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Commit 23c0c34

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LeetCode problem: 278. First Bad Version
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# 278. First Bad Version
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Difficulty: `Easy`
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Topics: `Binary Search`, `Interactive`
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You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of
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your product fails the quality check. Since each version is developed based on the previous version, all the versions
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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
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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
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the first bad version. You should minimize the number of calls to the API.
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**Example 1:**
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```text
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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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```
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**Example 2:**
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```text
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Input: n = 1, bad = 1
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Output: 1
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```
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**Constraints:**
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- `1 <= bad <= n <= 2^31 - 1`
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package com.borrelunde.leetcodesolutions.problem0278.firstbadversion;
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/**
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* This is the solution to the LeetCode problem: 278. First Bad Version
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*
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* @author Børre A. Opedal Lunde
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* @since 2024年02月18日
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*/
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public class Solution extends VersionControl {
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public int firstBadVersion(int n) {
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// Low and high boundaries for the binary search.
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int low = 1; // Versions start from 1, not 0.
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int high = n;
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// Binary search to find the first bad version.
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while (low <= high) {
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// Calculate middle and avoid integer overflow. Bitwise right shift
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// (>> 1) is used instead of dividing by 2.
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final int middle = low + ((high - low) >> 1);
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// Narrow the search based on the version check.
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if (isBadVersion(middle)) {
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// Adjust the search to the left half (exclude this version).
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high = middle - 1;
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} else {
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// Adjust the search to the right half (exclude this version).
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low = middle + 1;
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}
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}
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// At the end of the loop, low is the first bad version.
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return low;
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}
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}
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package com.borrelunde.leetcodesolutions.problem0278.firstbadversion;
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/**
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* This is the parent class that holds the isBadVersion API for the problem. It
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* is needed for the unit tests.
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*
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* @author Børre A. Opedal Lunde
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* @since 2024年02月18日
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*/
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public class VersionControl {
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private int badVersion;
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public void setBadVersion(final int bad) {
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this.badVersion = bad;
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}
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public boolean isBadVersion(int version) {
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return version >= badVersion;
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}
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}
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package com.borrelunde.leetcodesolutions.problem0278.firstbadversion;
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import org.junit.jupiter.api.DisplayName;
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import org.junit.jupiter.api.Test;
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import static org.junit.jupiter.api.Assertions.assertEquals;
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/**
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* This is the test to the LeetCode problem: 278. First Bad Version
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*
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* @author Børre A. Opedal Lunde
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* @since 2024年02月18日
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*/
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@DisplayName("First bad version")
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class SolutionTest {
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private final Solution solution = new Solution();
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@Test
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@DisplayName("Example one")
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void exampleOne() {
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int n = 5;
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int bad = 4;
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solution.setBadVersion(bad);
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int expected = bad;
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int actual = solution.firstBadVersion(n);
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assertEquals(expected, actual);
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}
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@Test
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@DisplayName("First version is bad")
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void firstVersionIsBad() {
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int n = 1;
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int bad = 1;
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solution.setBadVersion(bad);
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int expected = bad;
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int actual = solution.firstBadVersion(n);
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assertEquals(expected, actual);
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}
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@Test
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@DisplayName("Last version is bad")
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void lastVersionIsBad() {
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int n = Integer.MAX_VALUE;
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int bad = Integer.MAX_VALUE;
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solution.setBadVersion(bad);
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int expected = bad;
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int actual = solution.firstBadVersion(n);
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assertEquals(expected, actual);
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}
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@Test
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@DisplayName("Bad version is in the middle of many")
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void badVersionIsInTheMiddleOfMany() {
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int n = 1_000_000;
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int bad = 500_000;
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solution.setBadVersion(bad);
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int expected = bad;
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int actual = solution.firstBadVersion(n);
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assertEquals(expected, actual);
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}
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@Test
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@DisplayName("Bad version is minimum and n is maximum")
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void badVersionIsMinimumAndNIsMaximum() {
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int n = Integer.MAX_VALUE;
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int bad = 1;
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solution.setBadVersion(bad);
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int expected = bad;
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int actual = solution.firstBadVersion(n);
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assertEquals(expected, actual);
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}
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}

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