# Guess Number Higher or Lower | LeetCode | Solution Explained

374. Guess Number Higher or Lower LeetCode is based on the understanding of binary search, and how we can integrate the API provided with binary search. Problem Link.

## Guess Number: Problem Discussion

- input: n
- from the range of [1, n], the problem picks a number
- using the API
`guess(num)`

we can get it to return a`0`

if`num`

is the correct guess- -1 if
`num`

is lower than actual value - 1 if
`num`

is higher than actual value

## Observation

The key point is to realize the similarity of Guess Number structure with binary search. Binary search uses the same logic, but instead of using the `guess(num)`

API, we commonly use the `a[i]`

where `a`

is the array we are searching in.

## Binary Search

A quick review on binary search

Have `l`

and `r`

demarcating the ranges of the array (in this case its implicit – from the range [1,n]) we are trying to search in.

We then find the `m = (l+r)//2`

Be careful to consider overflow – use `m = l + (r-l)//2`

to get the same result in Java/C/C++.

At this point, we usually compare `m`

with `a[i]`

, but in this question, Guess Number, we can instead do `x = guess(m)`

.

- if
`x == 0`

return`m`

right away - if
`x > 0`

we have`m`

, the mid, too small. We increase the`l = m + 1`

, to mark that the array is in the second half - if
`x < 0`

we have`m`

too big, so we reduce the`r = m - 1`

so that in the next iteration,`(l + r)//2`

gives a smaller`m`

## Guess Number Code

# The guess API is already defined for you. # @param num, your guess # @return -1 if my number is lower, 1 if my number is higher, otherwise return 0 # def guess(num: int) -> int: class Solution: def guessNumber(self, n: int) -> int: l, r = 1, n while l < r: m = (l+r)//2 x = guess(m) if x == 0: return m elif x > 0: l = m + 1 else: r = m - 1 return l

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