Check Number
PROBLEM:- Given an array of length N and an integer x, you need to find if x is present in the array or not. Return true or false.
Do this recursively.
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
'true' or 'false'
Constraints :
1 <= N <= 10^3
3
9 8 10
8
Sample Output 1 :
true
3
9 8 10
2
Sample Output 2 :
false
SOLUTION: –
#include<iostream>
using namespace std;
bool checkNumber(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 false;
else{
if(input[size]==x)
return true;
else{
checkNumber(input,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;
if(checkNumber(input, n, x)) {
cout << "true" << endl;
}
else {
cout << "false" << endl;
}
}