LeetCode 3065 - Minimum Operations to Exceed Threshold Value I

LeetCode Problem 3065

Difficulty: 🟢 Easy
Topics: Array

Solution

Problem Understanding

The problem asks us to find the minimum number of operations needed to ensure that every element in an array is at least k. Each operation consists of removing the smallest element from the array. Essentially, we are incrementally deleting the smallest values until all remaining elements are greater than or equal to k.

The input is a 0-indexed integer array nums and a single integer k. The output is an integer representing the minimum number of removals required. The constraints are fairly small: the array length is at most 50, and individual elements and k can be as large as $10^9$. Importantly, the input guarantees that at least one element is already greater than or equal to k, which ensures a solution always exists.

Critical edge cases to note include arrays where all elements are already greater than or equal to k (resulting in 0 operations), arrays where only one element meets the threshold, and arrays where elements are already sorted or have duplicates of the smallest value.

Approaches

The brute-force approach would repeatedly find the smallest element and remove it until the remaining elements are all greater than or equal to k. This approach is correct because each operation guarantees that the current minimum element, which violates the condition, is eliminated. However, it is inefficient because finding the minimum repeatedly is $O(n)$ per operation, resulting in worst-case $O(n^2)$ for larger arrays.

The optimal approach relies on sorting the array once in ascending order. After sorting, all elements below k will appear at the beginning of the array. Counting how many elements are less than k gives the minimum number of removal operations directly, because each of those elements must be removed. Sorting is $O(n \log n)$, and the count operation is $O(n)$, resulting in a much faster solution than brute-force for larger arrays.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n^2)	O(1)	Repeatedly remove smallest elements
Optimal	O(n log n)	O(1) or O(n)	Sort array and count elements less than k

Algorithm Walkthrough

1. Sort the array nums in ascending order. Sorting ensures that all elements below k are grouped at the beginning of the array.
2. Initialize a counter operations to 0. This counter tracks how many elements must be removed.
3. Iterate through the sorted array from the start:

• For each element, if it is less than k, increment operations by 1.
• If an element is greater than or equal to k, stop iterating because all subsequent elements satisfy the condition.

4. Return the counter operations.

Why it works: Sorting ensures that all elements requiring removal are contiguous at the start of the array. By counting all elements less than k, we guarantee that removing exactly these elements will result in all remaining elements being greater than or equal to k. The invariant is that after removing operations elements, the remaining array contains only values $\ge k$.

Python Solution

from typing import List

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        nums.sort()
        operations = 0
        for num in nums:
            if num < k:
                operations += 1
            else:
                break
        return operations

The Python implementation first sorts nums. The for-loop iterates through the sorted array, incrementing operations for each element below k. Once an element is at least k, we stop because the remaining elements are all guaranteed to satisfy the threshold. This directly implements the algorithm described above.

Go Solution

import "sort"

func minOperations(nums []int, k int) int {
    sort.Ints(nums)
    operations := 0
    for _, num := range nums {
        if num < k {
            operations++
        } else {
            break
        }
    }
    return operations
}

The Go solution mirrors the Python approach. We use sort.Ints to sort the slice and iterate using a range loop. A counter tracks elements below k. The main difference from Python is handling slices and type safety, but the logic remains identical.

Worked Examples

Example 1: nums = [2,11,10,1,3], k = 10

Step	Sorted Array	Operations
Initial	[1,2,3,10,11]	0
1	2,3,10,11	1
2	3,10,11	2
3	10,11	3

Result: 3

Example 2: nums = [1,1,2,4,9], k = 1

Step	Sorted Array	Operations
Initial	[1,1,2,4,9]	0

All elements are $\ge 1$. Result: 0

Example 3: nums = [1,1,2,4,9], k = 9

Step	Sorted Array	Operations
Initial	[1,1,2,4,9]	0
Count elements < 9	1,1,2,4	4

Result: 4

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates the runtime; counting is O(n)
Space	O(1)	Sorting in-place (Python/Timsort) or O(n) depending on language implementation

Sorting is efficient for this small input size (max 50 elements), and counting elements is linear.

Test Cases

# Provided examples
assert Solution().minOperations([2,11,10,1,3], 10) == 3  # mixed array
assert Solution().minOperations([1,1,2,4,9], 1) == 0     # all elements >= k
assert Solution().minOperations([1,1,2,4,9], 9) == 4     # only one element >= k

# Edge cases
assert Solution().minOperations([10], 10) == 0           # single element exactly k
assert Solution().minOperations([1], 2) == 1            # single element below k
assert Solution().minOperations([5,5,5,5], 5) == 0      # all elements equal k
assert Solution().minOperations([1,2,3,4,5], 6) == 5    # all elements below k
assert Solution().minOperations([1,2,3,100], 50) == 3   # last element >= k

Test	Why
Mixed array	Validates normal operation
All elements >= k	Validates 0 operation case
Only one element >= k	Validates multiple removals
Single element = k	Minimum size edge case
Single element < k	Minimum size edge case needing removal
All elements = k	Edge case of equality
All elements < k	All elements need removal
Last element >= k	Ensures correct counting in presence of large numbers

Edge Cases

1. Single-element array: Arrays of length 1 could be a source of off-by-one errors. The code correctly handles it by checking the single element against k.
2. All elements equal to k: In this case, no operations are needed. The algorithm handles it by stopping iteration as soon as an element $\ge k$ is found.
3. Large values: Elements and k can be as large as $10^9$. The implementation does not perform arithmetic beyond comparisons, so integer overflow is not a concern in Python or Go for this problem.

This solution is robust, simple, and works efficiently across all input constraints.