LeetCode 3833: Count Dominant Indices

Meta Data

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Difficulty: easy First Attempt: 2026-02-08

• Submissions: Wrong Answer → Accepted

Approach

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nums[i] must exceed the average of the elements to its right (excluding itself). Scan right to left keeping their sum and count (Array Template 3). Compare averages by multiplying: nums[i] > s / c is nums[i] * c > s, avoiding floats and integer division. The last element has nothing to its right and does not count.

My Attempts

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Attempt 1 had two problems: the average used // integer division, and position i’s average included itself. For [4, 3, 5, 4] only 5 > 4 holds, so the answer is 1; it returned 2. Switching to / and comparing with the suffix average from i + 1 passed.

Solution

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class Solution:
    def dominantIndices(self, nums: List[int]) -> int:
        n, suffix, res = len(nums), 0, 0
        for i in range(n - 1, -1, -1):
            cnt = n - 1 - i
            if cnt and nums[i] * cnt > suffix:
                res += 1
            suffix += nums[i]
        return res

Complexity

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Time O(n), space O(1).

Mistakes to Avoid

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• The average excludes the element itself.
• Integer division rounds the average down and breaks the comparison; multiplying is safest.

Interview Explanation

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Scanning right to left with the sum of the right side, I check whether the element times the count on its right exceeds that sum, which is the same as beating the average without floats. Time O(n).