2380. Time Needed to Rearrange a Binary String

MediumStringDynamic ProgrammingSimulation
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Problem Description

The problem gives us a binary string s, which consists of '0's and '1's. We are then introduced to a procedure that takes place every second where every occurrence of the substring "01" is simultaneously replaced by "10". This process continues to happen every second until there are no more occurrences of "01" left in the string. Our task is to determine how many seconds it will take until the process can no longer continue, meaning the string has no occurrences of "01".

Intuition

To solve this problem, one could think about how the string evolves over time. Consider a "01" pattern. When you replace it with "10", the '0' effectively moves to the right. This process continues until all '0's have no '1's to their left. In other words, we want all the '0's in the string to be on the left side and all the '1's to be on the right side, and we count how many seconds it takes for this to happen.

To find the solution, we don't need to simulate the entire process. Instead, we keep track of how many '0's we have passed as we iterate through the string (cnt). When we encounter a '1', we know that if there is any '0' to the left of this '1' (which means cnt would be more than 0), we'll have a "01" pattern.

The key insight is that for each '1' found, if there was at least one '0' before it, we need at least one second for the leftmost '0' to move past this '1'. However, if we encounter another '1' while still having '0's in our count, we’ll need an additional second. Essentially, we continue incrementing our answer until there are no more '0's to move across '1's.

The variable ans keeps track of the maximum number of seconds needed so far. We either increment ans by 1 or update it to cnt if cnt is greater (which means we encountered a '1' with many '0's before it, increasing the required time). Once we’ve gone through the whole string, ans will contain the number of seconds needed to complete the process.

Learn more about Dynamic Programming patterns.

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Solution Approach

The solution uses a simple linear scan of the binary string, and it's based on the observation that every second, a '0' can move past a '1' if and only if there is a '0' to its immediate left.

Here's a detailed walk-through of the solution:

  1. Initialize ans to 0, which is used to store the maximum number of seconds required to get all '0's to the left side of all '1's in the string.

  2. Initialize cnt to 0, which is used to keep track of the number of '0's encountered while iterating through the string from left to right.

  3. Iterate through each character c in the binary string s:

    • If c is '0', this means we have one more '0' that might need to move rightward, so we increment cnt.
    • If c is '1' and cnt is greater than 0, this indicates there are '0's that need to move past this '1'. Therefore, we calculate the number of seconds required. This is the maximum of the current value of ans incremented by 1 (as we need at least one more second for a '0' to move past this '1') and cnt (which represents the bulk move of all the '0's we have encountered so far). We update ans to this calculated value.

  4. The loop continues until the end of the string. By the end of the loop, ans will hold the total number of seconds needed, as it captures the most time-consuming scenario of moving all '0's to the left of all '1's.

  5. Return ans as the result.

This approach doesn't require complex data structures or algorithms, simply iterating through the string and keeping track of two integer variables (ans and cnt). It efficiently arrives at the solution by focusing on when '1's are encountered and how many '0's are to their left—a fundamental aspect of how the process evolves over time. The pattern used here is essentially a greedy approach, optimizing for each '1' found in the string.

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Is the following code DFS or BFS?

1void search(Node root) {
2  if (!root) return;
3  visit(root);
4  root.visited = true;
5  for (Node node in root.adjacent) {
6    if (!node.visited) {
7      search(node);
8    }
9  }
10}

Example Walkthrough

Let's illustrate the solution approach with a small example. Consider the binary string s = "001011".

  1. Initialize ans to 0 and cnt to 0.

  2. Start iterating over the string from left to right:

    • First character is '0': increment cnt to 1.
    • Second character is '0': increment cnt to 2.
    • Third character is '1': cnt is greater than 0, so we have '0's that need to move past this '1'. We determine the temporary number of seconds which would be the maximum of ans+1 (0+1=1) and cnt (2). So we update ans to 2.
    • Fourth character is '0': increment cnt to 3.
    • Fifth character is '1': again cnt is greater than 0, indicating '0's need to move past this '1'. Calculate the maximum of ans+1 (2+1=3) and cnt (3). ans remains 3.
    • Sixth character is '1': cnt is still greater than 0, so we perform the same calculation: maximum of ans+1 (3+1=4) and cnt (3). ans is updated to 4.

  3. The iteration completes with a final ans of 4, which is the total number of seconds needed for the string to have no occurrences of "01".

Following the steps outlined in the solution approach, we've determined it will take 4 seconds for the string "001011" to become "000111", at which point the process cannot continue as there are no more occurrences of "01" left in the string.

Solution Implementation

1class Solution:
2    def secondsToRemoveOccurrences(self, s: str) -> int:
3        # Initialize a variable to keep track of the number of seconds.
4        seconds = 0
5        # Initialize a counter for the number of zeros seen so far.
6        zero_count = 0
7      
8        # Iterate over each character in the string.
9        for char in s:
10            if char == '0':
11                # Increment zero count on seeing a '0'.
12                zero_count += 1
13            elif zero_count:
14                # If a '1' is encountered and there was at least one '0' before it,
15                # increment seconds needed. It either takes one more second than the
16                # previous required seconds, or the number of observed zeros, whichever is larger.
17                # This is because you need at least as many seconds as the number of zeros
18                # you'll need to move past each '1' once.
19                seconds = max(seconds + 1, zero_count)
20      
21        # Return the total number of seconds needed to get all '1's to the right of all '0's.
22        return seconds
23
1class Solution {
2    public int secondsToRemoveOccurrences(String s) {
3        int secondsRequired = 0; // initializes the count of seconds required to remove all occurrences
4        int countZeros = 0; // initializes the count of '0's encountered that need to be moved
5
6        // Loop through each character in the string 's'
7        for (char character : s.toCharArray()) {
8          
9            if (character == '0') {
10                // Increment the count of '0's when a '0' is encountered
11                ++countZeros;
12            } else if (countZeros > 0) {
13                // If a '1' is encountered and there are '0's that need to be moved,
14                // we increment the seconds required. The logic behind this is that for
15                // each '1' we encounter after some '0's, we need at least one move to start
16                // moving all those '0's past this '1'. This essentially tracks the batches of movements required.
17              
18                // The max function ensures that if we have a consecutive batch of '0's followed by '1's larger
19                // than any previous batch, we use that larger value as the seconds required.
20                secondsRequired = Math.max(secondsRequired + 1, countZeros);
21            }
22        }
23      
24        // Return the number of seconds required to have no '01' occurrences in the string
25        return secondsRequired;
26    }
27}
28
1class Solution {
2public:
3    int secondsToRemoveOccurrences(string s) {
4        int maxSeconds = 0;  // This will keep track of the total seconds needed to remove all occurrences.
5        int zeroCount = 0;   // This will count the number of zeros we have seen so far.
6
7        // Iterate over each character in the string.
8        for (char currentChar : s) {
9            // If the current character is '0', increment the zeroCount.
10            if (currentChar == '0') {
11                ++zeroCount;
12            } 
13            // If it is '1' and we have seen at least one '0' before it,
14            // we need to perform a swap operation. This means that we need
15            // at least one second for each '1' after the first '0'.
16            else if (zeroCount) {
17                // The number of seconds required can be either one more than the number
18                // of seconds calculated so far or the number of zeros seen, whichever is larger.
19                // This accounts for the fact that we can move each '1' past all '0's seen so far
20                // by continuous swaps which take one second each.
21                maxSeconds = max(maxSeconds + 1, zeroCount);
22            }
23        }
24
25        // The value of maxSeconds is the total seconds required to remove all
26        // "01" occurrences in the string by swapping adjacent characters.
27        return maxSeconds;
28    }
29};
30
1// Define secondsToRemoveOccurrences as a global function that takes a string and returns a number.
2function secondsToRemoveOccurrences(s: string): number {
3    let maxSeconds = 0;  // This holds the total seconds needed to remove all "01" sequences.
4    let zeroCount = 0;   // This counts the number of '0's encountered up to the current position.
5
6    // Loop through each character in the string.
7    for (let i = 0; i < s.length; i++) {
8        // If the current character is '0', increment zeroCount.
9        if (s[i] === '0') {
10            zeroCount++;
11        } 
12        // If the current character is '1' and we have seen '0's before it.
13        else if (zeroCount > 0) {
14            // Increment maxSeconds or set it to zeroCount, whichever is larger.
15            // This handles the need to move '1' past all '0's seen so far, one second per swap.
16            maxSeconds = Math.max(maxSeconds + 1, zeroCount);
17        }
18    }
19
20    // Return the calculated maximum number of seconds to remove all "01" occurrences in the string.
21    return maxSeconds;
22}
23
24// Example Usage
25// const sequence: string = "001011";
26// console.log(secondsToRemoveOccurrences(sequence));  // Output will be the number of seconds required.
27
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What's the output of running the following function using the following tree as input? 

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

Time and Space Complexity

The given Python code defines a function secondsToRemoveOccurrences which takes a string s as its argument and calculates the number of seconds needed to remove all occurrences of the pattern "01" by repeatedly replacing it with "10" until no more occurrences are left.

Time Complexity

The time complexity of the code is O(n), where n is the length of the input string s. We can deduce this because the function contains a single loop that iterates through each character of the string exactly once. Inside the loop, it performs constant-time operations, including comparison, arithmetic operations, and assignment. There are no nested loops or recursive calls that would increase the complexity. Therefore, the time taken by the algorithm grows linearly with the size of the input string.

Space Complexity

The space complexity of the code is O(1). The function uses a fixed amount of extra space, regardless of the input size. The variables ans and cnt are used to keep track of the current state while iterating over the string, but no additional space that grows with the input size is allocated. The input string s is not modified, and no other data structures are used. Hence, the amount of memory used by the function remains constant, irrespective of the input size.

Learn more about how to find time and space complexity quickly using problem constraints.

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