Given an integer array, print all distinct elements in array. The given array may contain duplicates and the output should print every element only once. The given array is not sorted.
Examples:
Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}
Output: 12, 10, 9, 45, 2
Input: arr[] = {1, 2, 3, 4, 5}
Output: 1, 2, 3, 4, 5
Input: arr[] = {1, 1, 1, 1, 1}
Output: 1
A Simple Solution is to use twp nested loops. The outer loop picks an element one by one starting from the leftmost element. The inner loop checks if the element is present on left side of it. If present, then ignores the element, else prints the element. Following is the implementation of the simple algorithm.
C++
// C++ program to print all distinct elements in a given array #include <iostream> #include <algorithm> using namespace std; void printDistinct(int arr[], int n) { // Pick all elements one by one for (int i=0; i<n; i++) { // Check if the picked element is already printed int j; for (j=0; j<i; j++) if (arr[i] == arr[j]) break; // If not printed earlier, then print it if (i == j) cout << arr[i] << " "; } } // Driver program to test above function int main() { int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}; int n = sizeof(arr)/sizeof(arr[0]); printDistinct(arr, n); return 0; } |
Java
// Java program to print all distinct // elements in a given array import java.io.*; class GFG { static void printDistinct(int arr[], int n) { // Pick all elements one by one for (int i = 0; i < n; i++) { // Check if the picked element // is already printed int j; for (j = 0; j < i; j++) if (arr[i] == arr[j]) break; // If not printed earlier, // then print it if (i == j) System.out.print( arr[i] + " "); } } // Driver program public static void main (String[] args) { int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}; int n = arr.length; printDistinct(arr, n); } } // This code is contributed by vt_m |
Python3
# python program to print all distinct # elements in a given array def printDistinct(arr, n): # Pick all elements one by one for i in range(0, n): # Check if the picked element # is already printed d = 0 for j in range(0, i): if (arr[i] == arr[j]): d = 1 break # If not printed earlier, # then print it if (d == 0): print(arr[i]) # Driver program to test above function arr = [6, 10, 5, 4, 9, 120, 4, 6, 10] n = len(arr) printDistinct(arr, n) # This code is contributed by Sam007. |
C#
// C# program to print all distinct // elements in a given array using System; class GFG { static void printDistinct(int []arr, int n) { // Pick all elements one by one for (int i = 0; i < n; i++) { // Check if the picked element // is already printed int j; for (j = 0; j < i; j++) if (arr[i] == arr[j]) break; // If not printed earlier, // then print it if (i == j) Console.Write(arr[i] + " "); } } // Driver program public static void Main () { int []arr = {6, 10, 5, 4, 9, 120, 4, 6, 10}; int n = arr.Length; printDistinct(arr, n); } } // This code is contributed by Sam007. |
PHP
<?php // PHP program to print all distinct // elements in a given array function printDistinct($arr, $n) { // Pick all elements one by one for($i = 0; $i < $n; $i++) { // Check if the picked element // is already printed $j; for($j = 0; $j < $i; $j++) if ($arr[$i] == $arr[$j]) break; // If not printed // earlier, then print it if ($i == $j) echo $arr[$i] , " "; } } // Driver Code $arr = array(6, 10, 5, 4, 9, 120, 4, 6, 10); $n = sizeof($arr); printDistinct($arr, $n); // This code is contributed by nitin mittal ?> |
Output:
6 10 5 4 9 120
Time Complexity of above solution is O(n2). We can Use Sorting to solve the problem in O(nLogn) time. The idea is simple, first sort the array so that all occurrences of every element become consecutive. Once the occurrences become consecutive, we can traverse the sorted array and print distinct elements in O(n) time. Following is the implementation of the idea.
C++
// C++ program to print all distinct elements in a given array #include <iostream> #include <algorithm> using namespace std; void printDistinct(int arr[], int n) { // First sort the array so that all occurrences become consecutive sort(arr, arr + n); // Traverse the sorted array for (int i=0; i<n; i++) { // Move the index ahead while there are duplicates while (i < n-1 && arr[i] == arr[i+1]) i++; // print last occurrence of the current element cout << arr[i] << " "; } } // Driver program to test above function int main() { int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}; int n = sizeof(arr)/sizeof(arr[0]); printDistinct(arr, n); return 0; } |
Java
// Java program to print all distinct // elements in a given array import java.io.*; import java .util.*; class GFG { static void printDistinct(int arr[], int n) { // First sort the array so that // all occurrences become consecutive Arrays.sort(arr); // Traverse the sorted array for (int i = 0; i < n; i++) { // Move the index ahead while // there are duplicates while (i < n - 1 && arr[i] == arr[i + 1]) i++; // print last occurrence of // the current element System.out.print(arr[i] +" "); } } // Driver program public static void main (String[] args) { int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}; int n = arr.length; printDistinct(arr, n); } } // This code is contributed by vt_m |
C#
// C# program to print all distinct // elements in a given array using System; class GFG { static void printDistinct(int []arr, int n) { // First sort the array so that // all occurrences become consecutive Array.Sort(arr); // Traverse the sorted array for (int i = 0; i < n; i++) { // Move the index ahead while // there are duplicates while (i < n - 1 && arr[i] == arr[i + 1]) i++; // print last occurrence of // the current element Console.Write(arr[i] + " "); } } // Driver program public static void Main () { int []arr = {6, 10, 5, 4, 9, 120, 4, 6, 10}; int n = arr.Length; printDistinct(arr, n); } } // This code is contributed by Sam007. |
PHP
<?php // PHP program to print all distinct // elements in a given array function printDistinct( $arr, $n) { // First sort the array so // that all occurrences // become consecutive sort($arr); // Traverse the sorted array for ($i = 0; $i < $n; $i++) { // Move the index ahead // while there are duplicates while ($i < $n - 1 and $arr[$i] == $arr[$i + 1]) $i++; // print last occurrence // of the current element echo $arr[$i] , " "; } } // Driver Code $arr = array(6, 10, 5, 4, 9, 120, 4, 6, 10); $n = count($arr); printDistinct($arr, $n); // This code is contributed by anuj_67. ?> |
Output:
4 5 6 9 10 120
We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from left to right and keep track of visited elements in a hash table. Following is the implementation of the idea.
C++
/* CPP program to print all distinct elements of a given array */#include<bits/stdc++.h> using namespace std; // This function prints all distinct elements void printDistinct(int arr[],int n) { // Creates an empty hashset unordered_set<int> s; // Traverse the input array for (int i=0; i<n; i++) { // If not present, then put it in // hashtable and print it if (s.find(arr[i])==s.end()) { s.insert(arr[i]); cout << arr[i] << " "; } } } // Driver method to test above method int main () { int arr[] = {10, 5, 3, 4, 3, 5, 6}; int n=7; printDistinct(arr,n); return 0; } |
Java
/* Java program to print all distinct elements of a given array */import java.util.*; class Main { // This function prints all distinct elements static void printDistinct(int arr[]) { // Creates an empty hashset HashSet<Integer> set = new HashSet<>(); // Traverse the input array for (int i=0; i<arr.length; i++) { // If not present, then put it in hashtable and print it if (!set.contains(arr[i])) { set.add(arr[i]); System.out.print(arr[i] + " "); } } } // Driver method to test above method public static void main (String[] args) { int arr[] = {10, 5, 3, 4, 3, 5, 6}; printDistinct(arr); } } |
C#
// C# program to print all distinct // elements of a given array using System; using System.Collections.Generic; class GFG { // This function prints all // distinct elements public static void printDistinct(int[] arr) { // Creates an empty hashset HashSet<int> set = new HashSet<int>(); // Traverse the input array for (int i = 0; i < arr.Length; i++) { // If not present, then put it // in hashtable and print it if (!set.Contains(arr[i])) { set.Add(arr[i]); Console.Write(arr[i] + " "); } } } // Driver Code public static void Main(string[] args) { int[] arr = new int[] {10, 5, 3, 4, 3, 5, 6}; printDistinct(arr); } } // This code is contributed by Shrikant13 |
Output:
10 5 3 4 6
One more advantage of hashing over sorting is, the elements are printed in same order as they are in input array.
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