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.
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 <bits/stdc++.h>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 programint 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<Integer, Integer> hash = new HashMap<Integer, Integer>(); 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<Integer, Integer> x : hash.entrySet()) { if (x.getValue() < res_ind) { res_ind = x.getValue(); res = x.getKey(); } } return res;}// Driver Codepublic 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 Codepublic 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
<script>// JavaScript program to find last seen element in// an array.// Returns last seen element in arr[]function lastSeenElement(a, n){ // Store last occurrence index of // every element let hash = new Map(); for (let i = 0; i < n; i++) { hash.set(a[i], i); } // Find an element in hash with minimum // index value let res_ind = Number.MAX_SAFE_INTEGER, res = 0; for (let x of hash) { if (x[1] < res_ind) { res_ind = x[1]; res = x[0]; } } return res;}// Driver Code let a = [ 2, 1, 2, 2, 4, 1 ]; let n = a.length; document.write(lastSeenElement(a, n));// This code is contributed by gfgking</script> |
Output:
2
Time Complexity: O(n)
Auxiliary Space: O(n), since n extra space has been taken.
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