Given an array a[] with elements and the task is to print the last occurrences of the given elements in the list. Here we not only have to remove the duplicate elements but also we have to maintain the order of the occurrences of the elements in an array as per the last time they have occurred.
Like we have an array of 6 elements also containing some duplicate values i.e., {1,3, 2, 3, 1, 2} so the result should be in form of 3 1 2.
Example
Input: a[]={4,2,2,4,1,5,1} Output : 2 4 5 1
Algorithm
START Step 1-> Declare function void printelements(int a[], int n) Use STL unordered_map<int, int> ele Loop For int i=0 and i<n and i++ Set ele[a[i]]=i Loop For int i=0 and i<n and i++ IF ele[a[i]]=i Print a[i] End End Step 2 -> main() Declare array a[]={4,2,2,4,1,5,1} Declare int n=sizeof(a)/sizeof(a[0]) Call Function printelements(a,n) STOP
Example
#include <bits/stdc++.h> using namespace std; void printelements(int a[], int n) { unordered_map<int, int> ele; for (int i = 0; i < n; i++) ele[a[i]] = i; for (int i = 0; i < n; i++) { if (ele[a[i]] == i) cout << a[i] << " "; } } int main() { int a[] = { 4,2,2,4,1,5,1 }; int n = sizeof(a) / sizeof(a[0]); printelements(a, n); return 0; }
Output
if we run above program then it will generate following output
2 4 5 1