LeetCode #3634 — MEDIUM

Minimum Removals to Balance Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums and an integer k.

An array is considered balanced if the value of its maximum element is at most k times the minimum element.

You may remove any number of elements from nums without making it empty.

Return the minimum number of elements to remove so that the remaining array is balanced.

Note: An array of size 1 is considered balanced as its maximum and minimum are equal, and the condition always holds true.

Example 1:

Input: nums = [2,1,5], k = 2

Output: 1

Explanation:

Remove nums[2] = 5 to get nums = [2, 1].
Now max = 2, min = 1 and max <= min * k as 2 <= 1 * 2. Thus, the answer is 1.

Example 2:

Input: nums = [1,6,2,9], k = 3

Output: 2

Explanation:

Remove nums[0] = 1 and nums[3] = 9 to get nums = [6, 2].
Now max = 6, min = 2 and max <= min * k as 6 <= 2 * 3. Thus, the answer is 2.

Example 3:

Input: nums = [4,6], k = 2

Output: 0

Explanation:

Since nums is already balanced as 6 <= 4 * 2, no elements need to be removed.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. An array is considered balanced if the value of its maximum element is at most k times the minimum element. You may remove any number of elements from nums without making it empty. Return the minimum number of elements to remove so that the remaining array is balanced. Note: An array of size 1 is considered balanced as its maximum and minimum are equal, and the condition always holds true.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Sliding Window

Example 1

[2,1,5]
2

Example 2

[1,6,2,9]
3

Example 3

[4,6]
2

Step 02

Core Insight

What unlocks the optimal approach

Sort <code>nums</code> and use two pointers <code>i</code> and <code>j</code> so that the window's minimum is <code>nums[i]</code> and maximum is <code>nums[j]</code>.
Expand <code>j</code> while <code>nums[j] <= k * nums[i]</code> to maximize the balanced window; answer = <code>n - (j - i + 1)</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3634: Minimum Removals to Balance Array
class Solution {
    public int minRemoval(int[] nums, int k) {
        Arrays.sort(nums);
        int cnt = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            int j = n;
            if (1L * nums[i] * k <= nums[n - 1]) {
                j = Arrays.binarySearch(nums, nums[i] * k + 1);
                j = j < 0 ? -j - 1 : j;
            }
            cnt = Math.max(cnt, j - i);
        }
        return n - cnt;
    }
}

// Accepted solution for LeetCode #3634: Minimum Removals to Balance Array
func minRemoval(nums []int, k int) int {
	sort.Ints(nums)
	n := len(nums)
	cnt := 0
	for i := 0; i < n; i++ {
		j := n
		if int64(nums[i])*int64(k) <= int64(nums[n-1]) {
			target := int64(nums[i])*int64(k) + 1
			j = sort.Search(n, func(x int) bool {
				return int64(nums[x]) >= target
			})
		}
		cnt = max(cnt, j-i)
	}
	return n - cnt
}

# Accepted solution for LeetCode #3634: Minimum Removals to Balance Array
class Solution:
    def minRemoval(self, nums: List[int], k: int) -> int:
        nums.sort()
        cnt = 0
        for i, x in enumerate(nums):
            j = bisect_right(nums, k * x)
            cnt = max(cnt, j - i)
        return len(nums) - cnt

// Accepted solution for LeetCode #3634: Minimum Removals to Balance Array
impl Solution {
    pub fn min_removal(mut nums: Vec<i32>, k: i32) -> i32 {
        nums.sort();
        let mut cnt = 0;
        let n = nums.len();
        for i in 0..n {
            let mut j = n;
            let target = nums[i] as i64 * k as i64;
            if target <= nums[n - 1] as i64 {
                j = nums.partition_point(|&x| x as i64 <= target);
            }
            cnt = cnt.max(j - i);
        }
        (n - cnt) as i32
    }
}

// Accepted solution for LeetCode #3634: Minimum Removals to Balance Array
function minRemoval(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let cnt = 0;
    for (let i = 0; i < n; ++i) {
        let j = n;
        if (nums[i] * k <= nums[n - 1]) {
            const target = nums[i] * k + 1;
            j = _.sortedIndexBy(nums, target, x => x);
        }
        cnt = Math.max(cnt, j - i);
    }
    return n - cnt;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.