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 the 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 approach using System;class GFG{ // Function to print repeating // digits present in the number N static void printDup(string N) { // Stores the count of // unique digits // 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. |
Javascript
<script>// Javascript program for the above approach// Function to print repeating// digits present in the number Nfunction printDup(N){ // Stores the count of // unique digits var res = 0; // Map to store frequency // of each digit var cnt = Array(10); // Set count of all digits to 0 for (var i = 0; i < 10; i++) cnt[i] = 0; // Traversing the string for (var i = 0; i < N.length; i++) { // Convert the character // to equivalent integer var digit = (N[i] - '0'); // Increase the count of digit cnt[digit] += 1; } // Traverse the Map for (var i = 0; i < 10; i++) { // If frequency // of digit exceeds 1 if (cnt[i] > 1) document.write( i + " "); } document.write("<br>");}// Driver Codevar N = "45244336543";// Function call to print// repeating digits in NprintDup(N);</script> |
Output
3 4 5
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach 2: Using count().
Using count() we try to find the character whose occurrence is more than once, store it in a data structure, and sort it.
C++
#include <iostream>#include <string>#include <algorithm>#include <unordered_set>using namespace std;int main(){ // initializing number long long N = 45244336543; unordered_set<char> x; string s = to_string(N); // using count method for (char i : s) { if (count(s.begin(), s.end(), i) > 1 && x.find(i) == x.end()) { x.insert(i); } } string result; for (char i : x) { result += i; } sort(result.begin(), result.end()); cout << result << endl; return 0;}// This code is contributed by phasing17 |
Java
// Java code to implement the approachimport java.util.*;import java.util.stream.*;class GFG { // Driver code public static void main(String[] args) { // Initializing number long N = 45244336543L; // Creating hashset to store digits HashSet<Character> x = new HashSet<Character>(); // Converting number to string String s = Long.toString(N); // using groupby method Map<Character, Long> groups = s.chars() .mapToObj(c -> (char)c) .collect(Collectors.groupingBy( c -> c, Collectors.counting())); for (Map.Entry<Character, Long> group : groups.entrySet()) { if (group.getValue() > 1 && !x.contains(group.getKey())) { x.add(group.getKey()); } } String result = x.stream() .sorted() .map(Object::toString) .collect(Collectors.joining()); for (char r : result.toCharArray()) System.out.print(r + " "); }}// This code is contributed by phasing17 |
Python3
# Python implementation# of the above approach#initializing numberN = 45244336543x=[]s=str(N)for i in s: if(s.count(i)>1 and i not in x): x.append(i)x.sort()print(' '.join(x)) |
C#
using System;using System.Linq;using System.Collections.Generic;class GFG { static void Main(string[] args) { long N = 45244336543; HashSet<char> x = new HashSet<char>(); var s = N.ToString(); // using groupby method var groups = s.GroupBy(c => c); foreach (var group in groups) { if (group.Count() > 1 && !x.Contains(group.Key)) { x.Add(group.Key); } } var result = string.Concat(x.OrderBy(c => c)); Console.WriteLine(result); }}// This code is contributed by phasing17 |
Javascript
// JavaScript implementation// of the above approach// Initializing numberconst N = 45244336543;let x = new Set();const s = String(N);for (let i of s) if (s.split(i).length > 2) x.add(i); x = Array.from(x)x.sort(function(a, b){ return a - b;});console.log(x.join(' '));// This code is contributed by phasing17. |
Output
3 4 5
Time Complexity: O(n log n)
Auxiliary Space: O(n)
Approach 3: Using the Counter() function
Python3
# Python implementation# of the above approachfrom collections import Counter# initializing numberN = 45244336543x = []s = str(N)freq = Counter(s)for key, value in freq.items(): if value > 1: x.append(key)x.sort()print(' '.join(x)) |
Time Complexity: O(n log n)
Auxiliary Space: O(n)
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