How do you solve Count Dominant Indices on LeetCode?

Count Dominant Indices (LeetCode #3833) 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: Dominance requires comparing with averages of subsequent elements.

What is the time complexity of Count Dominant Indices?

The optimal solution for Count Dominant Indices runs in O(n) time and uses O(1) space. Single pass to calculate the total sum and another pass to check dominance leads to linear time complexity.

What is the brute force solution for Count Dominant Indices?

The brute force approach for Count Dominant Indices is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each element against the average of all elements to its right. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Dominant Indices?

Count Dominant Indices uses the following concepts: Array, Enumeration. The recommended patterns to study are: Array, Prefix Sum.

What are the interview tips for Count Dominant Indices?

Explain your thought process clearly. Consider edge cases like small arrays. Optimize before coding.

What are common mistakes when solving Count Dominant Indices?

Forgetting to exclude the last element from checks. Incorrectly calculating the average when elements are missing.

#3833

Count Dominant Indices

Easy

ArrayEnumerationArrayPrefix Sum

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)

Use a running sum to calculate the average efficiently without nested loops.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter and a running sum of elements to the right.
2Step 2: Traverse the array from right to left, updating the sum and checking dominance.
3Step 3: For each index, calculate the average using the running sum and the count of elements.

solution.py13 lines

1def countDominantIndices(nums):
2    count = 0
3    total_sum = 0
4    n = len(nums)
5    for i in range(n-1, 0, -1):
6        total_sum += nums[i]
7    right_count = n - 1
8    for i in range(n-1):
9        if nums[i] > total_sum / right_count:
10            count += 1
11        total_sum -= nums[i+1]
12        right_count -= 1
13    return count

ℹ

Complexity note: Single pass to calculate the total sum and another pass to check dominance leads to linear time complexity.

1Dominance requires comparing with averages of subsequent elements.
2Using a running sum can optimize average calculations.

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