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 10Input: N = 10, X = 2, Y = 20
Output: 0
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 <bits/stdc++.h>using namespace std;// Function to return the required value// that must be added to x so that// it is at least y percent of nint 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 codeint main(){ int n = 10, x = 2, y = 40; cout << minValue(n, x, y);} |
Java
// Java implementation of the approachimport 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 nstatic 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 codepublic 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 coden = 10; x = 2; y = 40print(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
<?php// php implementation of the approach// Function to return the required value// that must be added to x so that// it is at least y percent of nfunction minValue($n, $x, $y){ // Required value $val = ($y * $n) / 100; // If x is already >= y percent of n if ($x >= $val) return 0; else return (ceil($val) - $x);}// Driver code{ $n = 10; $x = 2; $y = 40; echo(minValue($n, $x, $y));}// This code is contributed by Code_Mech. |
Javascript
<script>// Javascript implementation of the approach// Function to return the required value// that must be added to x so that// it is at least y percent of nfunction minValue(n, x, y){ // Required value let val = (y * n) / 100; // If x is already >= y percent of n if (x >= val) return 0; else return (Math.ceil(val) - x);}// Driver codelet n = 10, x = 2, y = 40;document.write(minValue(n, x, y));// This code is contributed by souravmahato348</script> |
Output:
2
Time Complexity: O(1)
Auxiliary Space: O(1)
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