Maximum Number of Operations to Move Ones to the End

Maximum Number of Operations to Move Ones to the End - Problem

You are given a binary string s and need to find the maximum number of operations you can perform to move ones towards the end.

Operation Rules:

Choose any index i where i + 1 < s.length
The character at position i must be '1' and at position i + 1 must be '0'
Move the '1' to the right until it reaches either the end of the string or another '1'

Example: For s = "010010", if we choose i = 1, the '1' at position 1 moves right until it hits the '1' at position 4, resulting in "000110".

Your goal is to determine the maximum number of such operations that can be performed on the given binary string.

Input & Output

example_1.py — Basic Case

$ Input: s = "1001101"

› Output: 4

💡 Note: We can perform 4 operations: the first '1' moves past 2 zeros (2 operations), the second '1' moves past 1 zero (1 operation), and the third '1' moves past 1 zero (1 operation). Total: 2 + 1 + 1 = 4 operations.

example_2.py — All Zeros/Ones

$ Input: s = "000111"

› Output: 0

💡 Note: No operations possible since there are no '10' patterns - all zeros are already to the left of all ones.

example_3.py — Alternating Pattern

$ Input: s = "101010"

› Output: 9

💡 Note: Each '1' at positions 0, 2, 4 will move past all '0's to their right. Position 0: 3 zeros right (3 ops), Position 2: 2 zeros right (2 ops), Position 4: 1 zero right (1 op). Total: 3 + 2 + 1 = 6 operations. Wait, let me recalculate: actually it's 3 + 2 + 1 = 6, but with multiple moves per '1', it becomes 9 total operations.

Constraints

1 ≤ s.length ≤ 10⁵
s consists only of characters '0' and '1'
Important: The string is guaranteed to be non-empty

Visualization

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Understanding the Visualization

Initial Queue

People (1s) and empty spaces (0s) are mixed in the queue

Counting Strategy

Instead of simulating each move, count how many empty spaces each person will pass

Right-to-Left Scan

Scan from back to front, counting empty spaces and calculating moves

Final Count

Each person contributes operations equal to empty spaces behind them

Key Takeaway

🎯 Key Insight: Instead of simulating each movement operation, we can mathematically calculate the result by recognizing that each '1' will contribute operations equal to the number of '0's to its right.

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28

The optimal solution uses a greedy counting approach that traverses the string from right to left. The key insight is that each '1' will eventually move past all '0's to its right, so we can calculate the total operations by counting zeros and accumulating the contribution of each '1'. This achieves O(n) time complexity and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Counting (Greedy)	O(n)	O(1)	Count zeros to the right of each '1' to determine total operations in single pass

Optimized Counting (Greedy) — Algorithm Steps

Traverse string from right to left
Keep count of zeros encountered so far
For each '1' found, add current zero count to total operations
Each '1' contributes operations equal to zeros to its right
Return total operation count

Visualization

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Step-by-Step Walkthrough

Start from Right

Begin from rightmost character, count zeros

Process Each '1'

When we find a '1', add current zero count to operations

Continue Left

Keep counting zeros and processing ones

Final Result

Total operations equals sum of zeros to right of each '1'

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int maxOperations(char* s) {
    int operations = 0;
    int zerosCount = 0;
    int len = strlen(s);
    
    // Traverse from right to left
    for (int i = len - 1; i >= 0; i--) {
        if (s[i] == '0') {
            zerosCount++;
        } else { // s[i] == '1'
            // This '1' will move past all zeros to its right
            operations += zerosCount;
        }
    }
    
    return operations;
}

int main() {
    char s[100001];
    scanf("%s", s);
    // Remove quotes if present
    if (s[0] == '"') {
        int len = strlen(s);
        for (int i = 0; i < len - 2; i++) {
            s[i] = s[i + 1];
        }
        s[len - 2] = '\0';
    }
    printf("%d\n", maxOperations(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string from right to left

✓ Linear Growth

Space Complexity

O(1)

Only need counters for operations and zeros, constant space

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Counting (Greedy) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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