Least number to be added to or subtracted from N to make it a Perfect Square - GeeksforGeeks

20 May, 2021
Given a number N, find the minimum number that needs to be added to or subtracted from N, to make it a perfect square. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign.
Examples:  
Input: N = 14 Output: 2 Nearest perfect square before 14 = 9 Nearest perfect square after 14 = 16 Therefore, 2 needs to be added to 14 to get the closest perfect square.
Input: N = 18 Output: -2 Nearest perfect square before 18 = 16 Nearest perfect square after 18 = 25 Therefore, 2 needs to be subtracted from 18 to get the closest perfect square.  
Approach:  
Get the number.Find the square root of the number and convert the result as an integer.After converting the double value to an integer, it will contain the root of the perfect square above N, i.e. floor(square root(N)).Then find the square of this number, which will be the perfect square before N.Find the root of the perfect square after N, i.e. the ceil(square root(N)).Then find the square of this number, which will be the perfect square after N.Check whether the square of floor value is nearest to N or the ceil value.If the square of floor value is nearest to N, print the difference with a -sign. Else print the difference between the square of the ceil value and N with a + sign.
Get the number.
Find the square root of the number and convert the result as an integer.
After converting the double value to an integer, it will contain the root of the perfect square above N, i.e. floor(square root(N)).
Then find the square of this number, which will be the perfect square before N.
Find the root of the perfect square after N, i.e. the ceil(square root(N)).
Then find the square of this number, which will be the perfect square after N.
Check whether the square of floor value is nearest to N or the ceil value.
If the square of floor value is nearest to N, print the difference with a -sign. Else print the difference between the square of the ceil value and N with a + sign.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach #include <bits/stdc++.h>using namespace std; // Function to return the Least numberint nearest(int n){     // Get the perfect square    // before and after N    int prevSquare = sqrt(n);    int nextSquare = prevSquare + 1;    prevSquare = prevSquare * prevSquare;    nextSquare = nextSquare * nextSquare;     // Check which is nearest to N    int ans        = (n - prevSquare) < (nextSquare - n)              ? (prevSquare - n)              : (nextSquare - n);     // return the result    return ans;} // Driver codeint main(){    int n = 14;    cout << nearest(n) << endl;     n = 16;    cout << nearest(n) << endl;     n = 18;    cout << nearest(n) << endl;     return 0;}
// Java implementation of the approachclass GFG {             // Function to return the Least number    static int nearest(int n)    {             // Get the perfect square        // before and after N        int prevSquare = (int)Math.sqrt(n);        int nextSquare = prevSquare + 1;        prevSquare = prevSquare * prevSquare;        nextSquare = nextSquare * nextSquare;             // Check which is nearest to N        int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);             // return the result        return ans;    }         // Driver code    public static void main (String[] args)    {        int n = 14;        System.out.println(nearest(n));             n = 16;        System.out.println(nearest(n)) ;             n = 18;        System.out.println(nearest(n)) ;         }} // This code is contributed by AnkitRai01
# Python3 implementation of the approachfrom math import sqrt # Function to return the Least numberdef nearest(n) :     # Get the perfect square    # before and after N    prevSquare = int(sqrt(n));    nextSquare = prevSquare + 1;    prevSquare = prevSquare * prevSquare;    nextSquare = nextSquare * nextSquare;     # Check which is nearest to N    ans    = (prevSquare - n) if (n - prevSquare) < (nextSquare - n) else (nextSquare - n);     # return the result    return ans; # Driver codeif __name__ == "__main__" :     n = 14;    print(nearest(n)) ;     n = 16;    print(nearest(n));     n = 18;    print(nearest(n)); # This code is contributed by AnkitRai01
// C# implementation of the approachusing System; class GFG {             // Function to return the Least number    static int nearest(int n)    {             // Get the perfect square        // before and after N        int prevSquare = (int)Math.Sqrt(n);        int nextSquare = prevSquare + 1;        prevSquare = prevSquare * prevSquare;        nextSquare = nextSquare * nextSquare;             // Check which is nearest to N        int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);             // return the result        return ans;    }         // Driver code    public static void Main (string[] args)    {        int n = 14;        Console.WriteLine(nearest(n));             n = 16;        Console.WriteLine(nearest(n)) ;             n = 18;        Console.WriteLine(nearest(n)) ;         }} // This code is contributed by AnkitRai01
<script>// Javascript implementation of the above approach // Function to return the Least numberfunction nearest( n){    // Get the perfect square    // before and after N    var prevSquare = parseInt(Math.sqrt(n));    var nextSquare = prevSquare + 1;    prevSquare = prevSquare * prevSquare;    nextSquare = nextSquare * nextSquare;     // Check which is nearest to N     if((n - prevSquare) < (nextSquare - n))     {       ans = parseInt((prevSquare - n));     }     else       ans = parseInt((nextSquare - n));     // return the result    return ans;} var  n = 14;document.write( nearest(n) + "<br>"); n = 16;document.write(  nearest(n)  + "<br>"); n = 18;document.write(  nearest(n)  + "<br>");  // This code is contributed by SoumikMondal</script>
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