169 Majority Element - Easy

Problem:

Given an array nums of size n, return the majority element.

The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

Example 1:

Input: nums = [3,2,3] Output: 3

Example 2:

Input: nums = [2,2,1,1,1,2,2] Output: 2

Constraints:

n == nums.length
1 <= n <= 5 * 104
-109 <= nums[i] <= 109

Follow-up: Could you solve the problem in linear time and in O(1) space?

Problem Analysis:

High-Level Strategy:

The solution uses the Boyer-Moore Majority Vote Algorithm, which is a clever approach to find the majority element in a sequence without using extra memory.
The algorithm starts with a count initialized to 0 and a candidate initialized to None.
It iterates through the elements of the array:
- If count becomes 0, it updates the candidate to the current element because it means all occurrences of the previous candidate have been canceled out by occurrences of other elements so far.
- If the current element matches the candidate, count is incremented by 1; otherwise, count is decremented by 1.
At the end of the iteration, candidate will hold the majority element.

Complexity Analysis:

Time Complexity: The algorithm iterates through the array once, so the time complexity is O(n), where n is the number of elements in the input list nums.
Space Complexity: The algorithm uses only constant extra space, regardless of the size of the input array. Therefore, the space complexity is O(1).

Solutions:

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        count = 0
        candidate = None
        
        for num in nums:
            if count == 0:
                candidate = num
            count += (1 if num == candidate else -1)
        
        return candidate

Walter Teng.

169 Majority Element - Easy

Problem:

Problem Analysis:

Solutions:

Similar Questions