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  • week-7
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  • week-8
    • Kth Smallest Element in a BST
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  • Problem
  • Pseudocode
  • Solution
  • Time and Space Complexity
  1. week-2

First Bad Version

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Last updated 2 years ago

Problem

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.

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.

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.

Example 1:

Input: n = 5, bad = 4
Output: 4
Explanation:
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.

Example 2:

Input: n = 1, bad = 1
Output: 1

Pseudocode

false -> false -> false -> false -> true -> true

false = not faulty
true = fauty

the brute force/naive approach would be to go thorugh every version until the first true is returned.

this can be optimized by using a binary search, with a target of true && prev (i - 1) as false

Solution

  return function (n) {
    let firstBadVersion = null;
    let lo = 0;
    let hi = n;

    if (n === 1) {
      return n;
    }

    while (lo <= hi) {
      let mid = Math.floor(lo + (hi - lo) / 2);

      if (isBadVersion(mid) && !isBadVersion(mid - 1)) {
        return mid;
      }

      if (isBadVersion(mid)) {
        hi = mid - 1;
      }

      if (!isBadVersion(mid)) {
        lo = mid + 1;
      }
    }
  
    return firstBadVersion;
  };
};

Time and Space Complexity

Time

  • Binary search has a running time of O(log n)

  • Total - O(log n)

Space

  • Assignment of variables are constant time, no other variables stored - O(1)

  • Total - O(1)

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