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672. Bulb Switcher II

Description

There is a room with n bulbs labeled from 1 to n that all are turned on initially, and four buttons on the wall. Each of the four buttons has a different functionality where:

• Button 1: Flips the status of all the bulbs.
• Button 2: Flips the status of all the bulbs with even labels (i.e., 2, 4, ...).
• Button 3: Flips the status of all the bulbs with odd labels (i.e., 1, 3, ...).
• Button 4: Flips the status of all the bulbs with a label j = 3k + 1 where k = 0, 1, 2, ... (i.e., 1, 4, 7, 10, ...).

You must make exactly presses button presses in total. For each press, you may pick any of the four buttons to press.

Given the two integers n and presses, return the number of different possible statuses after performing all presses button presses.

 

Example 1:

Input: n = 1, presses = 1
Output: 2
Explanation: Status can be:
- [off] by pressing button 1
- [on] by pressing button 2

Example 2:

Input: n = 2, presses = 1
Output: 3
Explanation: Status can be:
- [off, off] by pressing button 1
- [on, off] by pressing button 2
- [off, on] by pressing button 3

Example 3:

Input: n = 3, presses = 1
Output: 4
Explanation: Status can be:
- [off, off, off] by pressing button 1
- [off, on, off] by pressing button 2
- [on, off, on] by pressing button 3
- [off, on, on] by pressing button 4

 

Constraints:

• 1 <= n <= 1000
• 0 <= presses <= 1000

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public int flipLights(int n, int presses) {
          int[] ops = new int[] {0b111111, 0b010101, 0b101010, 0b100100};
          Set<Integer> vis = new HashSet<>();
          n = Math.min(n, 6);
          for (int mask = 0; mask < 1 << 4; ++mask) {
              int cnt = Integer.bitCount(mask);
              if (cnt <= presses && cnt % 2 == presses % 2) {
                  int t = 0;
                  for (int i = 0; i < 4; ++i) {
                      if (((mask >> i) & 1) == 1) {
                          t ^= ops[i];
                      }
                  }
                  t &= ((1 << 6) - 1);
                  t >>= (6 - n);
                  vis.add(t);
              }
          }
          return vis.size();
      }
  }
  

• class Solution {
  public:
      int flipLights(int n, int presses) {
          n = min(n, 6);
          vector<int> ops = {0b111111, 0b010101, 0b101010, 0b100100};
          unordered_set<int> vis;
          for (int mask = 0; mask < 1 << 4; ++mask) {
              int cnt = __builtin_popcount(mask);
              if (cnt > presses || cnt % 2 != presses % 2) continue;
              int t = 0;
              for (int i = 0; i < 4; ++i) {
                  if (mask >> i & 1) {
                      t ^= ops[i];
                  }
              }
              t &= (1 << 6) - 1;
              t >>= (6 - n);
              vis.insert(t);
          }
          return vis.size();
      }
  };
  

• class Solution:
      def flipLights(self, n: int, presses: int) -> int:
          ops = (0b111111, 0b010101, 0b101010, 0b100100)
          n = min(n, 6)
          vis = set()
          for mask in range(1 << 4):
              cnt = mask.bit_count()
              if cnt <= presses and cnt % 2 == presses % 2:
                  t = 0
                  for i, op in enumerate(ops):
                      if (mask >> i) & 1:
                          t ^= op
                  t &= (1 << 6) - 1
                  t >>= 6 - n
                  vis.add(t)
          return len(vis)
  
  

• func flipLights(n int, presses int) int {
  	if n > 6 {
  		n = 6
  	}
  	ops := []int{0b111111, 0b010101, 0b101010, 0b100100}
  	vis := map[int]bool{}
  	for mask := 0; mask < 1<<4; mask++ {
  		cnt := bits.OnesCount(uint(mask))
  		if cnt <= presses && cnt%2 == presses%2 {
  			t := 0
  			for i, op := range ops {
  				if mask>>i&1 == 1 {
  					t ^= op
  				}
  			}
  			t &= 1<<6 - 1
  			t >>= (6 - n)
  			vis[t] = true
  		}
  	}
  	return len(vis)
  }
  

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