How do you solve Minimum Index of a Valid Split on LeetCode?

Minimum Index of a Valid Split (LeetCode #2780) 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: Understanding the concept of dominant elements is crucial.

What is the brute force solution for Minimum Index of a Valid Split?

The brute force approach for Minimum Index of a Valid Split is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible split point in the array and verifies if both resulting subarrays have the same dominant element. This is straightforward but inefficient, as it requires checking each split individually. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Index of a Valid Split?

Minimum Index of a Valid Split uses the following concepts: Array, Hash Table, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Index of a Valid Split?

Always clarify the problem statement and constraints. Think about edge cases, such as arrays with only two elements. Practice counting techniques and how to maintain counts efficiently.

What are common mistakes when solving Minimum Index of a Valid Split?

Not correctly identifying the dominant element. Failing to check both subarrays for dominance.

#2780

Minimum Index of a Valid Split

Medium

ArrayHash TableSortingHash 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)

The optimal solution uses a single pass to count the occurrences of the dominant element while iterating through the array. This allows us to efficiently determine if a valid split exists without needing to re-count elements repeatedly.

⚙️

Algorithm

4 steps

1Step 1: Count the total occurrences of the dominant element in the entire array.
2Step 2: Initialize a counter for the dominant element in the left subarray.
3Step 3: Iterate through the array, updating the left counter and calculating the right counter on-the-fly.
4Step 4: Check if the dominant element is valid in both subarrays at each split index.

solution.py11 lines

1def min_valid_split(nums):
2    total_count = nums.count(max(set(nums), key=nums.count))
3    left_count = 0
4    n = len(nums)
5    for i in range(n - 1):
6        if nums[i] == dominant:
7            left_count += 1
8        right_count = total_count - left_count
9        if left_count * 2 > (i + 1) and right_count * 2 > (n - i - 1):
10            return i
11    return -1

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a constant number of times, making it efficient. The space complexity is O(1) since we are using a fixed amount of extra space.

1Understanding the concept of dominant elements is crucial.
2Efficient counting techniques can significantly reduce time complexity.

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