给定一个数组arr []和一个整数K ，任务是找到一个长度为K的子数组，其总和为一个完美的平方。如果不存在这样的子数组，则打印-1 。否则，打印子数组。
注意：可能有多个子数组。打印其中任何一个。
例子：
Input: arr[] = {20, 34, 51, 10, 99, 87, 23, 45}, K = 3
Output: {10, 99, 87}
Explanation: 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.
天真的方法：解决问题的最简单方法是生成所有可能的大小为K的子数组，并检查所生成的任何子数组的总和是否为完美平方。如果发现为真，则打印该子数组。如果没有子数组满足条件，则打印“ -1” 。
时间复杂度： O(N * K)
辅助空间： O(K)
高效的方法：一种有效的方法是使用滑动窗口技术找到大小为K的连续子数组的和，然后使用Binary Search检查和是否为完美的正方形。以下是解决此问题的步骤：
- 计算前K个数组元素的总和，并将其存储在变量中，例如curr_sum 。
- 然后依次迭代其余的数组元素，并通过添加第i个元素并从数组中删除第(i – K)个元素来更新curr_sum 。
- 对于获得的curr_sum的每个值，请检查它是否是一个完美的平方数。
- 如果发现对于curr_sum的任何值都为true ，则打印相应的子数组。
- 否则，打印-1。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
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 arr, int k)
{
pair 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 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-JiPython3
# 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 prthe 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 29C#
// 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输出：
10 99 87
时间复杂度： O(N * log(sum))
辅助空间： O(1)