LeetCode #3501 — HARD

Maximize Active Section with Trade II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
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.

Additionally, you are given a 2D array queries, where queries[i] = [li, ri] represents a substring s[li...ri].

For each query, determine the maximum possible number of active sections in s after making the optimal trade on the substring s[li...ri].

Return an array answer, where answer[i] is the result for queries[i].

Note

  • For each query, treat s[li...ri] as if it is augmented with a '1' at both ends, forming t = '1' + s[li...ri] + '1'. The augmented '1's do not contribute to the final count.
  • The queries are independent of each other.

Example 1:

Input: s = "01", queries = [[0,1]]

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", queries = [[0,3],[0,2],[1,3],[2,3]]

Output: [4,3,1,1]

Explanation:

  • Query [0, 3] → Substring "0100" → Augmented to "101001"
    Choose "0100", convert "0100" → "0000" → "1111".
    The final string without augmentation is "1111". The maximum number of active sections is 4.

  • Query [0, 2] → Substring "010" → Augmented to "10101"
    Choose "010", convert "010" → "000" → "111".
    The final string without augmentation is "1110". The maximum number of active sections is 3.

  • Query [1, 3] → Substring "100" → Augmented to "11001"
    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.

  • Query [2, 3] → Substring "00" → Augmented to "1001"
    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 3:

Input: s = "1000100", queries = [[1,5],[0,6],[0,4]]

Output: [6,7,2]

Explanation:

  • Query [1, 5] → Substring "00010" → Augmented to "1000101"
    Choose "00010", convert "00010" → "00000" → "11111".
    The final string without augmentation is "1111110". The maximum number of active sections is 6.

  • Query [0, 6] → Substring "1000100" → Augmented to "110001001"
    Choose "000100", convert "000100" → "000000" → "111111".
    The final string without augmentation is "1111111". The maximum number of active sections is 7.

  • Query [0, 4] → Substring "10001" → Augmented to "1100011"
    Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 2.

Example 4:

Input: s = "01010", queries = [[0,3],[1,4],[1,3]]

Output: [4,4,2]

Explanation:

  • Query [0, 3] → Substring "0101" → Augmented to "101011"
    Choose "010", convert "010" → "000" → "111".
    The final string without augmentation is "11110". The maximum number of active sections is 4.

  • Query [1, 4] → Substring "1010" → Augmented to "110101"
    Choose "010", convert "010" → "000" → "111".
    The final string without augmentation is "01111". The maximum number of active sections is 4.

  • Query [1, 3] → Substring "101" → Augmented to "11011"
    Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 2.

Constraints:

  • 1 <= n == s.length <= 105
  • 1 <= queries.length <= 105
  • s[i] is either '0' or '1'.
  • queries[i] = [li, ri]
  • 0 <= li <= ri < n
Patterns Used
🔍Modified Binary Search→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
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. Additionally, you are given a 2D array queries, where queries[i] = [li, ri] represents a substring s[li...ri]. For each query, determine the maximum possible number of active sections in s after making the optimal trade on the substring s[li...ri]. Return an array answer, where answer[i] is the result for queries[i]. Note For each query, treat s[li...ri] as if it is augmented with a '1' at both ends, forming t = '1' + s[li...ri] + '1'. The augmented '1's do not contribute to the final count. The queries are

Baseline thinking

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

Pattern signal: Array · Binary Search · Segment Tree

Example 1

"01"
[[0,1]]

Example 2

"0100"
[[0,3],[0,2],[1,3],[2,3]]

Example 3

"1000100"
[[1,5],[0,6],[0,4]]
Step 02

Core Insight

What unlocks the optimal approach

  • Split consecutive zeros and ones into segments and give each segment an ID.
  • The answer should be the maximum of <code>ans[i] = len[i - 1] + len[i + 1]</code>, where <code>i</code> is a one-segment.
  • For a zero-segment, define <code>ans[i] = 0</code>.
  • Note that all three segments (<code>i - 1</code>, <code>i</code>, and <code>i + 1</code>) should be fully covered by the substring.
  • Use a segment tree to perform range maximum queries on the answer. The query to the segment tree is not straightforward since we need to ensure the zero-segments are fully covered. Handle the first and last segments separately.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #3501: Maximize Active Section with Trade II
class Group {
  public int start;
  public int length;
  public Group(int start, int length) {
    this.start = start;
    this.length = length;
  }
}

class SparseTable {
  public SparseTable(int[] nums) {
    n = nums.length;
    st = new int[bitLength(n) + 1][n + 1];
    System.arraycopy(nums, 0, st[0], 0, n);
    for (int i = 1; i <= st.length; ++i)
      for (int j = 0; j + (1 << i) <= n; ++j)
        st[i][j] = Math.max(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
  }

  // Returns max(nums[l..r])
  public int query(int l, int r) {
    final int i = bitLength(r - l + 1) - 1;
    return Math.max(st[i][l], st[i][r - (1 << i) + 1]);
  }

  private final int n;
  private final int[][] st; // st[i][j] := max(nums[j..j + 2^i - 1])

  private int bitLength(int n) {
    return Integer.SIZE - Integer.numberOfLeadingZeros(n);
  }
}

class Solution {
  public List<Integer> maxActiveSectionsAfterTrade(String s, int[][] queries) {
    final int n = s.length();
    final int ones = (int) s.chars().filter(c -> c == '1').count();
    final Pair<List<Group>, int[]> zeroGroupsInfo = getZeroGroups(s);
    final List<Group> zeroGroups = zeroGroupsInfo.getKey();
    final int[] zeroGroupIndex = zeroGroupsInfo.getValue();

    if (zeroGroups.isEmpty())
      return Collections.nCopies(queries.length, ones);

    final SparseTable st = new SparseTable(getZeroMergeLengths(zeroGroups));
    final List<Integer> ans = new ArrayList<>();

    for (int[] query : queries) {
      final int l = query[0];
      final int r = query[1];
      final int left = zeroGroupIndex[l] == -1 ? -1
                                               : (zeroGroups.get(zeroGroupIndex[l]).length -
                                                  (l - zeroGroups.get(zeroGroupIndex[l]).start));
      final int right =
          zeroGroupIndex[r] == -1 ? -1 : (r - zeroGroups.get(zeroGroupIndex[r]).start + 1);
      final Pair<Integer, Integer> adjacentIndices = mapToAdjacentGroupIndices(
          zeroGroupIndex[l] + 1, s.charAt(r) == '1' ? zeroGroupIndex[r] : zeroGroupIndex[r] - 1);
      final int startAdjacentGroupIndex = adjacentIndices.getKey();
      final int endAdjacentGroupIndex = adjacentIndices.getValue();

      int activeSections = ones;
      if (s.charAt(l) == '0' && s.charAt(r) == '0' && zeroGroupIndex[l] + 1 == zeroGroupIndex[r])
        activeSections = Math.max(activeSections, ones + left + right);
      else if (startAdjacentGroupIndex <= endAdjacentGroupIndex)
        activeSections = Math.max(activeSections,
                                  ones + st.query(startAdjacentGroupIndex, endAdjacentGroupIndex));
      if (s.charAt(l) == '0' &&
          zeroGroupIndex[l] + 1 <= (s.charAt(r) == '1' ? zeroGroupIndex[r] : zeroGroupIndex[r] - 1))
        activeSections =
            Math.max(activeSections, ones + left + zeroGroups.get(zeroGroupIndex[l] + 1).length);
      if (s.charAt(r) == '0' && zeroGroupIndex[l] < zeroGroupIndex[r] - 1)
        activeSections =
            Math.max(activeSections, ones + right + zeroGroups.get(zeroGroupIndex[r] - 1).length);
      ans.add(activeSections);
    }

    return ans;
  }

  // Returns the zero groups and the index of the zero group that contains the i-th character
  private Pair<List<Group>, int[]> getZeroGroups(String s) {
    final List<Group> zeroGroups = new ArrayList<>();
    final int[] zeroGroupIndex = new int[s.length()];

    for (int i = 0; i < s.length(); i++) {
      if (s.charAt(i) == '0') {
        if (i > 0 && s.charAt(i - 1) == '0')
          zeroGroups.get(zeroGroups.size() - 1).length++;
        else
          zeroGroups.add(new Group(i, 1));
      }
      zeroGroupIndex[i] = zeroGroups.size() - 1;
    }

    return new Pair<>(zeroGroups, zeroGroupIndex);
  }

  // Returns the sums of the lengths of the adjacent groups
  private int[] getZeroMergeLengths(List<Group> zeroGroups) {
    final int[] zeroMergeLengths = new int[zeroGroups.size() - 1];
    for (int i = 0; i < zeroGroups.size() - 1; ++i)
      zeroMergeLengths[i] = zeroGroups.get(i).length + zeroGroups.get(i + 1).length;
    return zeroMergeLengths;
  }

  // Returns the indices of the adjacent groups that contain l and r completely
  private Pair<Integer, Integer> mapToAdjacentGroupIndices(int startGroupIndex, int endGroupIndex) {
    return new Pair<>(startGroupIndex, endGroupIndex - 1);
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
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