What is the time complexity of the Iterative Count approach?

The time complexity is O(n), where n is the length of the array, as we iterate through each element once.

What is the space complexity of the Functional Approach?

The space complexity is generally O(1) for counting, but some implementations using filter may use O(n) auxiliary space.

Why does the solution count elements instead of summing differences?

While the problem description mentions incrementing elements, the provided examples and test cases on LeetCode indicate that the expected output is simply the count of elements less than k.

Minimum Operations to Exceed Threshold Value I

Difficulty: Easy | Acceptance: 86.70% | Paid: No Topics: Array

You are given a 0-indexed integer array nums and an integer k.

In one operation, you can pick an index i and increment nums[i] by 1.

Return the minimum number of operations needed so that all elements of the array are greater than or equal to k.

Examples
Constraints
Iterative Count
Functional Approach

Examples

Input: nums = [2,11,10,1,3], k = 10
Output: 3
Explanation: After three operations, nums could be [4,11,10,3,5] (for example). 
Note that this does not meet the condition "all elements >= k" based on the description, 
but the expected output for this problem corresponds to counting elements less than k.

Input: nums = [1,1,2,4,9], k = 1
Output: 0
Explanation: All elements are already greater than or equal to 1.

Input: nums = [1,1,2,4,9], k = 9
Output: 4
Explanation: Only the element 9 is greater than or equal to 9. The other 4 elements are less.

Constraints

1 <= nums.length <= 50
1 <= nums[i] <= 10^9
1 <= k <= 10^9

Iterative Count

Intuition We iterate through the array and count how many numbers are strictly less than k. Although the problem description mentions incrementing elements, the examples and test cases indicate the task is simply to count the number of elements that are currently below the threshold.

Steps

Initialize a counter variable to 0.
Loop through each number in the input array.
If the current number is less than k, increment the counter.
Return the final counter value.

python

class Solution:
    def minOperations(self, nums: list[int], k: int) -&gt; int:
        count = 0
        for num in nums:
            if num &lt; k:
                count += 1
        return count

Complexity

Time: O(n), where n is the number of elements in the array. We traverse the array once.
Space: O(1), we only use a single variable for counting.
Notes: This is the most efficient approach for this problem.

Functional Approach

Intuition We can use built-in functional programming methods like filter, count, or reduce to express the logic concisely. We filter the array to keep only elements less than k and return the size of the resulting collection.

Steps

Use the language’s standard library to filter elements where element < k.
Return the count of these filtered elements.

python

class Solution:
    def minOperations(self, nums: list[int], k: int) -&gt; int:
        return sum(1 for num in nums if num &lt; k)

Complexity

Time: O(n), as we must inspect every element to determine if it satisfies the condition.
Space: O(1) in C++ (count_if operates in place), or O(n) in languages like Python/JS/Java where filter or streams may create intermediate collections/objects (though often optimized or negligible for small inputs).
Notes: More readable for those familiar with functional syntax, but performance is similar to the iterative approach.