What is the brute force solution for Maximum Number of Operations to Move Ones to the End?

The brute force approach for Maximum Number of Operations to Move Ones to the End is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of '1' and '0' in the string and moving the '1' to the right until it can no longer move. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Operations to Move Ones to the End?

Maximum Number of Operations to Move Ones to the End uses the following concepts: String, Greedy, Counting. The recommended patterns to study are: Greedy, Counting.

What are the interview tips for Maximum Number of Operations to Move Ones to the End?

Always think about how to reduce the number of operations needed — can you count instead of simulate? Explain your thought process clearly; it shows understanding even if you make a mistake. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Maximum Number of Operations to Move Ones to the End?

Not considering the order of operations, which can lead to fewer total moves. Overcomplicating the problem by trying to simulate every move instead of counting.

#3228

Maximum Number of Operations to Move Ones to the End

Medium

StringGreedyCountingGreedyCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach counts the number of '1's and '0's, then calculates how many operations can be performed based on their positions. This method is efficient and runs in linear time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for '1's and a counter for operations.
2Step 2: Traverse the string and for each '1' found, add the number of '0's that have been seen so far to the operations counter.
3Step 3: Return the operations counter as the result.

solution.py10 lines

1# Full working Python code
2s = '1001101'
3count_ones = 0
4operations = 0
5for char in s:
6    if char == '1':
7        operations += count_ones
8    else:
9        count_ones += 1
10print(operations)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, counting '1's and calculating operations in a single pass. The space complexity is O(1) since we use a fixed amount of extra space.

1Performing operations on the leftmost '1' is optimal as it maximizes the number of moves.
2Counting the number of '0's encountered allows us to calculate potential moves efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.