How do you solve Minimum Removals to Balance Array on LeetCode?

Minimum Removals to Balance Array (LeetCode #3634) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting helps in efficiently finding the range of valid elements.

What is the time complexity of Minimum Removals to Balance Array?

The optimal solution for Minimum Removals to Balance Array runs in O(n log n) time and uses O(1) space. The sorting step dominates the complexity, followed by a linear scan with two pointers.

What is the brute force solution for Minimum Removals to Balance Array?

The brute force approach for Minimum Removals to Balance Array is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible subsets of the array to find the minimum removals needed to balance it. This is inefficient as it explores every combination. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Removals to Balance Array?

Minimum Removals to Balance Array uses the following concepts: Array, Binary Search, Sliding Window, Sorting. The recommended patterns to study are: Two Pointers, Sorting.

What are the interview tips for Minimum Removals to Balance Array?

Explain your thought process clearly. Use examples to illustrate your approach. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Removals to Balance Array?

Not sorting the array before applying the two-pointer technique. Confusing max and min conditions when checking balance.

#3634

Minimum Removals to Balance Array

Medium

ArrayBinary SearchSliding WindowSortingTwo PointersSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(1)

Sort the array and use a two-pointer technique to find the largest balanced window. The difference between the total length and this window gives the minimum removals.

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Algorithm

3 steps

1Step 1: Sort the array.
2Step 2: Initialize two pointers, i and j, both at the start of the array.
3Step 3: Expand j while nums[j] <= k * nums[i]; calculate removals as n - (j - i + 1).

solution.py11 lines

1def min_removals_optimal(nums, k):
2    nums.sort()
3    n = len(nums)
4    i, j, max_removals = 0, 0, 0
5    while j < n:
6        if nums[j] <= k * nums[i]:
7            j += 1
8        else:
9            i += 1
10        max_removals = max(max_removals, j - i)
11    return n - max_removals

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Complexity note: The sorting step dominates the complexity, followed by a linear scan with two pointers.

1Sorting helps in efficiently finding the range of valid elements.
2Using two pointers reduces the need for nested loops.

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