Junction Number is a number N if it can be written as K + SumOfDIgits(k) for at least two K.
101, 103, 105, 107, 109, 111, 113, 115….
Check if a number N is an Junction Number
Given a number N, the task is to check if N is an Junction Number or not. If N is an Junction Number then print “Yes” else print “No”.
Examples:
Input: N = 1106
Output: Yes
Explanation:
1106 = 1093 + (1+0+9+3) = 1102 + (1+1+0+2)
Input: N = 11
Output: No
Approach: Since Junction Number is a number N if it can be written as K + SumOfDIgits(k) for at least two K. So for all numbers from 1 to N in a loop we will check if i + sumOfdigitsof(i) equals to N or not. If equals to N print “Yes” else print “No”.
Below is the implementation of the above approach:
C++
// C++ implementation for the// above approach#include <bits/stdc++.h>using namespace std;// Function to find the // sum Of digits of Nint sum(int n){ // To store sum of N and // sumOfdigitsof(N) int sum = 0; while (n != 0) { // extracting digit int r = n % 10; sum = sum + r; n = n / 10; } return sum;}// Function to check Junction numbersbool isJunction(int n){ // To store count of ways n can be // represented as i + SOD(i) int count = 0; for (int i = 1; i <= n; i++) { if (i + sum(i) == n) count++; } return count >= 2;}// Driver Codeint main(){ int N = 111; // Function Call if (isJunction(N)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program for above approachclass GFG{ // Function to find the // sum Of digits of Nstatic int sum(int n){ // To store sum of N and // sumOfdigitsof(N) int sum = 0; while (n != 0) { // extracting digit int r = n % 10; sum = sum + r; n = n / 10; } return sum;}// Function to check Junction numbersstatic boolean isJunction(int n){ // To store count of ways n can be // represented as i + SOD(i) int count = 0; for (int i = 1; i <= n; i++) { if (i + sum(i) == n) count++; } return count >= 2;}// Driver Code public static void main(String[] args) { int N = 111; // Function Call if (isJunction(N)) System.out.print("Yes"); else System.out.print("No");} } // This code is contributed by Pratima Pandey |
Python3
# Python3 program for the above approachimport math# Function to find the # sum Of digits of Ndef sum1(n): # To store sum of N and # sumOfdigitsof(N) sum1 = 0 while (n != 0): # extracting digit r = n % 10 sum1 = sum1 + r n = n // 10 return sum1# Function to check Junction numbersdef isJunction(n): # To store count of ways n can be # represented as i + SOD(i) count = 0 for i in range(1, n + 1): if (i + sum1(i) == n): count = count + 1 return count >= 2# Driver Codeif __name__=='__main__': # Given Number n = 111 # Function Call if (isJunction(n) == 1): print("Yes") else: print("No")# This code is contributed by rock_cool |
C#
// C# program for above approachusing System;class GFG{ // Function to find the // sum Of digits of Nstatic int sum(int n){ // To store sum of N and // sumOfdigitsof(N) int sum = 0; while (n != 0) { // extracting digit int r = n % 10; sum = sum + r; n = n / 10; } return sum;}// Function to check Junction numbersstatic bool isJunction(int n){ // To store count of ways n can be // represented as i + SOD(i) int count = 0; for (int i = 1; i <= n; i++) { if (i + sum(i) == n) count++; } return count >= 2;}// Driver Code public static void Main() { int N = 111; // Function Call if (isJunction(N)) Console.Write("Yes"); else Console.Write("No");} } // This code is contributed by Nidhi Biet |
Javascript
<script>// Javascript program for above approach // Function to find the // sum Of digits of N function sum( n) { // To store sum of N and // sumOfdigitsof(N) let sum = 0; while (n != 0) { // extracting digit let r = n % 10; sum = sum + r; n = parseInt(n / 10); } return sum; } // Function to check Junction numbers function isJunction( n) { // To store count of ways n can be // represented as i + SOD(i) let count = 0; for ( i = 1; i <= n; i++) { if (i + sum(i) == n) count++; } return count >= 2; } // Driver Code let N = 111; // Function Call if (isJunction(N)) document.write("Yes"); else document.write("No");// This code contributed by Rajput-Ji</script> |
Output:
Yes
Time Complexity: O(n)
Space Complexity: O(1)
Reference: http://www.numbersaplenty.com/set/junction_number/
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