Given an array of length N and an integer x, you need to find and return the first index of integer x present in the array. Return -1 if it is not present in the array. First index means, the index of first occurrence of x in the input array. Do this recursively. Indexing in the array starts from 0.
First Index of Number
PROBLEM:-Given an array of length N and an integer x, you need to find and return the first index of integer x present in the array. Return -1 if it is not present in the array.First index means, the index of first occurrence of x in the input array.
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 :
first index or -1
Constraints :
1 <= N <= 10^3
Sample Input :
4
9 8 10 8
8
Sample Output :
1SOLUTION:-
#include<iostream>
using namespace std;
int firstIndex(int input[], int size, int x) {
/* Don't write main().
Don't read input, it is passed as function argument.
Return output and don't print it.
Taking input and printing output is handled automatically.
*/
if(size==0)
return -1;
else{
static int i=0;
if(input[0]==x)
return i;
else{
i++;
firstIndex(input+1,size-1,x);
}
}
}
int main(){
int n;
cin >> n;
int *input = new int[n];
for(int i = 0; i < n; i++) {
cin >> input[i];
}
int x;
cin >> x;
cout << firstIndex(input, n, x) << endl;
}