Elementos mínimos que se agregarán en un rango para que el recuento de elementos sea divisible por K

Dados tres enteros K , L y R (rango [L, R] ), la tarea es encontrar el número mínimo de elementos por los que se debe extender el rango para que el conteo de elementos en el rango sea divisible por K .

Ejemplos: 

Entrada: K = 3, L = 10, R = 10 
Salida: 2  La
cantidad de elementos en L a R es 1. 
Entonces, para hacerlo divisible por 3, increméntelo en 2.

Entrada: K = 5, L = 9, R = 12 
Salida:
 

Acercarse: 

  • Cuente el número total de elementos en el rango y guárdelo en una variable llamada count = R – L + 1 .
  • Ahora, el número mínimo de elementos que deben agregarse al rango será K – (recuento % K).

A continuación se muestra la implementación del enfoque anterior:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int minimumMoves(int k, int l, int r)
{
    // Total elements in the range
    int count = r - l + 1;
 
    // If total elements are already divisible by k
    if (count % k == 0)
        return 0;
 
    // Value that must be added to count
    // in order to make it divisible by k
    return (k - (count % k));
}
 
// Driver Program to test above function
int main()
{
    int k = 3, l = 10, r = 10;
    cout << minimumMoves(k, l, r);
 
return 0;
}

Java

// Java implementation of the approach
 
import java.io.*;
 
class GFG {
    
 static int minimumMoves(int k, int l, int r)
{
    // Total elements in the range
    int count = r - l + 1;
 
    // If total elements are already divisible by k
    if (count % k == 0)
        return 0;
 
    // Value that must be added to count
    // in order to make it divisible by k
    return (k - (count % k));
}
 
// Driver Program to test above function
    public static void main (String[] args) {
    int k = 3, l = 10, r = 10;
    System.out.print(minimumMoves(k, l, r));
    }
}
// This code is contributed
// by inder_verma..

Python3

# Python 3 implementation of the approach
 
def minimumMoves(k, l, r):
    # Total elements in the range
    count = r - l + 1
 
    # If total elements are already divisible by k
    if (count % k == 0):
        return 0
 
    # Value that must be added to count
    # in order to make it divisible by k
    return (k - (count % k))
 
# Driver Program to test above function
if __name__ == '__main__':
    k = 3
    l = 10
    r = 10
    print(minimumMoves(k, l, r))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG {
     
 
 
static int minimumMoves(int k, int l, int r)
{
    // Total elements in the range
    int count = r - l + 1;
 
    // If total elements are already divisible by k
    if (count % k == 0)
        return 0;
 
    // Value that must be added to count
    // in order to make it divisible by k
    return (k - (count % k));
}
 
// Driver Program to test above function
    public static void Main () {
    int k = 3, l = 10, r = 10;
    Console.WriteLine(minimumMoves(k, l, r));
    }
}
// This code is contributed
// by inder_verma..

PHP

<?php
// PHP implementation of the approach
 
function minimumMoves($k, $l, $r)
{
    // Total elements in the range
    $count = $r - $l + 1;
 
    // If total elements are already divisible by k
    if ($count % $k == 0)
        return 0;
 
    // Value that must be added to count
    // in order to make it divisible by k
    return ($k - ($count % $k));
}
 
// Driver Program to test above function
 
    $k = 3; $l = 10; $r = 10;
    echo minimumMoves($k, $l, $r);
// This code is contributed
// by inder_verma..
 
 
?>

Javascript

<script>
 
// Javascript implementation of the approach
function minimumMoves(k, l, r)
{
     
    // Total elements in the range
    let count = r - l + 1;
 
    // If total elements are already
    // divisible by k
    if (count % k == 0)
        return 0;
 
    // Value that must be added to count
    // in order to make it divisible by k
    return (k - (count % k));
}
 
// Driver code
let k = 3, l = 10, r = 10;
 
document.write(minimumMoves(k, l, r));
 
// This code is contributed by souravmahato348
 
</script>
Producción: 

2

 

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por Abdullah Aslam y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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