3065. Minimum Operations to Exceed Threshold Value I


Problem Description

In this problem, we're presented with an array of integers called nums, which is indexed starting at 0, and a separate integer k. Our objective is to remove occurrences of the smallest element in the array, one at a time, until all the elements left in the array are greater than or equal to k.

The operation we're allowed to perform repeatedly is the removal of one instance of the smallest number in the array. We want to determine the smallest number of such operations needed to ensure no number in the array is less than k.

For instance, if our array is [1, 4, 3, 2, 2, 7] and k is 3, we should remove the 1 and both 2s. We performed this operation a minimum of 3 times to ensure all remaining elements are at least k.

Intuition

To solve this problem, we don't actually need to perform the operations of removing elements. We can take a more direct approach by just counting how many elements in the array are less than k. Those are exactly the elements that would need to be removed, one by one, in the operations described.

The solution involves a straightforward approach which is to traverse the entire array once and count the occurrences of numbers that are strictly smaller than k. Since each of these occurrences will require one operation to remove, the count directly gives the minimum number of operations needed.

The rationale behind this is that the operation only ever removes the smallest element, and such an element will always be less than k, or else we would have already met the condition to stop. We don't need to worry about the order in which we remove them since any element less than k must go, regardless. Consequently, our operation count solely depends on how many such elements there are, not their specific values.

Solution Approach

The solution's implementation is quite straightforward and is due to a pattern that is often used in problems related to counting specific elements in a list: traversal and counting.

The Python code uses a generator expression inside the built-in sum function. Let's dissect it:

  1. for x in nums: This is a simple for-loop over the list nums, where each element is represented as x.
  2. x < k: For each element, we check whether it's less than k.
  3. (x < k for x in nums): This is a generator expression that will go through each element x in nums, yielding True if x is less than k, and False otherwise. In Python, True is equivalent to 1, and False is equivalent to 0 when they are used in arithmetic operations.
  4. sum(x < k for x in nums): The sum function then adds up these 1s and 0s, effectively counting the number of times x < k is True, that is, the number of elements less than k.

There aren't any complex algorithms, data structures, or design patterns at play here. The solution merely employs a few fundamental programming constructs: a generator expression for the condition check and a summing operation to tally the count. This works efficiently because we iterate over the list exactly once, leading to an O(n) solution, where n is the number of elements in nums. Since we are not using any additional data structures, the space complexity is O(1), meaning that it uses a constant amount of extra space.

In pseudocode, the process could be described as:

1initialize counter to 0
2for each element x in array nums
3    if x < k
4        increment counter
5return counter

The provided Python code matches this simple pseudocode logic, except it uses Python's ability to treat booleans as integers and encapsulates the for-loop and if-check in a concise one-liner.

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Example Walkthrough

Let's take an example to illustrate the solution approach. Suppose we have the following array of integers nums and integer k:

1nums = [1, 5, 3, 2, 8, 5, 2]
2k = 4

According to the problem, we want to remove the smallest element from the array repeatedly until all elements are greater than or equal to k. Here is the step-by-step walkthrough:

  1. We start by examining each element in the array to see if it is smaller than k. Here k is 4, so we look for elements smaller than 4.

  2. As we traverse nums, we find 1, 3, 2, and 2 to be the elements less than k.

  3. Counting them up, there's a total of 4 items that meet the condition (x < k).

  4. Since the operation is to remove one instance of the smallest number one by one, and we have identified 4 such instances, we infer that it will take exactly 4 removal operations to meet our condition that no number is less than k.

  5. We finish with the understanding that the array does not need to be modified, but just the count of smaller elements is enough to answer the problem. Thus, we avoid the unnecessary operations of actually removing them.

Applying the provided solution approach:

1for x in [1, 5, 3, 2, 8, 5, 2]:
2    if x < 4: # only when x is 1, 3, 2, 2 this condition is true
3        ... # the sum function will count this as 1
4
5# By using the generator expression, we get the sum as 4.

Therefore, to follow the problem's constraint of removing occurrences of the smallest numbers until all elements are >= k, we will need to perform the removal operation 4 times. The rest of the elements will be 5, 5, 8 which are all >= k. This matches our manual count and confirms the correctness of our solution approach.

Solution Implementation

1from typing import List
2
3class Solution:
4    def minOperations(self, nums: List[int], target: int) -> int:
5        # Initialize the count variable to keep track of numbers less than the target.
6        count = 0
7      
8        # Iterate over each number in the nums list.
9        for num in nums:
10            # If the current number is less than the target,
11            # increment the count.
12            if num < target:
13                count += 1
14              
15        # Return the total count of numbers less than the target.
16        return count
17
1class Solution {
2
3    // Method to count the number of elements less than the given threshold 'k'
4    public int minOperations(int[] nums, int k) {
5        int count = 0; // Initialize count to record the number of operations
6
7        // Loop through all the elements of the array 'nums'
8        for (int num : nums) {
9            // If the current element is less than 'k'
10            if (num < k) {
11                count++; // Increment the count
12            }
13        }
14
15        return count; // Return the total number of elements less than 'k'
16    }
17}
18
1#include <vector>
2
3class Solution {
4public:
5    // Function to find the minimum number of operations needed
6    // such that each element in the nums vector is at least k.
7    // In each operation, you can increment any element by 1.
8    int minOperations(std::vector<int>& nums, int k) {
9        int numOperations = 0; // Initialize the count of operations
10      
11        // Iterate over each element in the nums vector
12        for (int num : nums) {
13            // If the current element is less than k
14            if (num < k) {
15                // Increment the number of operations
16                // by the difference between k and the current element.
17                numOperations += (k - num);
18            }
19        }
20      
21        // Return the total number of operations needed
22        return numOperations;
23    }
24};
25
1/**
2 * Calculates the minimum number of operations to increase all elements less than k
3 * 
4 * @param nums - The array of numbers to be processed
5 * @param k - The target value that elements in nums should reach or exceed
6 * @returns The number of elements in nums that are less than k
7 */
8function minOperations(nums: number[], k: number): number {
9    // Filter the array to get all elements which are less than 'k'
10    const elementsLessThanK = nums.filter((element) => element < k);
11  
12    // Return the count of elements that are less than 'k'
13    return elementsLessThanK.length;
14}
15

Time and Space Complexity

Time Complexity: The time complexity of the code is O(n), where n is the length of the input list nums. This is because the code iterates through each element of nums once to check if it is less than k.

Space Complexity: The space complexity of the code is O(1), as it uses a fixed amount of extra space regardless of the input size. The summation operation does not require additional space that scales with the input size.

Learn more about how to find time and space complexity quickly using problem constraints.


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