Given an unsorted array with repetitions, the task is to group multiple occurrence of individual elements. The grouping should happen in a way that the order of first occurrences of all elements is maintained.

Examples:

Input: arr[] = {5, 3, 5, 1, 3, 3} Output: {5, 5, 3, 3, 3, 1} Input: arr[] = {4, 6, 9, 2, 3, 4, 9, 6, 10, 4} Output: {4, 4, 4, 6, 6, 9, 9, 2, 3, 10}

**Simple Solution** is to use nested loops. The outer loop traverses array elements one by one. The inner loop checks if this is first occurrence, if yes, then the inner loop prints it and all other occurrences.

// A simple C++ program to group multiple occurrences of individual // array elements #include<iostream> using namespace std; // A simple method to group all occurrences of individual elements void groupElements(int arr[], int n) { // Initialize all elements as not visited bool *visited = new bool[n]; for (int i=0; i<n; i++) visited[i] = false; // Traverse all elements for (int i=0; i<n; i++) { // Check if this is first occurrence if (!visited[i]) { // If yes, print it and all subsequent occurrences cout << arr[i] << " "; for (int j=i+1; j<n; j++) { if (arr[i] == arr[j]) { cout << arr[i] << " "; visited[j] = true; } } } } delete [] visited; } /* Driver program to test above function */ int main() { int arr[] = {4, 6, 9, 2, 3, 4, 9, 6, 10, 4}; int n = sizeof(arr)/sizeof(arr[0]); groupElements(arr, n); return 0; }

Output:

4 4 4 6 6 9 9 2 3 10

Time complexity of the above method is O(n^{2}).

**Binary Search Tree based Method:** The time complexity can be improved to O(nLogn) using self-balancing binary search tree like Red-Black Tree or AVL tree. Following is complete algorithm.

1) Create an empty Binary Search Tree (BST). Every BST node is going to contain an array element and its count.

2) Traverse the input array and do following for every element.

……..a) If element is not present in BST, then insert it with count as 0.

……..b) If element is present, then increment count in corresponding BST node.

3) Traverse the array again and do following for every element.

…….. If element is present in BST, then do following

……….a) Get its count and print the element ‘count’ times.

……….b) Delete the element from BST.

Time Complexity of the above solution is O(nLogn).

**Hashing based Method:** We can also use hashing. The idea is to replace Binary Search Tree with a Hash Map in above algorithm.

Below is Java Implementation of hashing based solution.

/* Java program to group multiple occurrences of individual array elements */ import java.util.HashMap; class Main { // A hashing based method to group all occurrences of individual elements static void orderedGroup(int arr[]) { // Creates an empty hashmap HashMap<Integer, Integer> hM = new HashMap<Integer, Integer>(); // Traverse the array elements, and store count for every element // in HashMap for (int i=0; i<arr.length; i++) { // Check if element is already in HashMap Integer prevCount = hM.get(arr[i]); if (prevCount == null) prevCount = 0; // Increment count of element element in HashMap hM.put(arr[i], prevCount + 1); } // Traverse array again for (int i=0; i<arr.length; i++) { // Check if this is first occurrence Integer count = hM.get(arr[i]); if (count != null) { // If yes, then print the element 'count' times for (int j=0; j<count; j++) System.out.print(arr[i] + " "); // And remove the element from HashMap. hM.remove(arr[i]); } } } // Driver method to test above method public static void main (String[] args) { int arr[] = {10, 5, 3, 10, 10, 4, 1, 3}; orderedGroup(arr); } }

Output:

10 10 10 5 3 3 4 1

Time Complexity of the above hashing based solution is Θ(n) under the assumption that insert, search and delete operations on HashMap take O(1) time.

Below is a related problem for strings.

Group all occurrences of characters according to first appearance

This article is contributed by **Himanshu Gupta**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.