How do you solve Count Subarrays With Majority Element I on LeetCode?

Count Subarrays With Majority Element I (LeetCode #3737) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Majority element must appear more than half the time in subarrays.

What is the time complexity of Count Subarrays With Majority Element I?

The optimal solution for Count Subarrays With Majority Element I runs in O(n) time and uses O(1) space. We only traverse the array once, leading to O(n) time complexity.

What is the brute force solution for Count Subarrays With Majority Element I?

The brute force approach for Count Subarrays With Majority Element I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible subarray and count the occurrences of the target. If the count exceeds half the length of the subarray, it's a valid majority subarray. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Subarrays With Majority Element I?

Clarify the definition of majority element. Discuss potential edge cases upfront. Explain your thought process while coding.

What are common mistakes when solving Count Subarrays With Majority Element I?

Not considering edge cases where target is absent. Incorrectly calculating majority condition.

#3737

Count Subarrays With Majority Element I

Medium

ArrayHash TableDivide and ConquerSegment TreeMerge SortCountingPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can use a two-pointer technique to efficiently count valid subarrays by maintaining a count of the target and adjusting the window size dynamically.

⚙️

Algorithm

3 steps

1Step 1: Initialize pointers and a count of target occurrences.
2Step 2: Expand the right pointer to include elements until the target count exceeds half the window size.
3Step 3: Count valid subarrays and adjust the left pointer to find new valid ranges.

solution.py13 lines

1def countSubarrays(nums, target):
2    count = 0
3    target_count = 0
4    left = 0
5    for right in range(len(nums)):
6        if nums[right] == target:
7            target_count += 1
8        while target_count > (right - left + 1) // 2:
9            count += (right - left + 1) - (right - left - target_count + 1)
10            if nums[left] == target:
11                target_count -= 1
12            left += 1
13    return count

ℹ

Complexity note: We only traverse the array once, leading to O(n) time complexity.

1Majority element must appear more than half the time in subarrays.
2Efficient counting can reduce time complexity significantly.

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