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
Algorithm:
Step 1: To solve this using recursion, make a recursion function with inputs, and a variable currIndex to traverse the input array.
Step2: Base Case:- If currIndex == size of the input array, return -1, i.e element not found.
Step3: Take input of next recursion call ,withcurrIndex incremented by 1 , in a variable ‘index’.
Step 4:
If(index == -1 && input[currIndex] == x)
Return currIndex
Else
Return index;
C++ Source Code/Function:
Output