C++ program to find presence of a number X in the array recursively

Algorithm:

To find the elements using recursion, we call upon a recursive function maintaining an output array, its size and a variable currIndex for traversal of the input array. The base case of this function will be if currIndex is equal to the size of Input array. i.e we have finished traversing the array. On every recursive call, the input element at currIndex is compared with ‘x’, and if it matches is then stored within the output array (and hence incrementing the output array size by 1). Next recursive call is made by incrementing the currIndex.

Source Code/Functions:

#include <bits/stdc++.h>
using namespace std;

void findIndices(int input[], int size, int x, int output[], int &k, int currIndex){
	if(currIndex==size){
		return;
	}
	if(input[currIndex]==x){
		output[k] = currIndex;
		k++;
	}
	findIndices(input,size,x,output,k,++currIndex);
}

int allIndices(int input[], int size, int x, int output[]){
	int k = 0;
	findIndices(input,size,x,output,k,0);
	return k;
}

int main(){
	int size = 5;
	int input[] = {9,8,10,8,8};
	int x = 8;
	int output[5];
	
	int outputSize = allIndices(input,size,x,output);
	for(int i =0;i<outputSize;i++){
		cout<<output[i]<<" ";
	}
	
	return 0;
}

Output

1 3 4