Find smallest perfect square number A such that N + A is also a perfect square number
Given a positive number N. The task is to find out the smallest perfect square number A such that N + A is also a perfect square number or return -1.
Examples:
Input: N = 3 Output: 1 Explanation: As 1 + 3 = 4 = 22 Input: N=1 Output: -1
Naive Approach:
Traverse M from {1, 2, 3, 4, 5…} and check whether (N + M * M) is a perfect square number or not.
Efficient Approach:
-
On observing, we have an equation like:
- N + (X * X) = (M * M) where N is given and M and X are unknown.
- We can rearrange it and get:
- N = (M * M) – (X * X)
- N = (M + X) * (M – X)
- Now we can see that for obtaining N, we need to find the factor of N.The factor of N can be obtained in O(N) time. But it can be optimized to O(N^1/2) by this method.
- Let the factor of N be a and b = (N / a). So, from the above equation a = (M – X) and b = (M + X), and after solving this we can obtain the value of X = (b – a)/2.
Below is the implementation of the above approach:
C++
// C++ code to find out the smallest // perfect square X which when added to N // yields another perfect square number. #include<bits/stdc++.h> using namespace std; long SmallestPerfectSquare(long N){ // X is the smallest perfect // square number long X = (long)1e9; long ans; // Loop from 1 to square root of N for(int i = 1; i < sqrt(N); i++) { // Condition to check whether i // is factor of N or not if (N % i == 0) { long a = i; long b = N / i; // Condition to check whether // factors satisfies the // equation or not if((b - a != 0) && ((b - a) % 2 == 0)) { // Stores minimum value X = min(X, (b - a) / 2); } } } // Return if X * X if X is not equal // to 1e9 else return -1 if (X != 1e9) ans = X * X; else ans = -1; return ans; } // Driver code int main() { long N = 3; cout << SmallestPerfectSquare(N); return 0; } // This code is contributed by AnkitRai01 |
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Java
// Java code to find out the smallest // perfect square X which when added to N // yields another perfect square number. public class GFG { static long SmallestPerfectSquare(long N) { // X is the smallest perfect // square number long X = (long)1e9; long ans; // Loop from 1 to square root of N for(int i = 1; i < Math.sqrt(N); i++){ // Condition to check whether i // is factor of N or not if (N % i == 0){ long a = i ; long b = N / i; // Condition to check whether // factors satisfies the // equation or not if ((b - a != 0) && ((b - a) % 2 == 0)){ // Stores minimum value X = Math.min(X, (b - a) / 2) ; } } } // Return if X * X if X is not equal // to 1e9 else return -1 if (X != 1e9) ans = X * X; else ans = -1; return ans; } // Driver code public static void main (String[] args){ long N = 3; System.out.println(SmallestPerfectSquare(N)) ; } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 code to find out the smallest # perfect square X which when added to N # yields another perfect square number. import math def SmallestPerfectSquare(N): # X is the smallest perfect # square number X = 1e9 # Loop from 1 to square root of N for i in range(1, int(math.sqrt(N)) + 1): # Condition to check whether i # is factor of N or not if N % i == 0: a = i b = N // i # Condition to check whether # factors satisfies the # equation or not if b - a != 0 and (b - a) % 2 == 0: # Stores minimum value X = min(X, (b - a) // 2) # Return if X * X if X is not equal # to 1e9 else return -1 return(X * X if X != 1e9 else -1) # Driver code if __name__ == "__main__" : N = 3 print(SmallestPerfectSquare(N)) |
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C#
// C# code to find out the smallest // perfect square X which when added to N // yields another perfect square number. using System; class GFG { static long SmallestPerfectSquare(long N) { // X is the smallest perfect // square number long X = (long)1e9; long ans; // Loop from 1 to square root of N for(int i = 1; i < Math.Sqrt(N); i++){ // Condition to check whether i // is factor of N or not if (N % i == 0) { long a = i; long b = N / i; // Condition to check whether // factors satisfies the // equation or not if ((b - a != 0) && ((b - a) % 2 == 0)) { // Stores minimum value X = Math.Min(X, (b - a) / 2); } } } // Return if X*X if X is not equal // to 1e9 else return -1 if (X != 1e9) ans = X * X; else ans = -1; return ans; } // Driver code public static void Main (string[] args) { long N = 3; Console.WriteLine(SmallestPerfectSquare(N)); } } // This code is contributed by AnkitRai01 |
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Output:
1
Time complexity: sqrt(N)
