found) {
console.log("-1");
}
}
let arr = [20, 34, 51, 10, 99, 87, 23, 45];
let K = 3;
subarrayHavingPerfectSquare(arr, K);
|
Output:
10 99 87
Time Complexity: O(n * sqrt(n)), where n is the size of the input array.
Auxiliary Space :O(1)