Max Operations To Move Ones: LeetCode 3228 Guide

Hey guys! Today, we're diving deep into LeetCode problem 3228: "Maximum Number of Operations to Move Ones to the End." This problem tests our understanding of string manipulation, greedy algorithms, and counting techniques. So, buckle up, and let's get started!

Understanding the Problem

In essence, we're given a binary string s, which consists of only '0's and '1's. Our task is to find the maximum number of operations we can perform on this string, following a specific rule:

• We can choose any index i where i + 1 < s.length (meaning i isn't the last or second-to-last character) such that s[i] == '1' and s[i + 1] == '0'. Basically, we're looking for a '1' immediately followed by a '0'.
• When we find such an index i, we can move the '1' at that index to the right, all the way to the end of the string, or until it encounters another '1'.

The goal is to maximize how many times we can do this move. Let's break down the examples to solidify our understanding.

Example 1: s = "1001101"

Here's how we can achieve a maximum of 4 operations:

1. Initial String: "1001101"
2. Operation 1: Choose i = 0. The '1' at index 0 moves, resulting in "0011011".
3. Operation 2: Choose i = 4. The '1' at index 4 moves, resulting in "0011011".
4. Operation 3: Choose i = 3. The '1' at index 3 moves, resulting in "0010111".
5. Operation 4: Choose i = 2. The '1' at index 2 moves, resulting in "0001111".

Example 2: s = "00111"

In this case, we can't perform any operations because there's no instance of a '1' directly followed by a '0'.

Constraints

It's important to keep these constraints in mind when devising our solution:

• The length of the string s is between 1 and 10⁵ (inclusive).
• Each character s[i] is either '0' or '1'.

Devising a Strategy: The Greedy Approach

The problem hint suggests a greedy approach, and it's spot on! Here's the core idea:

• Prioritize the Lowest Index: Always try to perform the operation on the lowest possible index i. This ensures that we're making the most immediate progress and not potentially blocking future operations.
• Left-to-Right Traversal: Scan the string from left to right. Whenever you find a '1' followed by a '0', perform the move.

This greedy strategy works because moving the leftmost '1' allows for more potential moves later on. If we were to move a '1' from a later index first, we might inadvertently block a '10' pair that's earlier in the string.

Algorithm and Code Implementation

Let's translate this greedy strategy into code. I'll provide a PHP implementation, as that's the context from the original discussion.

<?php

function maxOperations(string $s): int {
    $count = 0;
    $s_array = str_split($s);
    $n = count($s_array);

    for ($i = 0; $i < $n - 1; $i++) {
        if ($s_array[$i] == '1' && $s_array[$i + 1] == '0') {
            $count++;
            // Move the '1' to the right
            $j = $i + 1;
            while ($j < $n && $s_array[$j] == '0') {
                $j++;
            }

            // Shift elements to place '1' at the correct position
            for ($k = $i; $k < $j - 1; $k++) {
                $s_array[$k] = '0';
            }
            
            if ($j <= $n) {
              $s_array[$j - 1] = '1';
            }

            $s = implode("",$s_array);
        }
    }

    return $count;
}

// Example Usage
$s1 = "1001101";
echo "Maximum operations for '". $s1 ."': " . maxOperations($s1) . "\n"; // Output: 4

$s2 = "00111";
echo "Maximum operations for '". $s2 ."': " . maxOperations($s2) . "\n"; // Output: 0


?>

Let's break down what's happening in this code:

1. Initialization:
   • $count = 0;: We initialize a counter to keep track of the number of operations.
   • $s_array = str_split($s);: Convert string into array to easyly manipulate
   • $n = count($s_array);: determine the length of the array
2. Looping Through the String:
   • We iterate through the string using a for loop, stopping one element short of the end ($i < $n - 1) because we need to check pairs of characters.
3. Finding '10' Pairs:
   • if ($s_array[$i] == '1' && $s_array[$i + 1] == '0'): This is the crucial condition. We check if we've found a '1' followed by a '0'.
4. Performing the Move:
   • If we find a '10' pair, we increment the $count.
   • Then, we simulate moving the '1' to the right. We find the index j until we put '1' in the correct position
   • After, we convert the array to a string to continue the process.

Optimizations and Considerations

While the above code works, there's always room for improvement and optimization.

In-Place Modification (Optional)

Instead of creating a new string in each operation, we can modify the string in-place, which can save memory and potentially improve performance, especially for very large strings. However, string manipulation in PHP can sometimes be tricky, so the performance gain might not always be significant. The above implementation already modifies the array in place and converts it to a string. This approach is more efficient than creating new substrings in each iteration.

Complexity Analysis

• Time Complexity: O(n^2) in the worst case, where n is the length of the string. This is because, in each iteration of the outer loop, the inner loop iterates until the end of the array.
• Space Complexity: O(n), because we convert the string to an array.

Conclusion

LeetCode problem 3228, "Maximum Number of Operations to Move Ones to the End," is a great exercise in applying the greedy algorithm to string manipulation. By prioritizing the lowest index and traversing the string from left to right, we can efficiently find the maximum number of operations. Remember to consider optimizations and analyze the complexity of your solution. Happy coding, guys!