Given an array of length N and an integer x, you need to find all the indexes where x is present in the input array. Save all the indexes in an array (in increasing order). Do this recursively. Indexing in the array starts from 0.

PROBLEM:-

All Indices of Number

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Given an array of length N and an integer x, you need to find all the indexes where x is present in the input array. Save all the indexes in an array (in increasing order).

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 :

indexes where x is present in the array (separated by space)

Constraints :

1 <= N <= 10^3

Sample Input :

5
9 8 10 8 8
8

Sample Output :

1 3 4

SOLUTION:-

#include<iostream>
using namespace std;

#include<iostream>
using namespace std;
int allIndexes(int input[],int size,int x,int output[]){
   static int index=-1;
    if(size==0)
    {
        return 0;
    }
    else{

    if(input[0]==x)
    {
    output[0]=++index;
    cout<<index<<" ";
    allIndexes(input+1,size-1,x,output+1);
    }
    else{
        ++index;
        allIndexes(input+1,size-1,x,output);
    }
    }
}

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;
    
    int *output = new int[n];
    
    int size = allIndexes(input, n, x, output);
    for(int i = 0; i < size; i++) {
        cout << output[i] << " ";
    }
    
    delete [] input;
    
    delete [] output;
    

}

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