Gives you the highest power of x less then provided number n

highest_power

/*
  Problem, find the highest power less than a number n
  Iterative solution: O(n)
  Logarithmic solution: O(logn)
  Bitwise solution: O(logn)
*/
var highestPower = function(n, x) {
  var res = 0;
  for (var i = n; i >= 1; i--){
    var pow = Math.pow(x, i);
    if (pow < n) { res = pow; break;}
  }
  return res;
}

// Log solution
var highestPowerOf2Log = function(n) {
  var pow = parseInt(Math.log2(n))
  return Math.pow(2, pow);
}

// Using left shift operator
var highestPowerof2Bitwise = function(n) {
  var res = 1;
  for (var i = 0; i < n; i++){
    curr = 1 << i;
    if (curr > n) break;
    res = curr;
  }
  return res;
}

console.log('highest power iterative: ', highestPower(73, 2))
console.log('highest power logarithmic: ', highestPowerOf2Log(73))
console.log('highest power bitwise: ', highestPowerof2Bitwise(73))

/*
Left shifting explained:

.    9 (base 10): 00000000000000000000000000001001 (base 2)
                  --------------------------------
9 << 3 (base 10): 00000000000000000000000001001000 (base 2) = 72 (base 10)

Bitwise shifting any number x to the left by y bits yields x * 2 ^ y.
So e.g.: 9 << 3 translates to: 9 * (2 ^ 3) = 9 * (8) = 72.
*/