Question

Given a number n find the number of pairs (x,y) where both x and y are less then n and higest common factor (hcf) of x and y is 1)

Answer 1

Answer:

function countPairsBruteForce(X, Y, m, n){

   let ans = 0;

   for(let i=0; i<m; i++ ){

       for(let j=0;j<n;j++){

           if ((Math.pow(X[i], Y[j]) > Math.pow(Y[j], X[i]))){

               ans += 1;

            }

       }

   }

   return ans;

}

Step-by-step explanation:

This problem is recursive and can be broken into sub-problems. We start from the end of the given digit sequence. We initialize the total count of decodings as 0. We recur for two subproblems.

1) If the last digit is non-zero, recur for the remaining (n-1) digits and add the result to the total count.

2) If the last two digits form a valid character (or smaller than 27), recur for remaining (n-2) digits and add the result to the total count.