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LastIndexOfNumber.java
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60 lines (54 loc) · 1.74 KB
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package recursion1;
import java.util.Scanner;
/*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*/
public class LastIndexOfNumber {
private static int lastIndexOccurrence(int[] arr, int x, int startIndex) {
if (startIndex >= arr.length) {
return -1;
}
int smallAnswer = lastIndexOccurrence(arr, x, startIndex + 1);
if (smallAnswer != -1) {
return smallAnswer;
}
if (arr[startIndex] == x) {
return startIndex;
} else {
return -1;
}
}
public static int lastIndexOccurrence(int[] arr, int x) {
return lastIndexOccurrence(arr, x, 0);
}
public static int[] takeInput() {
Scanner scan = new Scanner(System.in);
int size = scan.nextInt();
int[] arr = new int[size];
for (int i = 0; i < size; i++) {
arr[i] = scan.nextInt();
}
return arr;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int[] arr = takeInput();
int x = scan.nextInt();
System.out.println(lastIndexOccurrence(arr, x));
}
}