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# Last seen array element (last appearance is earliest)

Given an array that might contain duplicates, find the element whose last appearance is latest.

Examples:

```Input :  arr[] = {10, 30, 20, 10, 20}
Output : 30
Explanation: Below are indexes of last
appearances of all elements (0 based indexes)
10 last occurs at index 3
30 last occurs at index 1
20 last occurs at index 2
The element whose last appearance earliest
is 30.

Input : arr[] = {20, 10, 20, 20, 40, 10}
Output : 20
Explanation:
Explanation: Below are indexes of last
appearances of all elements (0 based indexes)
20 last occurs at index 2
10 last occurs at index 5
40 last occurs at index 4
The element whose last appearance earliest
is 20.```
Recommended Practice

A naive approach is to take every element and iterate and compare its last occurrence with other elements. This requires two nested loops, and will take O(n*n) time.

An efficient approach is to use hashing. We store every elements index where it last occurred, and then iterate through all the possible elements and print the element with the least  index stored occurrence, as that will be the one which was last seen while traversing from left to right.

Implementation:

## C++

 `// C++ program to find last seen element in``// an array.``#include ``using` `namespace` `std;` `// Returns last seen element in arr[]``int` `lastSeenElement(``int` `a[], ``int` `n)``{``    ``// Store last occurrence index of``    ``// every element``    ``unordered_map<``int``, ``int``> hash;``    ``for` `(``int` `i = 0; i < n; i++)``        ``hash[a[i]] = i;` `    ``// Find an element in hash with minimum``    ``// index value``    ``int` `res_ind = INT_MAX, res;``    ``for` `(``auto` `x : hash)``    ``{``       ``if` `(x.second < res_ind)``       ``{``            ``res_ind = x.second;``            ``res = x.first;``       ``}``    ``}` `    ``return` `res;``}` `// driver program``int` `main()``{``    ``int` `a[] = { 2, 1, 2, 2, 4, 1 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);``    ``cout << lastSeenElement(a, n);``    ``return` `0;``}`

## Java

 `// Java program to find last seen element in``// an array.``import` `java.util.*;` `class` `GFG``{` `// Returns last seen element in arr[]``static` `int` `lastSeenElement(``int` `a[], ``int` `n)``{``    ``// Store last occurrence index of``    ``// every element``    ``HashMap hash = ``new` `HashMap();``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``hash.put(a[i], i);``    ``}` `    ``// Find an element in hash with minimum``    ``// index value``    ``int` `res_ind = Integer.MAX_VALUE, res = ``0``;``    ``for` `(Map.Entry x : hash.entrySet())``    ``{``        ``if` `(x.getValue() < res_ind)``        ``{``            ``res_ind = x.getValue();``            ``res = x.getKey();``        ``}  ``    ``}``    ``return` `res;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { ``2``, ``1``, ``2``, ``2``, ``4``, ``1` `};``    ``int` `n = a.length;``    ``System.out.println(lastSeenElement(a, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 program to find last seen``# element in an array.``import` `sys;` `# Returns last seen element in arr[]``def` `lastSeenElement(a, n):``    ` `    ``# Store last occurrence index of``    ``# every element``    ``hash` `=` `{}``    ` `    ``for` `i ``in` `range``(n):``        ``hash``[a[i]] ``=` `i``        ` `    ``# Find an element in hash with minimum``    ``# index value``    ``res_ind ``=` `sys.maxsize``    ``res ``=` `0``    ` `    ``for` `x, y ``in` `hash``.items():``        ``if` `y < res_ind:``            ``res_ind ``=` `y``            ``res ``=` `x``            ` `    ``return` `res` `# Driver code   ``if` `__name__``=``=``"__main__"``:``    ` `    ``a ``=` `[ ``2``, ``1``, ``2``, ``2``, ``4``, ``1` `]``    ``n ``=` `len``(a)``    ` `    ``print``(lastSeenElement(a,n))` `# This code is contributed by rutvik_56`

## C#

 `// C# program to find last seen element in``// an array.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Returns last seen element in arr[]``static` `int` `lastSeenElement(``int` `[]a, ``int` `n)``{``    ``// Store last occurrence index of``    ``// every element``    ``Dictionary<``int``,``               ``int``> hash = ``new` `Dictionary<``int``,``                                          ``int``>();``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if``(hash.ContainsKey(a[i]))``        ``{``            ``hash[a[i]] = i;``        ``}``        ``else``        ``{``            ``hash.Add(a[i], i);``        ``}``    ``}` `    ``// Find an element in hash with minimum``    ``// index value``    ``int` `res_ind = ``int``.MaxValue, res = 0;``    ``foreach``(KeyValuePair<``int``, ``int``> x ``in` `hash)``    ``{``        ``if` `(x.Value < res_ind)``        ``{``            ``res_ind = x.Value;``            ``res = x.Key;``        ``}``    ``}``    ``return` `res;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]a = { 2, 1, 2, 2, 4, 1 };``    ``int` `n = a.Length;``    ``Console.WriteLine(lastSeenElement(a, n));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`2`

Time Complexity: O(n)
Auxiliary Space: O(n), since n extra space has been taken.