A good way into this one is to ask: what do we need to remember so we never redo work blindly? For 319. Bulb Switcher, the solution in this repository is mainly a straightforward implementation solution.
Guide
What?
The first job is to translate the English prompt into state, transition, and stopping conditions. 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: straightforward implementation.
The notes already sitting in the source point us in the right direction:
- TLE
- 31 / 35 test cases passed.
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are bulbSwitch, state, getDivisorCount, sieveOfEratosthenes.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- 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:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
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(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//TLE
02//31 / 35 test cases passed.
03class Solution {
04public:
05 int bulbSwitch(int n) {
06 if(n == 1) return 1;
07
08 vector<bool> state(n, true);
09 for(int t = 2; t <= n; t++){
10 for(int i = t-1; i < n; i += t){
11 // cout << state[i] << " " << (~state[i]) << " " << !state[i] << endl;
12 //~: bitwise complement, !: not
13 //~1 -> -2, !1 -> 0, so we should use "!" here
14 state[i] = !state[i];
15 }
16 }
17
18 // for(int i = 0; i < n; i++){
19 // cout << state[i] << " ";
20 // }
21 // cout << endl;
22
23 return accumulate(state.begin(), state.end(), 0);
24 }
25};
26
27//count divisor
28//TLE
29//30 / 35 test cases passed.
30class Solution {
31public:
32 vector<int> divisorCount;
33
34 int getDivisorCount(int n){
35 if(divisorCount[n] == 0){
36 int count = 0; //1
37 double sqrt_n = sqrt(n);
38
39 for(int i = 1; i <= sqrt_n; i++){
40 if(n % i == 0){
41 if(i == sqrt_n) count++;
42 else count += 2;
43 }
44 }
45
46 divisorCount[n] = count;
47 }
48 return divisorCount[n];
49 };
50
51 int bulbSwitch(int n) {
52 //sum of divisor count from 1 to n
53 divisorCount = vector<int>(n+1, 0);
54 int ans = 0;
55
56 for(int i = 1; i <= n; i++){
57 int count = getDivisorCount(i);
58 // cout << i << " " << count << endl;
59 if(count % 2 == 1){
60 ans++;
61 }
62 }
63
64 return ans;
65 }
66};
67
68//Count Divisors of n in O(n^1/3)
69//TLE
70//27 / 35 test cases passed.
71//https://www.geeksforgeeks.org/count-divisors-n-on13/
72class Solution {
73public:
74 vector<int> primes;
75 vector<int> divisorCount;
76
77 void sieveOfEratosthenes(int n, vector<int>& primes){
78 vector<bool> isPrime = vector<bool>(n+1, true);
79 // vector<bool> isPrimeSquare = vector<bool>((long long)n*n+1, false);
80
81 //isPrime should be initialized as false
82 //isPrimeSquare should be initialized as true
83 isPrime[1] = false;
84
85 //mark prime's multiples as non-prime
86 for(int p = 2; p * p <= n; p++){
87 if(isPrime[p]){
88 //start from current prime's two times!
89 for(int i = p*2; i <= n; i+=p){
90 isPrime[i] = false;
91 }
92 }
93 }
94
95 for(int p = 2; p <= n; p++){
96 if(isPrime[p]){
97 // cout << p << " ";
98 // isPrimeSquare[p*p] = true;
99 primes.push_back(p);
100 }
101 }
102 // cout << endl;
103 }
104
105 int getDivisorCount(int n){
106 if(n == 1) return 1;
107 if(divisorCount[n] != 0) return divisorCount[n];
108
109 int org_n = n;
110
111 //split n into x*y
112
113 int count = 1;
114 /*
115 counting factors of x
116 if x = p^m + q^n,
117 then divisor count of x is (m+1)*(n+1)
118 */
119 for(int i = 0; pow(primes[i], 3) <= n; i++){
120 int pow_i = 1;
121 while(n % primes[i] == 0){
122 n /= primes[i];
123 pow_i++;
124 }
125 //now pow(primes[i], pow_i-1) becomes n
126
127 count *= pow_i;
128 }
129
130 // cout << "n: " << org_n << ". x: " << org_n/n << ", y: " << n << endl;
131
132 //counting factors of y
133
134 if(find(primes.begin(), primes.end(), n) != primes.end()){
135 //y is prime
136 count *= 2;
137 }else if(n/sqrt(n) == sqrt(n) && find(primes.begin(), primes.end(), sqrt(n)) != primes.end()){
138 //y is a prime's square
139 count *= 3;
140 }else if(n != 1){
141 //y is the product of two distinct prime numbers
142 count *= 4;
143 }
144
145 divisorCount[n] = count;
146
147 return count;
148 };
149
150 int bulbSwitch(int n) {
151 //sum of divisor count from 1 to n
152 divisorCount = vector<int>(n+1, 0);
153
154 sieveOfEratosthenes(n, primes);
155
156 int ans = 0;
157
158 for(int i = 1; i <= n; i++){
159 int count = getDivisorCount(i);
160 // cout << i << " " << count << endl;
161 if(count % 2 == 1){
162 ans++;
163 }
164 }
165
166 return ans;
167 }
168};
169
170//Math
171//https://leetcode.com/problems/bulb-switcher/discuss/77104/Math-solution..
172class Solution {
173public:
174 int bulbSwitch(int n) {
175 /*
176 we want to find how many i in the range [1...n] whose
177 divisor count is odd.
178 A number's divisor count is odd iff it's square.
179 So we want to find out how many i in the range [1...n] is square.
180 And the count of square root in the range [1...sqrt(n)] is same as the count of square in the range[1...n],
181 so we can just give sqrt(n) as our answer
182 */
183 return sqrt(n);
184 }
185};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.