Leetcode 1529. Bulb Switcher IV

Problem Explanation

We have a row of bulbs, which are initially all turned off. The bulbs are represented by a binary string where 0 character represents a bulb that is off, and 1 character represents a bulb that is on. We are given a target configuration for the bulbs, and the goal is to find out the minimum number of operations to reach the target configuration starting from the initial state.

An operation is defined as picking any bulb and flipping it and all the bulbs to its right, where flipping means to change the state of a bulb from off to on (0 to 1) or from on to off (1 to 0).

Approach

The initial state of the bulbs is all zeros. Our approach would be to iterate through the target configuration from left to right and keep track of our current state (0 or 1). If the bulb's target state is different from our current state, it means we need to flip starting from this light. In this case, we increment our flip count, and change our current state.

Working with an Example

Let's use target = "10111" as an example.

  1. Initially, ans = 0 (flip count) and state = 0 (current bulb state), str = "00000".
  2. We loop through each character in the target string.
    • For i = 0, target[0] = '1', which is different from state = 0, so we switch state and increment ans. Now, state = 1, ans = 1, str = "11111".
    • For i = 1, target[1] = '0', which is different from state = 1, so we switch state and increment ans. Now, state = 0, ans = 2, str = "10000".
    • For i = 2, target[2] = '1', which is different from state = 0, so we switch state and increment ans. Now, state = 1, ans = 3, str = "11111".
    • For i = 3 and i = 4, target[i] = '1' is the same as state = 1, so we don't need to do anything.
  3. Finally, we have reached the target configuration with a minimum of 3 flips.

Python Solution

1
2python
3class Solution:
4    def minFlips(self, target: str) -> int:
5        ans = 0
6        state = 0
7
8        for c in target:
9            if int(c) != state:
10                state = int(c)
11                ans += 1
12        return ans

Java Solution

1
2java
3public class Solution {
4    public int minFlips(String target) {
5        int ans = 0;
6        int state = 0;
7
8        for (char c : target.toCharArray()) {
9            if (c - '0' != state) {
10                state = c - '0';
11                ans++;
12            }
13        }
14        return ans;
15    }
16}

JavaScript Solution

1
2javascript
3class Solution {
4    minFlips(target) {
5        let ans = 0;
6        let state = 0;
7
8        for (const c of target) {
9            if (parseInt(c, 10) !== state) {
10                state = parseInt(c, 10);
11                ans++;
12            }
13        }
14        return ans;
15    }
16}

C++ Solution

1
2c++
3class Solution {
4public:
5    int minFlips(string target) {
6        int ans = 0;
7        int state = 0;
8
9        for (const char &c : target) {
10            if (c - '0' != state) {
11                state = c - '0';
12                ans++;
13            }
14        }
15        return ans;
16    }
17};

C# Solution

1
2csharp
3public class Solution {
4    public int MinFlips(string target) {
5        int ans = 0;
6        int state = 0;
7
8        foreach (char c in target) {
9            if (c - '0' != state) {
10                state = c - '0';
11                ans++;
12            }
13        }
14        return ans;
15    }
16}

In all of these solutions, we loop through the bulbs in target configuration and if at any point the state of bulb in the target configuration is not the same as our current state, we flip the bulb and any to the right of it. We keep track of the number of operations performed to reach the target configuration.This problem is a classic example of a greedy algorithm, where we make the local best choice at each step. The run-time complexity of our solution is O(n), where n is the length of the target string. This is because we need to perform one operation for each character in the target string. The space complexity is O(1), as we are using a fixed number of variables to keep track of our current state and the number of flips.

In conclusion, we have provided a problem statement and the pseudocode of a greedy algorithm to solve the problem. We have also implemented the algorithm in Python, JavaScript, Java, C++, and C#, providing a wide range for different programming backgrounds. The simplicity of the problem's logic combined with the stepwise nature of our solution makes it easy to understand and implement in a variety of languages. We hope that this problem and its solution helps sharpen your problem-solving and programming skills.

Remember, the best way to master coding problems is by practice and internalizing common patterns like greedy algorithms, dynamic programming, and more.

Happy coding!


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