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 integer, this will contain the root of the perfect square before 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++

`// C++ implementation of the approach ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the Least number ` `int` `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 code ` `int` `main() ` `{ ` ` ` `int` `n = 14; ` ` ` `cout << nearest(n) << endl; ` ` ` ` ` `n = 16; ` ` ` `cout << nearest(n) << endl; ` ` ` ` ` `n = 18; ` ` ` `cout << nearest(n) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `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` `; ` ` ` `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 ` |

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## Python3

`# Python3 implementation of the approach ` `from` `math ` `import` `sqrt ` ` ` `# Function to return the Least number ` `def` `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 code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `n ` `=` `14` `; ` ` ` `print` `(nearest(n)) ; ` ` ` ` ` `n ` `=` `16` `; ` ` ` `print` `(nearest(n)); ` ` ` ` ` `n ` `=` `18` `; ` ` ` `print` `(nearest(n)); ` ` ` `# This code is contributed by AnkitRai01 ` |

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## C#

`// C# implementation of the approach ` `using` `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 ` |

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**Output:**

2 0 -2

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