Print all repeating digits present in a given number in sorted order
Given an integer N, the task is to print all the repeating digits present in N in sorted order.
Examples:
Input: N = 45244336543
Output: 3 4 5
Explanation: The duplicate digits are 3 4 5Input: N = 987065467809
Output: 0 6 7 8 9
Approach: The idea is to use Hashing to store the frequency of digits of N and print those digits whose frequency exceeds 1.
- Initialize a HashMap, to store frequency of digits 0-9.
- Convert the integer to its equivalent string.
- Iterate over the characters of the string, and for each character, perform the following steps:
- Convert the current character to an integer using typecasting
- Increment the count of that digit in a HashMap.
- Traverse the HashMap and print the digits whose count exceeds 1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print repeating// digits present in the number Nvoid printDup(string N){ // Stores the count of // unique digits int res = 0; // Map to store frequency // of each digit int cnt[10]; // Set count of all digits to 0 for (int i = 0; i < 10; i++) cnt[i] = 0; // Traversing the string for (int i = 0; i < N.size(); i++) { // Convert the character // to equivalent integer int digit = (N[i] - '0'); // Increase the count of digit cnt[digit] += 1; } // Traverse the Map for (int i = 0; i < 10; i++) { // If frequency // of digit exceeds 1 if (cnt[i] > 1) cout << i << " "; } cout << endl;}// Driver Codeint main(){ string N = "45244336543"; // Function call to print // repeating digits in N printDup(N); return 0;}// This code is contributed by Kingash. |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to print repeating // digits present in the number N static void printDup(String N) { // Stores the count of // unique digits int res = 0; // Map to store frequency // of each digit int cnt[] = new int[10]; // Traversing the string for (int i = 0; i < N.length(); i++) { // Convert the character // to equivalent integer int digit = (N.charAt(i) - '0'); // Increase the count of digit cnt[digit] += 1; } // Traverse the Map for (int i = 0; i < 10; i++) { // If frequency // of digit exceeds 1 if (cnt[i] > 1) System.out.print(i + " "); } System.out.println(); } // Driver Code public static void main(String[] args) { String N = "45244336543"; // Function call to print // repeating digits in N printDup(N); }}// This code is contributed by Kingash. |
Python3
# Python implementation# of the above approach# Function to print repeating# digits present in the number Ndef printDup(N): # Stores the count of # unique digits res = 0 # Set count of all digits to 0 cnt = [0] * 10 # Convert integer to # equivalent string string = str(N) # Traversing the string for i in string: # Convert the character # to equivalent integer digit = int(i) # Increase the count of digit cnt[digit] += 1 # Traverse the Map for i in range(10): # If frequency # of digit is 1 if (cnt[i] > 1): # If count exceeds 1 print(i, end=" ")# Driver CodeN = 45244336543# Function call to print# repeating digits in NprintDup(N) |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to print repeating // digits present in the number N static void printDup(string N) { // Stores the count of // unique digits int res = 0; // Map to store frequency // of each digit int[] cnt = new int[10]; // Traversing the string for (int i = 0; i < N.Length; i++) { // Convert the character // to equivalent integer int digit = (N[i] - '0'); // Increase the count of digit cnt[digit] += 1; } // Traverse the Map for (int i = 0; i < 10; i++) { // If frequency // of digit exceeds 1 if (cnt[i] > 1) Console.Write(i + " "); } Console.WriteLine(); } // Driver code public static void Main() { string N = "45244336543"; // Function call to print // repeating digits in N printDup(N); }}// This code is contributed by code_hunt. |
Output:
3 4 5
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
