LeetCode Challenge Day 59 — 3228. Maximum Number of Operations to Move Ones to the End

Hey folks 👋

This is Day 59 of my LeetCode streak 🚀
Today’s problem is 3228. Maximum Number of Operations to Move Ones to the End — a medium-level greedy + counting problem based on understanding patterns inside a binary string.

Instead of simulating each operation, the trick is to observe how zero-block boundaries control the number of possible moves.

📌 Problem Statement

You are given a binary string s.
You may perform the following operation any number of times:

• Choose an index i such that s[i] == '1' and s[i + 1] == '0'
• Move that '1' to the right until it reaches:
  • the end of the string, or
  • another '1'

Return the maximum number of operations possible.

Example 1:

Input: "1001101"  
Output: 4

Example 2:

Input: "00111"  
Output: 0

💡 Intuition

Whenever a '1' is followed by a '0', it can produce one operation.
But instead of checking each pair, we notice:

👉 Only the last zero of each consecutive zero-block contributes to the final count.
Each such zero-block end allows all previously seen '1's to “jump” over it.

So we scan the string once and count operations based on block structure.

🔑 Approach

1. Traverse the string from left to right.
2. Maintain ones → number of '1's seen so far.
3. For each '0':
   • Check if it is the end of a zero-block
     (i.e., next char is '1' or we are at the last index).
   • If yes → add ones to the answer.
4. Return the final count.

This avoids any simulation and leverages pure counting.

⏱️ Complexity Analysis

• Time Complexity:
  ( O(n) )

• Space Complexity:
  ( O(1) )

🧑‍💻 Code (JavaScript)

/**
 * @param {string} s
 * @return {number}
 */
var maxOperations = function(s) {
    let ans = 0;
    let ones = 0;
    const n = s.length;

    for (let i = 0; i < n; ++i) {
        if (s[i] === '1') {
            ones += 1;
        } else { // s[i] === '0'
            // If this zero is the last in its zero-block
            if (i === n - 1 || s[i + 1] === '1') {
                ans += ones;
            }
        }
    }

    return ans;
};

🎯 Example Walkthrough

Input:
"1001101"

Zero-block ends at positions:

• index 2 → adds 1

• index 4 → adds 2

• index 5 → adds 4

Total = 4 operations

🎯 Reflection

This problem highlights how recognizing patterns and block structures in strings can eliminate the need for simulation and drastically simplify logic.

See you tomorrow for Day 60 — the grind continues! 💪
Happy Coding 👨‍💻✨