Given a number n, we need to find the sum of its digits such that:
If n < 10
digSum(n) = n
Else
digSum(n) = Sum(digSum(n))
Examples :
Input : 1234
Output : 1
Explanation : The sum of 1+2+3+4 = 10,
digSum(x) == 10
Hence ans will be 1+0 = 1
Input : 5674
Output : 4
A brute force approach is to sum all the digits until the sum < 10.
Flowchart:
Below is the brute force program to find the sum.
C++
// C++ program to find sum of// digits of a number until// sum becomes single digit.#include<bits/stdc++.h> using namespace std;int digSum(int n){ int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while(n > 0 || sum > 9) { if(n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum;}// Driver program to test the above functionint main(){ int n = 1234; cout << digSum(n); return 0;} |
Java
// Java program to find sum of// digits of a number until// sum becomes single digit.import java.util.*;public class GfG { static int digSum(int n) { int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || sum > 9) { if (n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum; } // Driver code public static void main(String argc[]) { int n = 1234; System.out.println(digSum(n)); }}// This code is contributed by Gitanjali. |
Python
# Python program to find sum of# digits of a number until# sum becomes single digit.import math # method to find sum of digits # of a number until sum becomes # single digitdef digSum( n): sum = 0 while(n > 0 or sum > 9): if(n == 0): n = sum sum = 0 sum += n % 10 n /= 10 return sum# Driver methodn = 1234print (digSum(n))# This code is contributed by Gitanjali. |
C#
// C# program to find sum of// digits of a number until// sum becomes single digit.using System;class GFG { static int digSum(int n) { int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || sum > 9) { if (n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum; } // Driver code public static void Main() { int n = 1234; Console.Write(digSum(n)); }}// This code is contributed by nitin mittal |
PHP
<?php// PHP program to find sum of// digits of a number until// sum becomes single digit.function digSum( $n){ $sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while($n > 0 || $sum > 9) { if($n == 0) { $n = $sum; $sum = 0; } $sum += $n % 10; $n = (int)$n / 10; } return $sum;}// Driver Code$n = 1234;echo digSum($n);// This code is contributed// by aj_36?> |
Javascript
<script>// Javascript program to find sum of // digits of a number until // sum becomes single digit. let n = 1234; //Function to get sum of digits function getSum(n) { let sum = 0; while (n > 0 || sum > 9) { if(n == 0) { n = sum; sum = 0; } sum = sum + n % 10; n = Math.floor(n / 10); } return sum; } //function call document.write(getSum(n)); //This code is contributed by Surbhi Tyagi</script> |
C
// C program to find sum of// digits of a number until// sum becomes single digit.#include<stdio.h>int digSum(int n){ int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while(n > 0 || sum > 9) { if(n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum;}// Driver program to test the above functionint main(){ int n = 1234; printf("%d",digSum(n)); return 0;} |
Output :
1
Time Complexity: O(log(n)).
Auxiliary Space: O(1)
So, another challenge is “Could you do it without any loop/recursion in O(1) runtime?”
YES!!
There exists a simple and elegant O(1) solution for this too. The answer is given simply:-
If n == 0
return 0;
If n % 9 == 0
digSum(n) = 9
Else
digSum(n) = n % 9
How does the above logic works?
The logic behind this approach is :
To check if a number is divisible by 9, add the digits of the number and check if the sum is divisible by 9 or not. If yes, is the case, the number is divisible by 9, otherwise, it’s not.
let’s take 27 i.e (2+7 = 9) hence divisible by 9.
If a number n is divisible by 9, then the sum of its digit until the sum becomes a single digit is always 9. For example,
Let, n = 2880
Sum of digits = 2 + 8 + 8 = 18: 18 = 1 + 8 = 9
Therefore,
A number can be of the form 9x or 9x + k. For the first case, the answer is always 9. For the second case, and is always k which is the remainder left.
The problem is widely known as the digit root problem.
You may find this Wikipedia article useful. -> https://en.wikipedia.org/wiki/Digital_root
Below is the implementation of the above idea :
C++
#include<bits/stdc++.h> using namespace std;int digSum(int n){ if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9);}// Driver program to test the above functionint main(){ int n = 9999; cout<<digSum(n); return 0;} |
Java
import java.io.*;class GFG { static int digSum(int n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Driver program to test the above function public static void main (String[] args) { int n = 9999; System.out.println(digSum(n)); }}// This code is contributed by anuj_67. |
Python3
def digSum(n): if (n == 0): return 0 if (n % 9 == 0): return 9 else: return (n % 9)# Driver program to test the above functionn = 9999print(digSum(n))# This code is contributed by# Smitha Dinesh Semwal |
C#
using System;class GFG{ static int digSum(int n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Driver Code public static void Main () { int n = 9999; Console.Write(digSum(n)); }}// This code is contributed by aj_36 |
PHP
<?phpfunction digSum($n){ if ($n == 0) return 0; return ($n % 9 == 0) ? 9 : ($n % 9);}// Driver program to test the above function$n = 9999;echo digSum($n);//This code is contributed by anuj_67.?> |
Javascript
<script> function digSum(n){ if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9);}// Driver coden = 9999;document.write(digSum(n));// This code is contributed by code_hunt</script> |
Output:
9
Time Complexity: O(1)
Auxiliary Space: O(1)
Related Post :
https://www.geeksforgeeks.org/digital-rootrepeated-digital-sum-given-integer/
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