Given an array arr[] consisting of N integers, the task is to print all unique digits of the number formed by concatenating all array elements in the order of their occurrence after excluding leading zeroes.
Examples:
Input: arr[] = {122, 474, 612, 932}
Output: 7 6 9 3
Explanation:
The number formed by concatenating array elements is “122474612932”.
Unique digits present in the number are 7 6 9 3 (in the order of their occurrence).Input: arr[]={0, 912, 231, 14}
Output: 9 3 4
Explanation:
The number formed by concatenating array elements is “091223114″.
Final number obtained after removal of leading 0s is “91223114”.
Unique digits present in the number are 9 3 4 (in the order of their occurrence).
Approach: The idea is to convert all array elements to their equivalent strings and concatenate those strings and use Hashing to find unique digits present in the number obtained.
Follow the steps below to solve the problem.
- Traverse the array arr[] and convert each array element to its equivalent string and concatenate all the strings in a variable, say S.
- Convert string S to equivalent integer (say N) using typecasting (removing leading 0s)
- Initialize a Hash Table of size 10, to store the frequency of digits [0, 9].
- Initialize an empty list, say lis
- Now, for each digit of number N, increment the count of that index in the Hash Table.
- For each digit of number N, perform the following operations:
- Check if it is visited or not
- If the number is not visited and if its frequency is 1, then append this digit to lis. Make the value of that index as visited.
- Reverse list lis and print it
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 long unique elementsvoid printUnique(vector<long long> lis){ // Reverse the list reverse(lis.begin(),lis.end()); // Traverse the list for(long long i:lis) cout << i << " ";}// Function which check for// all unique digitsvoid checkUnique(string st){ // Stores the final number vector<long long> lis; // Stores the count of // unique digits long long res = 0; // Converting string to long longer // to remove leading zeros long long N = stoll(st); // Stores count of digits vector<long long> cnt(10, 0), cnt1(10, 0); // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N long long rem = N % 10; // Increase the count // of the last digit cnt[rem] += 1; // Remove the last digit of N N = N /10; } // Converting string to long longer again N = stoll(st); // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N long long rem = N % 10; // If the value of this digit // is not visited if(cnt1[rem] == 0) { // If its frequency is 1 (unique) if(cnt[rem] == 1) lis.push_back(rem); } // Mark the digit visited cnt1[rem] = 1; // Remove the last digit of N N = N /10; } // Passing this list to print long // the reversed list printUnique(lis);}// Function to concatenate array elementsvoid combineArray(vector<long long> lis){ // Stores the concatenated number string st = ""; // Traverse the array for (long long el : lis) { // Convert to equivalent string string ee = to_string(el); // Concatenate the string st = st + ee; } // Passing string to checkUnique function checkUnique(st);}// Driver Codeint main(){ vector<long long> arr = {122, 474, 612, 932}; // Function call to prlong long unique // digits present in the // concatenation of array elements combineArray(arr); return 0;}// This code is contributed by mohit kumar 29. |
Java
// Java program for the above approachimport java.util.Arrays;class GFG { public static void main(String[] args) { long[] arr = { 122, 474, 612, 932 }; combineArray(arr); } // Function to print long unique elements static void printUnique(long[] lis) { // Reverse the list for (int i = 0; i < lis.length / 2; i++) { long temp = lis[i]; lis[i] = lis[lis.length - i - 1]; lis[lis.length - i - 1] = temp; } // Traverse the list for (long i : lis) { if (i != 0) { System.out.print(i + " "); } } } // Function which check for // all unique digits static void checkUnique(String str) { // Stores the final number long[] lis = new long[str.length()]; // Stores the count of // unique digits long res = 0; // Converting string to long to remove leading zeros long N = Long.parseLong(str); // Stores count of digits long[] cnt = new long[10]; // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N long rem = N % 10; // Increase the count // of the last digit cnt[(int)rem]++; // Remove the last digit of N N /= 10; } // Converting string to long longer again N = Long.parseLong(str); // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N long rem = N % 10; if (cnt[(int)rem] != 'v') { if (cnt[(int)rem] == 1) { lis[(int)res++] = rem; } } cnt[(int)rem] = 'v'; N /= 10; } // Function to print long unique elements printUnique(lis); } // Function to concatenate array elements static void combineArray(long[] lis) { String str = ""; for (long el : lis) { str += String.valueOf(el); } checkUnique(str); }}// This code is contributed by phasing17 |
Python3
# Python implementation# of above approach# Function to print unique elementsdef printUnique(lis): # Reverse the list lis.reverse() # Traverse the list for i in lis: print(i, end =" ")# Function which check for# all unique digitsdef checkUnique(string): # Stores the final number lis = [] # Stores the count of # unique digits res = 0 # Converting string to integer # to remove leading zeros N = int(string) # Stores count of digits cnt = [0] * 10 # Iterate over the digits of N while (N > 0): # Retrieve the last digit of N rem = N % 10 # Increase the count # of the last digit cnt[rem] += 1 # Remove the last digit of N N = N // 10 # Converting string to integer again N = int(string) # Iterate over the digits of N while (N > 0): # Retrieve the last digit of N rem = N % 10 # If the value of this digit # is not visited if(cnt[rem] != 'visited'): # If its frequency is 1 (unique) if(cnt[rem] == 1): lis.append(rem) # Mark the digit visited cnt[rem] = 'visited' # Remove the last digit of N N = N // 10 # Passing this list to print # the reversed list printUnique(lis)# Function to concatenate array elementsdef combineArray(lis): # Stores the concatenated number string = "" # Traverse the array for el in lis: # Convert to equivalent string el = str(el) # Concatenate the string string = string + el # Passing string to checkUnique function checkUnique(string)# Driver Code# Inputarr = [122, 474, 612, 932]# Function call to print unique# digits present in the# concatenation of array elementscombineArray(arr) |
C#
using System;using System.Linq;class Program{ static void Main(string[] args) { long[] arr = { 122, 474, 612, 932 }; CombineArray(arr); } // Function to print long unique elements static void PrintUnique(long[] lis) { // Reverse the list Array.Reverse(lis); // Traverse the list foreach (long i in lis) { if (i != 0) Console.Write(i + " "); } } // Function which check for // all unique digits static void CheckUnique(string str) { // Stores the final number long[] lis = new long[str.Length]; // Stores the count of // unique digits long res = 0; // Converting string to long longer // to remove leading zeros long N = long.Parse(str); // Stores count of digits long[] cnt = new long[10]; // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N long rem = N % 10; // Increase the count // of the last digit cnt[rem]++; // Remove the last digit of N N /= 10; } // Converting string to long longer again N = long.Parse(str); // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N long rem = N % 10; if (cnt[rem] != 'v') { if (cnt[rem] == 1) { lis[res++] = rem; } } cnt[rem] = 'v'; N /= 10; } // Function to print long unique elements PrintUnique(lis); } // Function to concatenate array elements static void CombineArray(long[] lis) { string str = ""; foreach (long el in lis) { str += el.ToString(); } CheckUnique(str); }}// This code is contributed by phasing17. |
Javascript
<script>// Javascript program for the above approach// Function to print long unique elementsfunction printUnique(lis){ // Reverse the list lis.reverse(); // Traverse the list for(var i=0; i<lis.length; i++) { document.write(lis[i]+" ") }}// Function which check for// all unique digitsfunction checkUnique(st){ // Stores the final number var lis = []; // Stores the count of // unique digits var res = 0; // Converting string to long longer // to remove leading zeros var N = parseInt(st); // Stores count of digits var cnt = Array(10).fill(0); var cnt1 = Array(10).fill(0); // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N var rem = N % 10; // Increase the count // of the last digit cnt[rem] += 1; // Remove the last digit of N N = parseInt(N /10); } // Converting string to long longer again N = parseInt(st); // Iterate over the digits of N while (N > 0) { // Retrieve the last digit of N var rem = N % 10; // If the value of this digit // is not visited if(cnt1[rem] == 0) { // If its frequency is 1 (unique) if(cnt[rem] == 1) lis.push(rem); } // Mark the digit visited cnt1[rem] = 1; // Remove the last digit of N N = parseInt(N /10); } // Passing this list to print long // the reversed list printUnique(lis);}// Function to concatenate array elementsfunction combineArray(lis){ // Stores the concatenated number var st = ""; // Traverse the array for(var i =0; i< lis.length; i++) { // Convert to equivalent string var ee = (lis[i].toString()); // Concatenate the string st = st + ee; } // Passing string to checkUnique function checkUnique(st);}// Driver Codevar arr = [122, 474, 612, 932];// Function call to print long unique// digits present in the// concatenation of array elementscombineArray(arr);</script> |
Output:
7 6 9 3
Time Complexity: O(N)
Auxiliary Space: O(N)
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