# Minimum value to be added to X such that it is at least Y percent of N

• Last Updated : 23 Jun, 2022

Given three integers N, X and Y, the task is to find the minimum integer that should be added to X to make it at least Y percent of N.

Examples:

Input: N = 10, X = 2, Y = 40
Output:
Adding 2 to X gives 4 which is 40% of 10

Input: N = 10, X = 2, Y = 20
Output:
X is already 20% of 10

Approach: Find val = (N * Y) / 100 which is the Y percent of N. Now in order for X to be equal to val, val – X must be added to X only if X < val.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the required value` `// that must be added to x so that` `// it is at least y percent of n` `int` `minValue(``int` `n, ``int` `x, ``int` `y)` `{`   `    ``// Required value` `    ``float` `val = (y * n) / 100;`   `    ``// If x is already >= y percent of n` `    ``if` `(x >= val)` `        ``return` `0;` `    ``else` `        ``return` `(``ceil``(val) - x);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 10, x = 2, y = 40;` `    ``cout << minValue(n, x, y);` `}`

## Java

 `// Java implementation of the approach` `import` `java.lang.Math;`   `class` `GFG ` `{` `    `  `// Function to return the required value` `// that must be added to x so that` `// it is at least y percent of n` `static` `int` `minValue(``int` `n, ``int` `x, ``int` `y)` `{`   `    ``// Required value` `    ``float` `val = (y * n) / ``100``;`   `    ``// If x is already >= y percent of n` `    ``if` `(x >= val)` `        ``return` `0``;` `    ``else` `        ``return` `(``int``)(Math.ceil(val)-x);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``10``, x = ``2``, y = ``40``;` `    ``System.out.println(minValue(n, x, y));` `}` `}`   `// This code is contributed by Code_Mech.`

## Python3

 `import` `math`   `# Function to return the required value ` `# that must be added to x so that ` `# it is at least y percent of n ` `def` `minValue(n, x, y):`   `    ``# Required value` `    ``val ``=` `(y ``*` `n)``/``100`   `    ``# If x is already >= y percent of n ` `    ``if` `x >``=` `val:` `        ``return` `0` `    ``else``:` `        ``return` `math.ceil(val) ``-` `x`   `# Driver code` `n ``=` `10``; x ``=` `2``; y ``=` `40` `print``(minValue(n, x, y))`     `# This code is contributed by Shrikant13`

## C#

 `// C# implementation of the approach ` `using` `System;`   `class` `GFG ` `{ ` `    `  `    ``// Function to return the required value ` `    ``// that must be added to x so that ` `    ``// it is at least y percent of n ` `    ``static` `int` `minValue(``int` `n, ``int` `x, ``int` `y) ` `    ``{ ` `    `  `        ``// Required value ` `        ``float` `val = (y * n) / 100; ` `    `  `        ``// If x is already >= y percent of n ` `        ``if` `(x >= val) ` `            ``return` `0; ` `        ``else` `            ``return` `(``int``)(Math.Ceiling(val)-x) ; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 10, x = 2, y = 40; ` `        ``Console.WriteLine((``int``)minValue(n, x, y)); ` `    ``} ` `} `   `// This code is contributed by Ryuga.`

## PHP

 `= y percent of n` `    ``if` `(``\$x` `>= ``\$val``)` `        ``return` `0;` `    ``else` `        ``return` `(``ceil``(``\$val``) - ``\$x``);` `}`   `// Driver code` `{` `    ``\$n` `= 10; ``\$x` `= 2; ``\$y` `= 40;` `    ``echo``(minValue(``\$n``, ``\$x``, ``\$y``));` `}`   `// This code is contributed by Code_Mech.`

## Javascript

 ``

Output:

`2`

Time Complexity: O(1)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up
Related Articles