This is one of those problems where the clean idea matters more than the amount of code. For 278. First Bad Version, the solution in this repository is mainly a binary search solution.
Guide
What?
The code is easier to read if we treat it as a controlled search through possible states. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: binary search, two pointers, sliding window.
The notes already sitting in the source point us in the right direction:
- The API isBadVersion is defined for you.
- bool isBadVersion(int version);
Guide
When?
Reach for this shape when a direct simulation would work logically but waste time revisiting the same information. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are isBadVersion, firstBadVersion.
Guide
Why?
The solution works because it narrows the problem until every update has a clear reason to exist.
- A map keeps the lookup side cheap; the code pays a little memory to avoid repeated searching.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Start from the smallest reliable state.
- Expand one legal move at a time.
- Cache, count, or merge information as soon as it becomes settled.
- Let the final stored value answer the original question.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(logN), space: O(1)
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//Runtime: 0 ms, faster than 100.00% of C++ online submissions for First Bad Version.
02//Memory Usage: 6.3 MB, less than 100.00% of C++ online submissions for First Bad Version.
03// The API isBadVersion is defined for you.
04// bool isBadVersion(int version);
05
06class Solution {
07public:
08 int firstBadVersion(int n) {
09 //n is 1-based
10 if(n == 1) return 1;
11 //first -1 to avoid overflow
12 int left = 1, right = n;
13 int mid = left + (right-left)/2;
14 //-1: meaningless, 0: false, 1: true
15 unordered_map<int, bool> cache;
16 // cout << n+1 << " space allocated." << endl;
17
18 do{
19 // cout << left << ", " << mid << ", " << right << endl;
20 cache[mid] = isBadVersion(mid);
21 if(cache[mid]){
22 //search left part
23 //7->4, 6->3
24 right = mid-1;
25 }else{
26 //search right part
27 left = mid+1;
28
29 }
30 mid = left + (right-left)/2;
31 // if(mid == 1) break; //mid[1-1] is meaningless
32 }while((left <= right) && !(cache.find(mid-1)!= cache.end() && cache.find(mid)!= cache.end() && !cache[mid-1] && cache[mid]));
33
34 return mid;
35 }
36};
37
38//binary search
39//Runtime: 0 ms, faster than 100.00% of C++ online submissions for First Bad Version.
40//Memory Usage: 6.2 MB, less than 100.00% of C++ online submissions for First Bad Version.
41//time: O(logN), space: O(1)
42// The API isBadVersion is defined for you.
43// bool isBadVersion(int version);
44
45class Solution {
46public:
47 int firstBadVersion(int n) {
48 int left = 1, right = n;
49 while(left < right){
50 int mid = left + (right-left)/2;
51 // cout << left << ", " << mid << ", " << right << endl;
52 if(isBadVersion(mid)){
53 /*
54 when right is equal to left+1,
55 mid is also left,
56 we should jump out (left < right = mid breaks),
57 because we have found
58 the first bad version which is mid,
59 and mid is equal to left
60 */
61 right = mid;
62 }else{
63 left = mid+1;
64 }
65 }
66 return left;
67 }
68};
69
70//binary seach, different stop condition
71//Runtime: 0 ms, faster than 100.00% of C++ online submissions for First Bad Version.
72//Memory Usage: 5.9 MB, less than 100.00% of C++ online submissions for First Bad Version.
73// The API isBadVersion is defined for you.
74// bool isBadVersion(int version);
75
76class Solution {
77public:
78 int firstBadVersion(int n) {
79 int left = 1, right = n;
80 while(left <= right){
81 int mid = left + (right-left)/2;
82 // cout << left << ", " << mid << ", " << right << endl;
83 if(isBadVersion(mid)){
84 /*
85 when right is equal to left+1,
86 mid is also left,
87 we should jump out (left = mid <= right = mid-1 breaks),
88 because we have found
89 the first bad version which is mid,
90 and mid is equal to left
91 */
92 right = mid-1;
93 }else{
94 left = mid+1;
95 }
96 }
97 return left;
98 }
99};Cost