Given an integer N, the task is to check whether N contains a digit D such that it is the average of all other digits present in N.
Examples:
Input: N = 132
Output: Yes
Explanation:
Since, (1 + 3)/2 = 2.Input: N = 436
Output: No
Explanation:
No such digit exists which is the average of all other digits.
Approach:
To solve the problem, follow the steps below:
- Store the sum and count of digits of N.
- Store the digits in a Set .
- If sum%count is 0 and the integer sum/count is present in the Set, that is sum/count is a digit of N, then print Yes. Otherwise print No.
Below is the implementation of the above approach:
C++
// C++ Program to check whether a// given number contains a digit// which is the average of all// other digits#include <bits/stdc++.h>using namespace std;// Function which checks if a// digits exists in n which// is the average of all other digitsvoid check(int n){ set<int> digits; int temp = n; int sum = 0; int count = 0; while (temp > 0) { // Calculate sum of // digits in n sum += temp % 10; // Store the digits digits.insert(temp % 10); // Increase the count // of digits in n count++; temp = temp / 10; } // If average of all digits // is an integer if (sum % count == 0 // If the average is a digit // in n && digits.find(sum / count) != digits.end()) cout << "Yes" << endl; // Otherwise else cout << "No" << endl;}// Driver Codeint main(){ int n = 42644; check(n);} |
Java
// Java program to check whether a// given number contains a digit// which is the average of all// other digitsimport java.util.*;class GFG{// Function which checks if a// digits exists in n which// is the average of all other digitsstatic void check(int n){ HashSet<Integer> digits = new HashSet<Integer>(); int temp = n; int sum = 0; int count = 0; while (temp > 0) { // Calculate sum of // digits in n sum += temp % 10; // Store the digits digits.add(temp % 10); // Increase the count // of digits in n count++; temp = temp / 10; } // If average of all digits // is an integer if (sum % count == 0 && digits.contains(sum / count)) System.out.print("Yes" + "\n"); // Otherwise else System.out.print("No" + "\n");}// Driver Codepublic static void main(String[] args){ int n = 42644; check(n);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to check whether a given # number contains a digit which is# the average of all other digits# Function which checks if a # digits exists in n which is # the average of all other digits def check (n): digits = set() temp = n Sum = 0 count = 0 while(temp > 0): # Calculate sum of # digits in n Sum += temp % 10 # Store digits digits.add(temp % 10) # Increase the count of # digits in n count += 1 temp = temp // 10 # If average of all digits is integer if ((Sum % count == 0) and ((int)(Sum / count) in digits)): print("Yes") else: print("No")# Driver coden = 42644# Function callingcheck(n) # This code is contributed by himanshu77 |
C#
// C# program to check whether a given // number contains a digit which is the // average of all other digitsusing System;using System.Collections.Generic;class GFG{// Function which checks if a// digits exists in n which// is the average of all other digitsstatic void check(int n){ HashSet<int> digits = new HashSet<int>(); int temp = n; int sum = 0; int count = 0; while (temp > 0) { // Calculate sum of // digits in n sum += temp % 10; // Store the digits digits.Add(temp % 10); // Increase the count // of digits in n count++; temp = temp / 10; } // If average of all digits // is an integer if (sum % count == 0 && digits.Contains(sum / count)) Console.Write("Yes" + "\n"); // Otherwise else Console.Write("No" + "\n");}// Driver Codepublic static void Main(String[] args){ int n = 42644; check(n);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript Program to check whether a// given number contains a digit// which is the average of all// other digits// Function which checks if a// digits exists in n which// is the average of all other digitsfunction check(n){ var digits = new Set(); var temp = n; var sum = 0; var count = 0; while (temp > 0) { // Calculate sum of // digits in n sum += temp % 10; // Store the digits digits.add(temp % 10); // Increase the count // of digits in n count++; temp = parseInt(temp / 10); } // If average of all digits // is an integer if (sum % count == 0 // If the average is a digit // in n && digits.has(sum / count)) document.write( "Yes" ); // Otherwise else document.write( "No" );}// Driver Codevar n = 42644;check(n);</script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1) because constant space for the set is required.
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