LastIndexOfANumberInAnArray_Question.java

// Last Index Of a Number in an Array - Question
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// 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

import java.util.Scanner;

public class LastIndexOfANumberInAnArray_Question {

    public static void takeInput(int input[], int n, Scanner s) {
        System.out.println("Enter values of the array : ");
        for (int i = 0; i < n; i++) {
            input[i] = s.nextInt();
        }
    }

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        System.out.print("Enter length of the array : ");
        int n = s.nextInt();
        int input[] = new int[n];
        takeInput(input, n, s);
        System.out.println("Enter x : ");
        int x = s.nextInt();
        int ans = lastIndex(input, x);
        System.out.println("Output : " + ans);
        s.close();
    }

    public static int lastIndex(int input[], int x) {
        if (input.length == 0) {
            return -1;
        }
        int arr[] = new int[input.length - 1];
        for (int i = 1; i < input.length; i++) {
            arr[i - 1] = input[i];
        }
        int k = lastIndex(arr, x);

        if (k != -1) {
            return k + 1;
        } else {
            if (input[0] == x) {
                return 0;
            } else {
                return -1;
            }

        }
    }
}