Round the given number to nearest multiple of 10

Given a positive integer n, round it to nearest whole number having zero as last digit.
Examples: 
 

Input : 4722
Output : 4720

Input : 38
Output : 40

Input : 10
Output: 10


Approach: 
Let’s round down the given number n to the nearest integer which ends with 0 and store this value in a variable a. 
a = (n / 10) * 10. So, the round up n (call it b) is b = a + 10.
If n – a > b – n then the answer is b otherwise the answer is a.
Below is the implementation of the above approach: 
 

C++

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// CPP program to round the given
// integer to a whole number
// which ends with zero.
#include <bits/stdc++.h>
using namespace std;
 
// function to round the number
int round(int n)
{
    // Smaller multiple
    int a = (n / 10) * 10;
     
    // Larger multiple
    int b = a + 10;
 
    // Return of closest of two
    return (n - a > b - n)? b : a;
}
 
// driver function
int main()
{
    int n = 4722;
    cout << round(n) << endl;
    return 0;
}

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Java

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// JAVA Code for Round the given number
// to nearest multiple of 10
import java.util.*;
 
class GFG {
     
    // function to round the number
    static int round(int n)
    {
        // Smaller multiple
        int a = (n / 10) * 10;
          
        // Larger multiple
        int b = a + 10;
      
        // Return of closest of two
        return (n - a > b - n)? b : a;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int n = 4722;
         System.out.println(round(n));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

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Python3

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# Python3 code to round the given
# integer to a whole number
# which ends with zero.
 
# function to round the number
def round( n ):
 
    # Smaller multiple
    a = (n // 10) * 10
     
    # Larger multiple
    b = a + 10
     
    # Return of closest of two
    return (b if n - a > b - n else a)
 
# driver code
n = 4722
print(round(n))
 
# This code is contributed by "Sharad_Bhardwaj".

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C#

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// C# Code for Round the given number
// to nearest multiple of 10
using System;
 
class GFG {
     
    // function to round the number
    static int round(int n)
    {
        // Smaller multiple
        int a = (n / 10) * 10;
         
        // Larger multiple
        int b = a + 10;
     
        // Return of closest of two
        return (n - a > b - n)? b : a;
    }
     
    // Driver program
    public static void Main()
    {
        int n = 4722;
        Console.WriteLine(round(n));
    }
}
 
// This code is contributed by Vt_m.

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PHP

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<?php
// PHP program to round the given integer
// to a whole number which ends with zero.
 
// function to round the number
function roundFunation($n)
{
    // Smaller multiple
    $a = (int)($n / 10) * 10;
     
    // Larger multiple
    $b = ($a + 10);
 
    // Return of closest of two
    return ($n - $a > $b - $n) ? $b : $a;
}
 
// Driver Code
$n = 4722;
echo roundFunation($n), "\n";
     
// This code is contributed by ajit
?>

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Output

4720

Output: 
 

4720


Another method if n is large:
The above method is good only for Integer or Long MAX value. if the input length is greater then the int or long-range above method does not work.



We can solve the problem using String.

Java

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// Java code for above approach
import java.io.*;
 
class GFG
{
   
  // Program to round the number to the
  // nearest number having one's digit 0
  public static String round(String s, int n)
  {
    char[] c=s.toCharArray();
 
    // last character is 0 then return the
    // original string
    if(c[n-1]=='0')
      return s;
     
    // if last character is
    // 1 or 2 or 3 or 4 or 5 make it 0
    else if(c[n-1] == '1' || c[n-1] == '2' ||
            c[n-1] == '3' || c[n-1] == '4' ||
            c[n-1] == '5' )
    {
      c[n-1]='0';
      return new String(c);
    }
    else
    {
      c[n-1]='0';
       
      // process carry
      for(int i = n - 2 ; i >= 0 ; i--)
      {
        if(c[i] == '9')
          c[i]='0';
        else
        {
          int t= c[i] - '0' + 1;
          c[i]=(char)(48+t);
          break;
        }
      }
    }
 
    String s1=new String(c);
 
    if(s1.charAt(0) == '0')
      s1="1"+s1;
     
    // return final string
    return s1;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    String s="5748965412485599999874589965999";
    int n=s.length();
     
    // Function Call
    System.out.println(round(s,n));
 
  }
}

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Output

5748965412485599999874589966000

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