How do you solve Online Majority Element In Subarray on LeetCode?

Online Majority Element In Subarray (LeetCode #1157) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q log n) time complexity and O(n) space complexity. Key insight: A majority element appears more than half the time in a subset, which can be leveraged for efficient searching.

What is the time complexity of Online Majority Element In Subarray?

The optimal solution for Online Majority Element In Subarray runs in O(n + q log n) time and uses O(n) space. The preprocessing step takes O(n) to build the index map, and each query takes O(log n) for binary search, leading to an overall complexity of O(n + q log n).

What is the brute force solution for Online Majority Element In Subarray?

The brute force approach for Online Majority Element In Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each subarray for the majority element by counting occurrences of each element within the specified range. This is simple but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n + q log n).

What are the interview tips for Online Majority Element In Subarray?

Always think about preprocessing data for repeated queries to optimize performance. Explain your thought process clearly; it helps interviewers understand your approach. Practice writing clean and efficient code, as readability is often as important as functionality.

What are common mistakes when solving Online Majority Element In Subarray?

Not considering edge cases where the threshold is equal to the size of the subarray. Failing to preprocess the data effectively, leading to inefficient query handling.

#1157

Online Majority Element In Subarray

Hard

ArrayBinary SearchDesignBinary Indexed TreeSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n + q log n)Space O(n)

Using a combination of a hashmap and a binary search approach allows us to efficiently count occurrences and quickly find majority elements in subarrays.

⚙️

Algorithm

3 steps

1Step 1: Preprocess the array to create a hashmap that stores the indices of each element.
2Step 2: For each query, use binary search to find the leftmost and rightmost indices of the queried element within the specified range.
3Step 3: Count the occurrences of the element using the indices and check if it meets the threshold.

solution.py18 lines

1from collections import defaultdict
2from bisect import bisect_left, bisect_right
3
4class MajorityChecker:
5    def __init__(self, arr):
6        self.arr = arr
7        self.indices = defaultdict(list)
8        for i, num in enumerate(arr):
9            self.indices[num].append(i)
10
11    def query(self, left, right, threshold):
12        for num in self.indices:
13            indices = self.indices[num]
14            left_index = bisect_left(indices, left)
15            right_index = bisect_right(indices, right) - 1
16            if right_index - left_index + 1 >= threshold:
17                return num
18        return -1

ℹ

Complexity note: The preprocessing step takes O(n) to build the index map, and each query takes O(log n) for binary search, leading to an overall complexity of O(n + q log n).

1A majority element appears more than half the time in a subset, which can be leveraged for efficient searching.
2Using data structures like maps and binary search can significantly reduce the time complexity for repeated queries.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.