Leetcode #1150: Check If a Number Is Majority Element in a Sorted Array
In this guide, we solve Leetcode #1150 Check If a Number Is Majority Element in a Sorted Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an integer array nums sorted in non-decreasing order and an integer target, return true if target is a majority element, or false otherwise. A majority element in an array nums is an element that appears more than nums.length / 2 times in the array.
Quick Facts
- Difficulty: Easy
- Premium: Yes
- Tags: Array, Binary Search
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums = [2,4,5,5,5,5,5,6,6], target = 5
Output: true
Explanation: The value 5 appears 5 times and the length of the array is 9.
Thus, 5 is a majority element because 5 > 9/2 is true.
Python Solution
class Solution:
def isMajorityElement(self, nums: List[int], target: int) -> bool:
left = bisect_left(nums, target)
right = bisect_right(nums, target)
return right - left > len(nums) // 2
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.