A Number N is said to be Self Number if it can not be written as M + sum of digits of M for any M.
The first few Self numbers are:
1, 3, 5, 7, 9, 20, 31, 42…………….
Check if N is a Self number
Given an integer N, the task is to find if this number is Self number or not.
Examples:
Input: N = 3
Output: Yes
Explanation:
1 + sumofDigits(1) = 2
2 + sumofDigits(2) = 4
3 + sumofDigits(3) = 6
Hence 3 can not be written as
m + sum of digits of m for any m.
Input: N = 4
Output: No
2 + sumodDigits(2) = 4
Approach: The idea is to iterate from 1 to N and for each number check that sum of its value and sum of its digit is equal to N or not. If yes then the number is not a self number. Otherwise, the number is a self number.
For Example:
if N = 3
// Check for every number
// from 1 to N
1 + sumofDigits(1) = 1
2 + sumofDigits(2) = 4
3 + sumofDigits(3) = 6
Hence 3 can not be written as
M + sum of digits of M for any M.
Below is the implementation of the above approach:
Example :
C++
// C++ implementation to check if the// number is a self number or not#include <bits/stdc++.h>using namespace std;// Function to find the sum of// digits of a number Nint getSum(int n){ int sum = 0; while (n != 0) { sum = sum + n % 10; n = n / 10; } return sum;}// Function to check for Self numberbool isSelfNum(int n){ for (int m = 1; m <= n; m++) { if (m + getSum(m) == n) return false; } return true;}// Driver codeint main(){ int n = 20; if (isSelfNum(n)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
// Java implementation to check if the// number is a self number or notclass GFG{ // Function to find the sum // of digits of a number Nstatic int getSum(int n){ int sum = 0; while (n != 0) { sum = sum + n % 10; n = n / 10; } return sum;}// Function to check for Self numberstatic boolean isSelfNum(int n){ for(int m = 1; m <= n; m++) { if (m + getSum(m) == n) return false; } return true;}// Driver codepublic static void main(String[] args){ int n = 20; if (isSelfNum(n)) { System.out.println("Yes"); } else { System.out.println("No"); }}}// This code is contributed by Ritik Bansal |
Python3
# Python3 implementation to check if the# number is a self number or not# Function to find the sum of# digits of a number Ndef getSum(n): sum1 = 0; while (n != 0): sum1 = sum1 + n % 10; n = n // 10; return sum1;# Function to check for Self numberdef isSelfNum(n): for m in range(1, n + 1): if (m + getSum(m) == n): return False; return True;# Driver coden = 20;if (isSelfNum(n)): print("Yes");else: print("No");# This code is contributed by Code_Mech |
C#
// C# implementation to check if the// number is a self number or notusing System;class GFG{ // Function to find the sum // of digits of a number Nstatic int getSum(int n){ int sum = 0; while (n != 0) { sum = sum + n % 10; n = n / 10; } return sum;}// Function to check for Self numberstatic bool isSelfNum(int n){ for(int m = 1; m <= n; m++) { if (m + getSum(m) == n) return false; } return true;}// Driver codepublic static void Main(){ int n = 20; if (isSelfNum(n)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by Code_Mech |
Javascript
<script>// Javascript implementation to check if the// number is a self number or not // Function to find the sum // of digits of a number N function getSum( n) { let sum = 0; while (n != 0) { sum = sum + n % 10; n = parseInt(n / 10); } return sum; } // Function to check for Self number function isSelfNum( n) { for ( let m = 1; m <= n; m++) { if (m + getSum(m) == n) return false; } return true; } // Driver code let n = 20; if (isSelfNum(n)) { document.write("Yes"); } else { document.write("No"); } // This code is contributed by aashish1995 </script> |
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)
References: https://en.wikipedia.org/wiki/Self_number
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