98 lines
2.2 KiB
Markdown
98 lines
2.2 KiB
Markdown
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# Leetcode First-Bad-Version
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#### 2022-07-09 09:52
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> ##### Algorithms:
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> #algorithm #binary_search
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> ##### Data structures:
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> #DS #array
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#binary_search
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```
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##### Links:
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- [Link to problem](https://leetcode.com/problems/first-bad-version/)
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___
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### Problem
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You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
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Suppose you have `n` versions `[1, 2, ..., n]` and you want to find out the first bad one, which causes all the following ones to be bad.
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You are given an API `bool isBadVersion(version)` which returns whether `version` is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
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#### Examples
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**Example 1:**
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**Input:** n = 5, bad = 4
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**Output:** 4
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**Explanation:**
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call isBadVersion(3) -> false
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call isBadVersion(5) -> true
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call isBadVersion(4) -> true
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Then 4 is the first bad version.
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**Example 2:**
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**Input:** n = 1, bad = 1
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**Output:** 1
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#### Constraints
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- `1 <= bad <= n <= 231 - 1`
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### Thoughts
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> [!summary]
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> This is a #binary_search problem
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Note that [[Leetcode First-Bad-Version#Constraints]], n can be 2**31, which means there might be integer overflow.
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To address that, according to [[Binary Search Algorithm#How to implement Binary search]], use `mid = l + (r - l) / 2`
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I use a `first` variable to keep track of the first bad version.
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### Solution
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```cpp
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// The API isBadVersion is defined for you.
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// bool isBadVersion(int version);
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class Solution {
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public:
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int firstBadVersion(int n) {
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// variant of BS
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// 1-indexed
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int r = n;
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int l = 1;
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int mid;
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int first = n;
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do {
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mid = l + (r - l) / 2;
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if (isBadVersion(mid)) {
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first = min(n, mid);
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// Search left
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r = mid - 1;
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} else {
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l = mid + 1;
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}
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} while (l <= r);
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return first;
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}
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};
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```
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