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

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: 2
Adding 2 to X gives 4 which is 40% of 10

Input: N = 10, X = 2, Y = 20
Output: 0
X is already 20% of 10

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. `

Output:

```2
```

Time Complexity: O(1)

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