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

Count Subarrays With Majority Element II (LeetCode #3739) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) 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 II?

The optimal solution for Count Subarrays With Majority Element II runs in O(n) time and uses O(n) space. We traverse the array once and use a hashmap for prefix sums, leading to linear time complexity.

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

The brute force approach for Count Subarrays With Majority Element II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray and count occurrences of the target. If it appears more than half the time, count it as valid. The optimal Optimal Solution approach improves this to O(n).

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

Explain your thought process clearly. Consider edge cases and constraints. Practice similar problems to build confidence.

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

Not considering cases where target does not appear in nums. Overlooking the need for prefix sum adjustments.

#3739

Count Subarrays With Majority Element II

Hard

ArrayHash TableDivide and ConquerSegment TreeMerge SortPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Transform the array to +1/-1 based on the target, then use prefix sums to count valid subarrays efficiently.

⚙️

Algorithm

3 steps

1Step 1: Convert nums to arr where arr[i] = 1 if nums[i] == target else -1.
2Step 2: Build prefix sums and count occurrences of each prefix sum in a hashmap.
3Step 3: For each prefix sum, check how many times it has been seen before to find valid subarrays.

solution.py10 lines

1def countMajoritySubarrays(nums, target):
2    arr = [1 if x == target else -1 for x in nums]
3    prefix_count = {0: 1}
4    count, prefix_sum = 0, 0
5    for num in arr:
6        prefix_sum += num
7        if prefix_sum > 0:
8            count += prefix_count.get(prefix_sum - 1, 0)
9        prefix_count[prefix_sum] = prefix_count.get(prefix_sum, 0) + 1
10    return count

ℹ

Complexity note: We traverse the array once and use a hashmap for prefix sums, leading to linear time complexity.

1Majority element must appear more than half the time in subarrays.
2Transforming the problem can simplify counting occurrences.

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