LeetCode #278 — EASY

First Bad Version

Build confidence with an intuition-first walkthrough focused on binary search fundamentals.

The Problem

Problem Statement

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

Constraints:

1 <= bad <= n <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Binary Search

Example 1

5
4

Example 2

1
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #278: First Bad Version
/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int l = 1, r = n;
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (isBadVersion(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #278: First Bad Version
/**
 * Forward declaration of isBadVersion API.
 * @param   version   your guess about first bad version
 * @return 	 	      true if current version is bad
 *			          false if current version is good
 * func isBadVersion(version int) bool;
 */

func firstBadVersion(n int) int {
	l, r := 1, n
	for l < r {
		mid := (l + r) >> 1
		if isBadVersion(mid) {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

# Accepted solution for LeetCode #278: First Bad Version
# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:


class Solution:
    def firstBadVersion(self, n: int) -> int:
        l, r = 1, n
        while l < r:
            mid = (l + r) >> 1
            if isBadVersion(mid):
                r = mid
            else:
                l = mid + 1
        return l

// Accepted solution for LeetCode #278: First Bad Version
// The API isBadVersion is defined for you.
// isBadVersion(version:i32)-> bool;
// to call it use self.isBadVersion(version)

impl Solution {
    pub fn first_bad_version(&self, n: i32) -> i32 {
        let (mut l, mut r) = (1, n);
        while l < r {
            let mid = l + (r - l) / 2;
            if self.isBadVersion(mid) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}

// Accepted solution for LeetCode #278: First Bad Version
/**
 * The knows API is defined in the parent class Relation.
 * isBadVersion(version: number): boolean {
 *     ...
 * };
 */

var solution = function (isBadVersion: any) {
    return function (n: number): number {
        let [l, r] = [1, n];
        while (l < r) {
            const mid = (l + r) >>> 1;
            if (isBadVersion(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
};

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.