« Last Index of Number in the array.

Problem

Last Index of Number in array

Given an array of length N and an integer x, you need to find and return the last index of integer x present in the array. Return -1 if it is not present in the array. Last index means - if x is present multiple times in the array, return the index at which x comes last in the array. You should start traversing your array from 0, not from (N - 1). Do this recursively. Indexing in the array starts from 0.

Input Format :

Line 1 : An Integer N i.e. size of array

Line 2 : N integers which are elements of the array, separated by spaces

Line 3 : Integer x

Output Format :

last index or -1

Constraints :

1 <= N <= 10^3

Sample Input :

4 9 8 10 8 8

Sample Output :

3

Solution

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1#include <iostream>
2#include <iomanip>
3#include <algorithm>
4#include <string>
5#include <cstring>
6#include <vector>
7#include <cmath>
8#include <map>
9#include <climits>
10// climits for INT_MIN
11#include <unordered_map>
12using namespace std;
13
14int lastIndex(int input[], int size, int x)
15{
16    if (size == 1)
17    {
18        return (input[0] == x) ? 0 : -1;
19    }
20    int ans = lastIndex(input + 1, size - 1, x);
21    if (ans >= 0)
22    {
23        return ans + 1;
24    }
25    else
26    {
27        return (input[0] == x) ? 0 : -1;
28    }
29}
30int main()
31{
32    int n;
33    cin >> n;
34
35    int *input = new int[n];
36
37    for (int i = 0; i < n; i++)
38    {
39        cin >> input[i];
40    }
41
42    int x;
43
44    cin >> x;
45
46    cout << lastIndex(input, n, x) << endl;
47    return 0;
48}