Minimum Difference Between Largest and Smallest Value in Three Moves

Medium

Asked by:

25 views

Topics:

ArraysGreedy Algorithms

You are given an integer array nums.

In one move, you can choose one element of nums and change it to any value.

Return the minimum difference between the largest and smallest value of nums after performing at most three moves.

Example 1:

Input: nums = [5,3,2,4]
Output: 0
Explanation: We can make at most 3 moves.
In the first move, change 2 to 3. nums becomes [5,3,3,4].
In the second move, change 4 to 3. nums becomes [5,3,3,3].
In the third move, change 5 to 3. nums becomes [3,3,3,3].
After performing 3 moves, the difference between the minimum and maximum is 3 - 3 = 0.

Example 2:

Input: nums = [1,5,0,10,14]
Output: 1
Explanation: We can make at most 3 moves.
In the first move, change 5 to 0. nums becomes [1,0,0,10,14].
In the second move, change 10 to 0. nums becomes [1,0,0,0,14].
In the third move, change 14 to 1. nums becomes [1,0,0,0,1].
After performing 3 moves, the difference between the minimum and maximum is 1 - 0 = 1.
It can be shown that there is no way to make the difference 0 in 3 moves.

Example 3:

Input: nums = [3,100,20]
Output: 0
Explanation: We can make at most 3 moves.
In the first move, change 100 to 7. nums becomes [3,7,20].
In the second move, change 20 to 7. nums becomes [3,7,7].
In the third move, change 3 to 7. nums becomes [7,7,7].
After performing 3 moves, the difference between the minimum and maximum is 7 - 7 = 0.

Constraints:

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

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 is the size limit of the input array `nums`?
Can the values in the array `nums` be negative?
Are duplicate values allowed within the array `nums`?
If after three moves, all elements can be made equal, should I return 0?
Is there a guaranteed minimum size for the input array, or could it be empty or have fewer than 4 elements?

Brute Force Solution

Approach

The brute force approach to minimizing the difference between the largest and smallest values after three moves involves trying every possible combination of moves. We essentially explore all the possible resulting lists after making exactly three changes and pick the one with the smallest difference between its maximum and minimum values.

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

Consider making the first three elements equal to the last element. Note the difference between the largest and smallest number in the resulting list.
Now, consider making the last three elements equal to the first element. Note the difference between the largest and smallest number in the resulting list.
Consider making the first element, last element and middle element equal to some random element. Note the difference between the largest and smallest number in the resulting list.
For every possible combination of three changes to the list, calculate the new difference between the largest and smallest element.
Compare all those differences you calculated.
The smallest difference found among all the possible changes is your answer.

Code Implementation

def min_difference_three_moves_brute_force(numbers):
    list_length = len(numbers)

    # If the list has 3 or less elements, all can be changed
    if list_length <= 3:
        return 0

    minimum_difference = float('inf')

    # Consider making the first three elements equal to the last element.
    modified_list_first_three = numbers[:] 
    for index in range(min(3, list_length)): 
        modified_list_first_three[index] = numbers[list_length - 1]

    minimum_difference = min(minimum_difference, max(modified_list_first_three) - min(modified_list_first_three))

    # Consider making the last three elements equal to the first element
    modified_list_last_three = numbers[:] 
    for index in range(max(0, list_length - 3), list_length): 
        modified_list_last_three[index] = numbers[0]
    minimum_difference = min(minimum_difference, max(modified_list_last_three) - min(modified_list_last_three))

    # Try all possible combinations of three changes, this is the core of the brute force
    for first_index in range(list_length):
        for second_index in range(first_index, list_length):
            for third_index in range(second_index, list_length):
                modified_list = numbers[:]
                modified_list[first_index] = numbers[0] #arbitrary assignment, required by prompt
                modified_list[second_index] = numbers[0]

                #We must update this minimum value every iteration of modification. 
                modified_list[third_index] = numbers[0]
                current_difference = max(modified_list) - min(modified_list)
                minimum_difference = min(minimum_difference, current_difference)

    # Exhaustive search done, return minimum value
    return minimum_difference

Big(O) Analysis

Time Complexity

O(n^3) – The described brute force approach involves considering all possible combinations of three moves on the input array of size n. For each of the three moves, we potentially iterate through the array to find elements to change or to calculate the new maximum and minimum. The number of combinations can be considered to be on the order of n^3. Therefore, the time complexity of this approach is O(n^3) as the algorithm essentially explores a three-dimensional space of possible changes.

Space Complexity

O(1) – The described brute force approach iterates through various combinations of three changes to the input list, but it doesn't appear to create any auxiliary data structures that scale with the input size N (where N is the length of the input list). The algorithm calculates the difference between the largest and smallest numbers for each combination, but it only stores the current minimum difference found so far. Therefore, the auxiliary space required is constant.

Optimal Solution

Approach

The key insight is that we only need to consider removing the largest and smallest elements. We don't need to check every possible combination. By removing elements from either end, we can efficiently minimize the difference between the remaining largest and smallest values.

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

First, if the number of elements is small enough (4 or less), removing three of them will leave only one element, making the difference zero. So, just return zero in that case.
If there are more than four elements, think about the possible moves. You can remove three of the largest numbers, three of the smallest numbers, two largest and one smallest, or one largest and two smallest.
Find the difference between the largest and smallest numbers after each of these four scenarios.
Removing three largest leaves the smallest number as is, and the new largest will be the fourth smallest element.
Removing three smallest leaves the largest number as is, and the new smallest will be the fourth largest element.
Removing two largest and one smallest, find the difference between the third smallest element and the second largest element.
Removing one largest and two smallest, find the difference between the second smallest element and the third largest element.
Finally, pick the smallest difference from those four possibilities, and that's your answer.

Code Implementation

def min_difference(nums):
    nums_length = len(nums)

    # If the array has 4 or fewer elements, return 0.
    if nums_length <= 4:
        return 0

    nums.sort()

    # Calculate the difference after removing three largest elements.
    case_1 = nums[nums_length - 4] - nums[0]

    # Calculate the difference after removing three smallest elements.
    case_2 = nums[nums_length - 1] - nums[3]

    # Calculate difference after removing two largest and one smallest.
    case_3 = nums[nums_length - 3] - nums[1]

    # Calculate difference after removing one largest and two smallest.
    case_4 = nums[nums_length - 2] - nums[2]

    # Return the minimum difference among all possible cases.
    return min(case_1, case_2, case_3, case_4)

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this solution is sorting the input array of size n. The sorting operation, which is implicitly or explicitly required to find the largest and smallest elements to remove, typically takes O(n log n) time using efficient algorithms like merge sort or quicksort. The remaining operations, such as calculating the differences between specific elements after the potential removals, take constant time, O(1). Therefore, the overall time complexity is driven by the sorting step, resulting in O(n log n).

Space Complexity

O(1) – The algorithm's space complexity is constant because it only uses a fixed number of variables to store intermediate results and does not create any data structures that scale with the input size, N (the number of elements in the input array). The algorithm calculates differences and keeps track of the minimum difference so far using a few variables, irrespective of the number of elements in the input array. Therefore, the space used remains constant regardless of the size of the input array.

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 if the array is null or empty as no operations are possible.
Array with length less than or equal to 3	If array length is <= 3, we can make 3 moves to make all elements equal, so the difference is 0.
Array with all elements being identical	The minimum difference will always be 0 as making all elements equal results in 0 difference.
Array with very large or very small numbers (potential integer overflow)	Use long integers to prevent overflow when calculating the difference between the largest and smallest values.
Array is sorted in ascending or descending order	The solution should correctly identify and manipulate the largest and smallest elements regardless of the initial order.
Array contains duplicate minimum and maximum values	The removal of values should correctly consider the duplicates when choosing elements to remove.
Maximum-sized input array (performance considerations)	Optimize the solution to use efficient sorting algorithms with time complexity of O(n log n) to ensure it scales efficiently for large inputs.
Array with a single very large value and all other values small	The solution correctly identifies and removes the largest element in the three moves.

Empty or null input array

How to Handle:

Return 0 if the array is null or empty as no operations are possible.

Array with length less than or equal to 3

How to Handle:

If array length is <= 3, we can make 3 moves to make all elements equal, so the difference is 0.

Array with all elements being identical

How to Handle:

The minimum difference will always be 0 as making all elements equal results in 0 difference.

Array with very large or very small numbers (potential integer overflow)

How to Handle:

Use long integers to prevent overflow when calculating the difference between the largest and smallest values.

Array is sorted in ascending or descending order

How to Handle:

The solution should correctly identify and manipulate the largest and smallest elements regardless of the initial order.

Array contains duplicate minimum and maximum values

How to Handle:

The removal of values should correctly consider the duplicates when choosing elements to remove.

Maximum-sized input array (performance considerations)

How to Handle:

Optimize the solution to use efficient sorting algorithms with time complexity of O(n log n) to ensure it scales efficiently for large inputs.

Array with a single very large value and all other values small

How to Handle:

The solution correctly identifies and removes the largest element in the three moves.