Minimum Operations to Exceed Threshold Value I

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Arrays

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

In one operation, you can remove one occurrence of the smallest element of nums.

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

Example 1:

Input: nums = [2,11,10,1,3], k = 10
Output: 3
Explanation: After one operation, nums becomes equal to [2, 11, 10, 3].
After two operations, nums becomes equal to [11, 10, 3].
After three operations, nums becomes equal to [11, 10].
At this stage, all the elements of nums are greater than or equal to 10 so we can stop.
It can be shown that 3 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10.

Example 2:

Input: nums = [1,1,2,4,9], k = 1
Output: 0
Explanation: All elements of the array are greater than or equal to 1 so we do not need to apply any operations on nums.

Example 3:

Input: nums = [1,1,2,4,9], k = 9
Output: 4
Explanation: only a single element of nums is greater than or equal to 9 so we need to apply the operations 4 times on nums.

Constraints:

1 <= nums.length <= 50
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹
The input is generated such that there is at least one index i such that nums[i] >= k.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the numbers within the input array?
Can the threshold value ever be negative or zero?
What should I return if no amount of operations can exceed the threshold?
Are there any size limitations for the input array?
Is the operation guaranteed to result in an integer value, or might there be floating-point calculations involved?

Brute Force Solution

Approach

We want to find the smallest number of actions needed to make a number in a collection bigger than a target. The brute force way is to try every possible sequence of actions until we find one that works.

Here's how the algorithm would work step-by-step:

Start by considering each number in the collection one at a time.
For each number, simulate performing one action on it.
Check if the result of the action is now bigger than the target. If it is, then we have found one possible sequence and note the number of operations (which is 1 in this case).
If the number is still not bigger than the target, try performing a second action on the number.
Check if the result of the two actions is now bigger than the target. If it is, then we have found one possible sequence of two actions. Note the operations (which is 2 in this case)
Continue doing this for each possible number of actions (3, 4, and so on) until we find an action that gets us over the threshold.
For each number in the collection, we will have recorded the minimum number of operations needed to make that number bigger than the target. Take the smallest number from all of those numbers and that is the overall minimum number of operations.
If it is not possible to make any of the numbers bigger than the target no matter how many actions we perform, then we cannot achieve the target with the actions performed.

Code Implementation

def min_operations_to_exceed_threshold(numbers, threshold):
    minimum_operations = float('inf')

    for number in numbers:
        operations = 0
        current_value = number

        while operations <= 100: #Limit number of operations
            operations += 1
            current_value *= 2 #Simulate the operation

            if current_value > threshold:
                # Found a sequence. Check if it's minimum
                minimum_operations = min(minimum_operations, operations)
                break

    # If no number exceeds the threshold, return -1
    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

O(n*k) – The algorithm iterates through each of the n numbers in the input array. For each number, it performs up to k operations (where k is the number of operations needed to exceed the threshold, which is related to the threshold and the values in the input array). Inside the loop for each number, there's a constant-time check to see if the threshold is exceeded. Therefore, the time complexity is O(n*k) where n is the number of elements in the input array and k is the maximum number of operations applied to each element.

Space Complexity

O(1) – The provided plain English explanation suggests an iterative approach where we're processing each number in the collection and performing actions until a threshold is exceeded. We are only keeping track of the minimum operations across the numbers, and the number of operations for each number. No data structures dependent on the input size (N, the number of elements in the collection) are used to store intermediate results, visited nodes, or recursive call stacks. Therefore, the auxiliary space used is constant, regardless of the input size N.

Optimal Solution

Approach

The key is to find the two smallest numbers and repeatedly combine them until you reach or exceed the target value. This strategy focuses on efficient combination rather than exploring all possibilities.

Here's how the algorithm would work step-by-step:

Find the two smallest numbers in the set.
Add the two smallest numbers together.
Replace the two smallest numbers with the result of their addition.
Repeat this process of finding the two smallest and adding them until the smallest number in the set is greater than or equal to the target value.
Count how many times you had to add the smallest numbers together; this is the minimum number of operations needed.

Code Implementation

def min_operations_to_exceed_threshold(numbers, threshold_value):
    operations_count = 0

    while True:
        # Sort to easily find the two smallest numbers
        numbers.sort()

        if numbers[0] >= threshold_value:
            break

        # Add the two smallest numbers
        sum_of_smallest = numbers[0] + numbers[1]

        # Replace the two smallest with their sum
        numbers[0] = sum_of_smallest
        numbers.pop(1)
        operations_count += 1

    return operations_count

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is repeatedly finding the two smallest numbers in the set. While we could naively scan the array each time to find the minimum two elements in O(n) time, sorting the set allows us to easily identify the smallest elements. Using an efficient sorting algorithm like merge sort or heap sort initially takes O(n log n) time. After sorting, we iterate combining the two smallest elements which takes O(n) time at worst. Since the sorting dominates the combination operation, the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm primarily modifies the input array in place. It only needs to store a few variables to keep track of the two smallest numbers and the operation count. The amount of extra memory used remains constant regardless of the input size N (the number of elements in the set), therefore the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array nums	Return 0 as no operations are needed to exceed threshold with no elements.
Input array nums contains a single element that is already greater than the threshold	Return 0 as no operations are needed because the threshold is already exceeded.
All elements in nums are 0 and threshold is a positive value	Return the length of the array minus one, as all elements need to be combined
Threshold is 0 or negative and nums contains only positive integers	Return 0 as initial elements exceeds threshold.
Input array contains very large numbers that might cause integer overflow during summation.	Use a data type (e.g., long) that can accommodate larger sums or check for potential overflow before adding.
Input array has extremely large length and the threshold value is huge	Optimize for time complexity, as a brute-force approach might timeout, by processing from smallest to largest.
Array contains negative numbers, zero and positive numbers. The threshold is a large positive number.	Sum the largest numbers first, to reduce the total operations needed.
The array has some negative values that when summed actually helps exceed the threshold quicker	Always combine the two smallest values; the combination might still be negative but closer to the threshold

Empty input array nums

How to Handle:

Return 0 as no operations are needed to exceed threshold with no elements.

Input array nums contains a single element that is already greater than the threshold

How to Handle:

Return 0 as no operations are needed because the threshold is already exceeded.

All elements in nums are 0 and threshold is a positive value

How to Handle:

Return the length of the array minus one, as all elements need to be combined

Threshold is 0 or negative and nums contains only positive integers

How to Handle:

Return 0 as initial elements exceeds threshold.

Input array contains very large numbers that might cause integer overflow during summation.

How to Handle:

Use a data type (e.g., long) that can accommodate larger sums or check for potential overflow before adding.

Input array has extremely large length and the threshold value is huge

How to Handle:

Optimize for time complexity, as a brute-force approach might timeout, by processing from smallest to largest.

Array contains negative numbers, zero and positive numbers. The threshold is a large positive number.

How to Handle:

Sum the largest numbers first, to reduce the total operations needed.

The array has some negative values that when summed actually helps exceed the threshold quicker

How to Handle:

Always combine the two smallest values; the combination might still be negative but closer to the threshold