Leetcode | Solution of Guess Number Higher or Lower in JavaScript

May 11th, 2020
3 min read

In this post, we will solve Guess Number Higher or Lower from leetcode and compute it's time and space complexities. Let's begin.

Problem Statement

The question can be found at leetcode Guess Number Higher or Lower problem.

The problem states that we need to guess the correct number between 1 to n which the problem has picked based on the following rules

Constraints and challenges

We are given a guess API which returns

  • 1 if the picked number is higher than our guessed number
  • 0 if it is equal to the guessed number
  • -1 if the guessed number is lower than the picked number


The problem is nothing but a binary search. In the series, 1...n find a given number. I wish I had more to explain here, but it is a simple binary search. If you don't recall the algorithm, please read about it and then you won't need to read further emoji-smile

Here is the solution is you are still reading this.

var guessNumber = function (n) {

  if (guess(n) === 0) return n

  let low = 1, high = n;
  while (n > 0) {
    const pick = (parseInt(high + low) / 2);
    const res = guess(pick);

    if (res === 0) {
      return pick
    if (res === -1) {
      high = pick;
    if (res === 1) {
      low = pick;

First, take care of an edge case. If n is the picked number we return it first thing in the program. Next, we apply a binary search to find the number.

Here are the stats on submission

Status: Accepted
Runtime: 48ms
Memory: 33.7MB

Time and space complexity

Time complexity

The time complexity here is logarithmic because the input size decreases by half every time. O(log n).

Space complexity

We are not using any extra space except for a couple of variables. So space complexity is constant, O(1).


So, we solved the Guess Number Higher or Lower problem by using binar search and calculated the time and space complexities.

I hope you enjoyed solving this question. This is it for this one, complete source code for this post can be found on my Github Repo. Will see you in the next one.

There you go guys, you made it to end of the post. Subscribe to my youtube channel for regular updates. Follow me on twitter, drop me a mail or leave a comment here if you still have any doubts and I will try my best to help you out. Thanks

Stay tuned and see you around :)