LeetCode #1157 — HARD

Online Majority Element In Subarray

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

Solve on LeetCode
The Problem

Problem Statement

Design a data structure that efficiently finds the majority element of a given subarray.

The majority element of a subarray is an element that occurs threshold times or more in the subarray.

Implementing the MajorityChecker class:

  • MajorityChecker(int[] arr) Initializes the instance of the class with the given array arr.
  • int query(int left, int right, int threshold) returns the element in the subarray arr[left...right] that occurs at least threshold times, or -1 if no such element exists.

Example 1:

Input
["MajorityChecker", "query", "query", "query"]
[[[1, 1, 2, 2, 1, 1]], [0, 5, 4], [0, 3, 3], [2, 3, 2]]
Output
[null, 1, -1, 2]

Explanation
MajorityChecker majorityChecker = new MajorityChecker([1, 1, 2, 2, 1, 1]);
majorityChecker.query(0, 5, 4); // return 1
majorityChecker.query(0, 3, 3); // return -1
majorityChecker.query(2, 3, 2); // return 2

Constraints:

  • 1 <= arr.length <= 2 * 104
  • 1 <= arr[i] <= 2 * 104
  • 0 <= left <= right < arr.length
  • threshold <= right - left + 1
  • 2 * threshold > right - left + 1
  • At most 104 calls will be made to query.
Patterns Used
🔍Modified Binary Search→🏗Design Patterns→📐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: Design a data structure that efficiently finds the majority element of a given subarray. The majority element of a subarray is an element that occurs threshold times or more in the subarray. Implementing the MajorityChecker class: MajorityChecker(int[] arr) Initializes the instance of the class with the given array arr. int query(int left, int right, int threshold) returns the element in the subarray arr[left...right] that occurs at least threshold times, or -1 if no such element exists.

Baseline thinking

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

Pattern signal: Array · Binary Search · Design · Segment Tree

Example 1

["MajorityChecker","query","query","query"]
[[[1,1,2,2,1,1]],[0,5,4],[0,3,3],[2,3,2]]
Step 02

Core Insight

What unlocks the optimal approach

  • What's special about a majority element ?
  • A majority element appears more than half the length of the array number of times.
  • If we tried a random index of the array, what's the probability that this index has a majority element ?
  • It's more than 50% if that array has a majority element.
  • Try a random index for a proper number of times so that the probability of not finding the answer tends to zero.
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 #1157: Online Majority Element In Subarray
class Node {
    int l, r;
    int x, cnt;
}

class SegmentTree {
    private Node[] tr;
    private int[] nums;

    public SegmentTree(int[] nums) {
        int n = nums.length;
        this.nums = nums;
        tr = new Node[n << 2];
        for (int i = 0; i < tr.length; ++i) {
            tr[i] = new Node();
        }
        build(1, 1, n);
    }

    private void build(int u, int l, int r) {
        tr[u].l = l;
        tr[u].r = r;
        if (l == r) {
            tr[u].x = nums[l - 1];
            tr[u].cnt = 1;
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

    public int[] query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return new int[] {tr[u].x, tr[u].cnt};
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (r <= mid) {
            return query(u << 1, l, r);
        }
        if (l > mid) {
            return query(u << 1 | 1, l, r);
        }
        var left = query(u << 1, l, r);
        var right = query(u << 1 | 1, l, r);
        if (left[0] == right[0]) {
            left[1] += right[1];
        } else if (left[1] >= right[1]) {
            left[1] -= right[1];
        } else {
            right[1] -= left[1];
            left = right;
        }
        return left;
    }

    private void pushup(int u) {
        if (tr[u << 1].x == tr[u << 1 | 1].x) {
            tr[u].x = tr[u << 1].x;
            tr[u].cnt = tr[u << 1].cnt + tr[u << 1 | 1].cnt;
        } else if (tr[u << 1].cnt >= tr[u << 1 | 1].cnt) {
            tr[u].x = tr[u << 1].x;
            tr[u].cnt = tr[u << 1].cnt - tr[u << 1 | 1].cnt;
        } else {
            tr[u].x = tr[u << 1 | 1].x;
            tr[u].cnt = tr[u << 1 | 1].cnt - tr[u << 1].cnt;
        }
    }
}

class MajorityChecker {
    private SegmentTree tree;
    private Map<Integer, List<Integer>> d = new HashMap<>();

    public MajorityChecker(int[] arr) {
        tree = new SegmentTree(arr);
        for (int i = 0; i < arr.length; ++i) {
            d.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
        }
    }

    public int query(int left, int right, int threshold) {
        int x = tree.query(1, left + 1, right + 1)[0];
        int l = search(d.get(x), left);
        int r = search(d.get(x), right + 1);
        return r - l >= threshold ? x : -1;
    }

    private int search(List<Integer> arr, int x) {
        int left = 0, right = arr.size();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr.get(mid) >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

/**
 * Your MajorityChecker object will be instantiated and called as such:
 * MajorityChecker obj = new MajorityChecker(arr);
 * int param_1 = obj.query(left,right,threshold);
 */
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.

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