How do you solve Maximum Score After Binary Swaps on LeetCode?

Maximum Score After Binary Swaps (LeetCode #3781) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Swapping '1's left maximizes score.

What is the time complexity of Maximum Score After Binary Swaps?

The optimal solution for Maximum Score After Binary Swaps runs in O(n log n) time and uses O(n) space. Using a priority queue allows us to efficiently manage the values we can swap, leading to a linearithmic time complexity.

What is the brute force solution for Maximum Score After Binary Swaps?

The brute force approach for Maximum Score After Binary Swaps is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible swaps to see which combination gives the highest score. This involves checking every possible pair of indices where a swap can occur. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Score After Binary Swaps?

Maximum Score After Binary Swaps uses the following concepts: Array, String, Greedy, Heap (Priority Queue). The recommended patterns to study are: Greedy, Heap.

What are the interview tips for Maximum Score After Binary Swaps?

Explain your thought process clearly. Use examples to illustrate your approach. Practice coding under time constraints.

What are common mistakes when solving Maximum Score After Binary Swaps?

Not considering all possible swaps. Forgetting to check edge cases with no '1's.

#3781

Maximum Score After Binary Swaps

Medium

ArrayStringGreedyHeap (Priority Queue)GreedyHeap

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

We can maximize the score by moving '1's leftwards to cover the highest values in nums. This can be done in a single pass using a greedy approach.

⚙️

Algorithm

3 steps

1Step 1: Initialize score and a priority queue to store values of nums at '1' positions.
2Step 2: Traverse the string s, adding nums[i] to the score for '1's and pushing nums[i] to the queue for '0's before '1's.
3Step 3: Pop from the queue to maximize the score as we encounter '1's.

solution.py13 lines

1import heapq
2
3def maxScore(nums, s):
4    score = 0
5    max_heap = []
6    for i in range(len(s)):
7        if s[i] == '1':
8            score += nums[i]
9            if max_heap:
10                score += -heapq.heappop(max_heap)
11        else:
12            heapq.heappush(max_heap, -nums[i])
13    return score

ℹ

Complexity note: Using a priority queue allows us to efficiently manage the values we can swap, leading to a linearithmic time complexity.

1Swapping '1's left maximizes score.
2Using a priority queue helps efficiently manage potential swaps.

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