LeetCode 3228: Maximum Number of Operations to Move Ones to the End Solution

Master LeetCode problem 3228 (Maximum Number of Operations to Move Ones to the End), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3228. Maximum Number of Operations to Move Ones to the End

Medium

String Greedy Counting

Problem Explanation

Explanation

To solve this problem, we can iterate through the given binary string s and count the number of operations we can perform. Whenever we encounter a situation where s[i] == '1' and s[i + 1] == '0', we increment the count of operations. The maximum number of operations we can perform is the number of such situations we encounter.

Algorithm

Initialize a variable operations to 0.
Iterate through the binary string s.
If s[i] == '1' and s[i + 1] == '0', increment operations.
Return the final value of operations.

Time Complexity

The time complexity of this algorithm is O(n), where n is the length of the input binary string s.

Space Complexity

The space complexity is O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int maxOperations(String s) {
        int operations = 0;
        for (int i = 0; i < s.length() - 1; i++) {
            if (s.charAt(i) == '1' && s.charAt(i + 1) == '0') {
                operations++;
            }
        }
        return operations;
    }
}

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