LeetCode #3781 — MEDIUM

Maximum Score After Binary Swaps

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums of length n and a binary string s of the same length.

Initially, your score is 0. Each index i where s[i] = '1' contributes nums[i] to the score.

You may perform any number of operations (including zero). In one operation, you may choose an index i such that 0 <= i < n - 1, where s[i] = '0', and s[i + 1] = '1', and swap these two characters.

Return an integer denoting the maximum possible score you can achieve.

Example 1:

Input: nums = [2,1,5,2,3], s = "01010"

Output: 7

Explanation:

We can perform the following swaps:

Swap at index i = 0: "01010" changes to "10010"
Swap at index i = 2: "10010" changes to "10100"

Positions 0 and 2 contain '1', contributing nums[0] + nums[2] = 2 + 5 = 7. This is maximum score achievable.

Example 2:

Input: nums = [4,7,2,9], s = "0000"

Output: 0

Explanation:

There are no '1' characters in s, so no swaps can be performed. The score remains 0.

Constraints:

n == nums.length == s.length
1 <= n <= 10⁵
1 <= nums[i] <= 10⁹
s[i] is either '0' or '1'

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of length n and a binary string s of the same length. Initially, your score is 0. Each index i where s[i] = '1' contributes nums[i] to the score. You may perform any number of operations (including zero). In one operation, you may choose an index i such that 0 <= i < n - 1, where s[i] = '0', and s[i + 1] = '1', and swap these two characters. Return an integer denoting the maximum possible score you can achieve.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[2,1,5,2,3]
"01010"

Example 2

[4,7,2,9]
"0000"

Step 02

Core Insight

What unlocks the optimal approach

The problem can be solved greedily
The operation only allows <code>'1'</code>s to move backward
Going from left to right, maintain a priority queue (max heap)
When at a <code>'1'</code>, pop the top of the priority queue if it is not empty

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3781: Maximum Score After Binary Swaps
class Solution {
    public long maximumScore(int[] nums, String s) {
        long ans = 0;
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        for (int i = 0; i < nums.length; i++) {
            int x = nums[i];
            char c = s.charAt(i);
            pq.offer(x);
            if (c == '1') {
                ans += pq.poll();
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3781: Maximum Score After Binary Swaps
func maximumScore(nums []int, s string) int64 {
	var ans int64
	pq := &hp{}
	heap.Init(pq)
	for i, x := range nums {
		pq.push(x)
		if s[i] == '1' {
			ans += int64(pq.pop())
		}
	}
	return ans
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

# Accepted solution for LeetCode #3781: Maximum Score After Binary Swaps
class Solution:
    def maximumScore(self, nums: List[int], s: str) -> int:
        ans = 0
        pq = []
        for x, c in zip(nums, s):
            heappush(pq, -x)
            if c == "1":
                ans -= heappop(pq)
        return ans

// Accepted solution for LeetCode #3781: Maximum Score After Binary Swaps
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3781: Maximum Score After Binary Swaps
// class Solution {
//     public long maximumScore(int[] nums, String s) {
//         long ans = 0;
//         PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
//         for (int i = 0; i < nums.length; i++) {
//             int x = nums[i];
//             char c = s.charAt(i);
//             pq.offer(x);
//             if (c == '1') {
//                 ans += pq.poll();
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3781: Maximum Score After Binary Swaps
function maximumScore(nums: number[], s: string): number {
    let ans = 0;
    const pq = new MaxPriorityQueue<number>();

    for (let i = 0; i < nums.length; i++) {
        const x = nums[i];
        const c = s[i];
        pq.enqueue(x);
        if (c === '1') {
            ans += pq.dequeue()!;
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.