How hard is LeetCode 1157. Online Majority Element In Subarray?

LeetCode 1157. Online Majority Element In Subarray is rated Hard on LeetCode.

What topics does LeetCode 1157. Online Majority Element In Subarray cover?

LeetCode 1157. Online Majority Element In Subarray is tagged Design, Binary Indexed Tree, Segment Tree, Array and Binary Search on LeetCode.

Online Majority Element In Subarray — LeetCode 1157 Python Solution

HardDesignBinary Indexed TreeSegment TreeArrayBinary Search

Problem: #1157
Pattern: Monotonic Stack
Reading time: 12 min
Source: leetcode.com

The problem

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.

Example

Input: ["MajorityChecker", "query", "query", "query"]
Output: [null, 1, -1, 2]
Explanation: MajorityChecker majorityChecker = new MajorityChecker([1, 1, 2, 2, 1, 1]);

Python solution

Python

class Node:
    __slots__ = ("l", "r", "x", "cnt")

    def __init__(self):
        self.l = self.r = 0
        self.x = self.cnt = 0


class SegmentTree:
    def __init__(self, nums):
        self.nums = nums
        n = len(nums)
        self.tr = [Node() for _ in range(n << 2)]
        self.build(1, 1, n)

    def build(self, u, l, r):
        self.tr[u].l, self.tr[u].r = l, r
        if l == r:
            self.tr[u].x = self.nums[l - 1]
            self.tr[u].cnt = 1
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)
        self.pushup(u)

    def query(self, u, l, r):
        if self.tr[u].l >= l and self.tr[u].r <= r:
            return self.tr[u].x, self.tr[u].cnt
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        if r <= mid:
            return self.query(u << 1, l, r)
        if l > mid:
            return self.query(u << 1 | 1, l, r)
        x1, cnt1 = self.query(u << 1, l, r)
        x2, cnt2 = self.query(u << 1 | 1, l, r)
        if x1 == x2:
            return x1, cnt1 + cnt2
        if cnt1 >= cnt2:
            return x1, cnt1 - cnt2
        else:
            return x2, cnt2 - cnt1

    def pushup(self, u):
        if self.tr[u << 1].x == self.tr[u << 1 | 1].x:
            self.tr[u].x = self.tr[u << 1].x
            self.tr[u].cnt = self.tr[u << 1].cnt + self.tr[u << 1 | 1].cnt
        elif self.tr[u << 1].cnt >= self.tr[u << 1 | 1].cnt:
            self.tr[u].x = self.tr[u << 1].x
            self.tr[u].cnt = self.tr[u << 1].cnt - self.tr[u << 1 | 1].cnt
        else:
            self.tr[u].x = self.tr[u << 1 | 1].x
            self.tr[u].cnt = self.tr[u << 1 | 1].cnt - self.tr[u << 1].cnt


class MajorityChecker:
    def __init__(self, arr: List[int]):
        self.tree = SegmentTree(arr)
        self.d = defaultdict(list)
        for i, x in enumerate(arr):
            self.d[x].append(i)

    def query(self, left: int, right: int, threshold: int) -> int:
        x, _ = self.tree.query(1, left + 1, right + 1)
        l = bisect_left(self.d[x], left)
        r = bisect_left(self.d[x], right + 1)
        return x if r - l >= threshold else -1


# Your MajorityChecker object will be instantiated and called as such:
# obj = MajorityChecker(arr)
# param_1 = obj.query(left,right,threshold)

Complexity

Measure	Complexity
Time	O(log n) or O(n log n)
Space	O(n) auxiliary

Pattern: Monotonic Stack

Answer "what is the next greater element" for every position in one pass. LeetCode 1157. Online Majority Element In Subarray is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.

The monotonic stack guide has the Python template for the pattern and the 225 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 1157. Online Majority Element In Subarray?: LeetCode 1157. Online Majority Element In Subarray is rated Hard on LeetCode.
What topics does LeetCode 1157. Online Majority Element In Subarray cover?: LeetCode 1157. Online Majority Element In Subarray is tagged Design, Binary Indexed Tree, Segment Tree, Array and Binary Search on LeetCode.