LeetCode #3499 — MEDIUM

Maximize Active Section with Trade I

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

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The Problem

Problem Statement

You are given a binary string s of length n, where:

'1' represents an active section.
'0' represents an inactive section.

You can perform at most one trade to maximize the number of active sections in s. In a trade, you:

Convert a contiguous block of '1's that is surrounded by '0's to all '0's.
Afterward, convert a contiguous block of '0's that is surrounded by '1's to all '1's.

Return the maximum number of active sections in s after making the optimal trade.

Note: Treat s as if it is augmented with a '1' at both ends, forming t = '1' + s + '1'. The augmented '1's do not contribute to the final count.

Example 1:

Input: s = "01"

Output: 1

Explanation:

Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 1.

Example 2:

Input: s = "0100"

Output: 4

Explanation:

String "0100" → Augmented to "101001".
Choose "0100", convert "101001" → "100001" → "111111".
The final string without augmentation is "1111". The maximum number of active sections is 4.

Example 3:

Input: s = "1000100"

Output: 7

Explanation:

String "1000100" → Augmented to "110001001".
Choose "000100", convert "110001001" → "110000001" → "111111111".
The final string without augmentation is "1111111". The maximum number of active sections is 7.

Example 4:

Input: s = "01010"

Output: 4

Explanation:

String "01010" → Augmented to "1010101".
Choose "010", convert "1010101" → "1000101" → "1111101".
The final string without augmentation is "11110". The maximum number of active sections is 4.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a binary string s of length n, where: '1' represents an active section. '0' represents an inactive section. You can perform at most one trade to maximize the number of active sections in s. In a trade, you: Convert a contiguous block of '1's that is surrounded by '0's to all '0's. Afterward, convert a contiguous block of '0's that is surrounded by '1's to all '1's. Return the maximum number of active sections in s after making the optimal trade. Note: Treat s as if it is augmented with a '1' at both ends, forming t = '1' + s + '1'. The augmented '1's do not contribute to the final count.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"01"

Example 2

"0100"

Example 3

"1000100"

Step 02

Core Insight

What unlocks the optimal approach

Split the string into several zero-one segments.
For each one-segment, if it has two neighbors (i.e., it is surrounded by two zero-segments), the total sum of their lengths is one of the candidates for <code>delta</code>.
Find the maximum <code>delta</code> and add it to the total number of ones in the string.

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 #3499: Maximize Active Section with Trade I
class Solution {
    public int maxActiveSectionsAfterTrade(String s) {
        int n = s.length();
        int ans = 0, i = 0;
        int pre = Integer.MIN_VALUE, mx = 0;

        while (i < n) {
            int j = i + 1;
            while (j < n && s.charAt(j) == s.charAt(i)) {
                j++;
            }
            int cur = j - i;
            if (s.charAt(i) == '1') {
                ans += cur;
            } else {
                mx = Math.max(mx, pre + cur);
                pre = cur;
            }
            i = j;
        }

        ans += mx;
        return ans;
    }
}

// Accepted solution for LeetCode #3499: Maximize Active Section with Trade I
func maxActiveSectionsAfterTrade(s string) (ans int) {
	n := len(s)
	pre, mx := math.MinInt, 0

	for i := 0; i < n; {
		j := i + 1
		for j < n && s[j] == s[i] {
			j++
		}
		cur := j - i
		if s[i] == '1' {
			ans += cur
		} else {
			mx = max(mx, pre+cur)
			pre = cur
		}
		i = j
	}

	ans += mx
	return
}

# Accepted solution for LeetCode #3499: Maximize Active Section with Trade I
class Solution:
    def maxActiveSectionsAfterTrade(self, s: str) -> int:
        n = len(s)
        ans = i = 0
        pre, mx = -inf, 0
        while i < n:
            j = i + 1
            while j < n and s[j] == s[i]:
                j += 1
            cur = j - i
            if s[i] == "1":
                ans += cur
            else:
                mx = max(mx, pre + cur)
                pre = cur
            i = j
        ans += mx
        return ans

// Accepted solution for LeetCode #3499: Maximize Active Section with Trade I
// 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 #3499: Maximize Active Section with Trade I
// class Solution {
//     public int maxActiveSectionsAfterTrade(String s) {
//         int n = s.length();
//         int ans = 0, i = 0;
//         int pre = Integer.MIN_VALUE, mx = 0;
// 
//         while (i < n) {
//             int j = i + 1;
//             while (j < n && s.charAt(j) == s.charAt(i)) {
//                 j++;
//             }
//             int cur = j - i;
//             if (s.charAt(i) == '1') {
//                 ans += cur;
//             } else {
//                 mx = Math.max(mx, pre + cur);
//                 pre = cur;
//             }
//             i = j;
//         }
// 
//         ans += mx;
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3499: Maximize Active Section with Trade I
function maxActiveSectionsAfterTrade(s: string): number {
    let n = s.length;
    let [ans, mx] = [0, 0];
    let pre = Number.MIN_SAFE_INTEGER;

    for (let i = 0; i < n; ) {
        let j = i + 1;
        while (j < n && s[j] === s[i]) {
            j++;
        }
        let cur = j - i;
        if (s[i] === '1') {
            ans += cur;
        } else {
            mx = Math.max(mx, pre + cur);
            pre = cur;
        }
        i = j;
    }

    ans += mx;
    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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.