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Minimize the Maximum Difference of Pairs

#586 Most AskedMedium

Minimize the Maximum Difference of Pairs #586 Most Asked

Medium

Amazon

17 views

Topics:

ArraysBinary Search

You are given a 0-indexed integer array nums and an integer p. Find p pairs of indices of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure no index appears more than once amongst the p pairs.

Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x.

Return the minimum maximum difference among all p pairs. We define the maximum of an empty set to be zero.

Example 1:

Input: nums = [10,1,2,7,1,3], p = 2
Output: 1
Explanation: The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. 
The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.

Example 2:

Input: nums = [4,2,1,2], p = 1
Output: 0
Explanation: Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹
0 <= p <= (nums.length)/2

Try coding it in LeetCode

Minimize the Maximum Difference of Pairs

#586 Most AskedMedium

Minimize the Maximum Difference of Pairs #586 Most Asked

Medium

Amazon

17 views

Topics:

ArraysBinary Search

Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x.

Return the minimum maximum difference among all p pairs. We define the maximum of an empty set to be zero.

Example 1:

Input: nums = [10,1,2,7,1,3], p = 2
Output: 1
Explanation: The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. 
The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.

Example 2:

Input: nums = [4,2,1,2], p = 1
Output: 0
Explanation: Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹
0 <= p <= (nums.length)/2

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 constraints on the size of the input array `nums` and the value of `p`?
Can the elements in the `nums` array be negative, zero, or floating-point numbers?
If it's not possible to form `p` pairs, what should the function return?
Are duplicate numbers allowed in the `nums` array, and if so, how should they be handled?
Is it guaranteed that `p` will always be less than or equal to `nums.length / 2`?

Brute Force Solution

Approach

The brute force strategy for this problem involves exploring all possible pairings of numbers. We will try every combination of pairs to find the one that minimizes the largest difference between any of the chosen pairs. It's like trying every possible group until we find the 'best' one.

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

First, think about all the different ways you can pick pairs of numbers from the entire set of numbers.
Start by picking the first two numbers as a pair, then calculate the difference between them.
Next, pick the first number and the third number as a pair, calculate the difference, and so on, trying every possible pairing with the first number.
Then, move on to the second number and pair it with every number that comes after it, again calculating the differences.
Continue doing this until you've considered every possible pair of numbers.
Now, consider groups of multiple pairs. For each group of pairs, find the largest difference among all the pairs in that group.
Keep track of that largest difference for each possible group of pairs.
Finally, compare all the largest differences you've found from all the groups, and pick the smallest one. That's your answer.

Code Implementation

def minimize_maximum_difference_of_pairs_brute_force(numbers, number_of_pairs):
    number_of_elements = len(numbers)
    if number_of_elements < 2 * number_of_pairs:
        return -1

    minimum_maximum_difference = float('inf')

    # Iterate through all possible combinations of pairs.
    for i in range(1 << number_of_elements):
        pairs = []
        current_pair = []
        number_of_selected_elements = 0

        for j in range(number_of_elements):
            if (i >> j) & 1:
                number_of_selected_elements += 1
                current_pair.append(numbers[j])

                if len(current_pair) == 2:
                    pairs.append(abs(current_pair[0] - current_pair[1]))
                    current_pair = []

        # Ensure the correct number of pairs was selected.
        if len(pairs) == number_of_pairs and number_of_selected_elements == 2 * number_of_pairs:
            
            maximum_difference_in_current_pairs = max(pairs)

            #Keep track of minimum of the maximum differences
            minimum_maximum_difference = min(minimum_maximum_difference, maximum_difference_in_current_pairs)

    if minimum_maximum_difference == float('inf'):
        return -1
    return minimum_maximum_difference

Big(O) Analysis

Time Complexity

O(n!) – The described brute force solution iterates through all possible pairings of numbers from the input array of size n. The number of ways to choose 'k' pairs from 'n' elements explodes combinatorially, necessitating checking all possible combinations of pairs and then finding the largest difference within each combination. Evaluating all these combinations to minimize the maximum difference leads to a factorial time complexity, specifically O(n!). Therefore, this approach becomes extremely inefficient as 'n' grows.

Space Complexity

O(1) – The provided brute force algorithm calculates differences between pairs and tracks the largest difference seen so far. It does not utilize any auxiliary data structures that scale with the input size N (where N is the number of elements). The algorithm only uses a constant amount of extra memory for storing a few variables like indices and the minimum maximum difference, irrespective of the size of the input array. Therefore, the space complexity is constant.

Optimal Solution

Approach

The core idea is to find the smallest possible maximum difference between pairs. We can do this efficiently by first sorting the numbers and then using a technique to 'guess' the maximum difference and check if we can form enough pairs with that guess.

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

First, arrange the numbers in order from smallest to largest. This makes it easier to pair them later.
Think of a range of possible answers. The smallest possible answer is zero (if two numbers are the same), and the largest is the difference between the biggest and smallest numbers.
Pick a number from that range as a 'guess' for the maximum difference we're allowed between pairs.
Go through the sorted numbers and try to create as many pairs as possible where the difference between the numbers in each pair is no more than our 'guess'.
If we can create enough pairs using our 'guess', it means the actual minimum possible maximum difference is probably equal to or smaller than our guess. If we can't create enough pairs, it means our guess was too small.
Adjust our 'guess' based on whether we could form enough pairs. If we made enough pairs, we try a smaller guess. If not, we try a larger guess.
Keep repeating the guessing and checking process, narrowing down the range of possible answers until we find the smallest possible maximum difference that allows us to create the required number of pairs.

Code Implementation

def minimize_the_maximum_difference_of_pairs(numbers, number_of_pairs):
    numbers.sort()

    def can_form_pairs(maximum_difference):
        pairs_formed = 0
        index = 0
        while index < len(numbers) - 1:
            if numbers[index + 1] - numbers[index] <= maximum_difference:
                pairs_formed += 1
                index += 2
            else:
                index += 1
        return pairs_formed >= number_of_pairs

    left_bound = 0
    right_bound = numbers[-1] - numbers[0]
    minimum_maximum_difference = right_bound

    # Binary search to find the minimum maximum difference
    while left_bound <= right_bound:
        middle = (left_bound + right_bound) // 2

        # Check if we can form the required number of pairs
        if can_form_pairs(middle):
            minimum_maximum_difference = middle

            # Try to minimize the maximum difference further
            right_bound = middle - 1
        else:

            # Need a larger difference to form enough pairs
            left_bound = middle + 1

    return minimum_maximum_difference

Big(O) Analysis

Time Complexity

O(n log n) – The solution first sorts the input array of size n, which takes O(n log n) time. After sorting, a binary search is performed on the range of possible differences (from 0 to the maximum difference), which takes O(log m) time where m is the range of potential differences (at most the difference between max and min of the input array which is still bounded by the input values). Inside the binary search, the algorithm iterates through the sorted array once to count the number of valid pairs for the given difference, taking O(n) time. Therefore, the overall time complexity is dominated by the sorting and the binary search with the pair checking, giving O(n log n) + O(log m) * O(n). Since m is bounded, the final runtime is O(n log n).

Space Complexity

O(1) – The algorithm sorts the input array in-place, so it doesn't require extra space proportional to the input size N. It uses a few variables to store the left and right boundaries for the binary search, and a counter to track the number of pairs formed. The number of such variables is constant and independent of the input size N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 if p is 0, otherwise raise an IllegalArgumentException because it's impossible to form any pairs.
Input array with less than 2p elements (nums.length < 2p)	Return the smallest possible integer if p is 0, otherwise raise an IllegalArgumentException because not enough elements exist to form p pairs.
p is 0	Return 0 since no pairs are required, minimizing the maximum difference is trivially achieved.
Array contains duplicate values	The algorithm based on sorting and binary search handles duplicates correctly, as the difference between duplicates will be 0.
Array contains negative numbers	The absolute difference calculation \|nums[i] - nums[j]\| correctly handles negative numbers.
Array contains very large numbers that might lead to integer overflow when calculating the difference	Use long data type for intermediate calculations of differences to avoid overflow.
All numbers in the array are the same	The binary search will converge to 0 quickly, as all differences will be 0.
p is equal to nums.length / 2	The algorithm should handle the case where we need to create the maximum possible number of pairs efficiently.

Empty input array (nums is null or has length 0)

How to Handle:

Return 0 if p is 0, otherwise raise an IllegalArgumentException because it's impossible to form any pairs.

Input array with less than 2*p elements (nums.length < 2*p)

How to Handle:

Return the smallest possible integer if p is 0, otherwise raise an IllegalArgumentException because not enough elements exist to form p pairs.

p is 0

How to Handle:

Return 0 since no pairs are required, minimizing the maximum difference is trivially achieved.

Array contains duplicate values

How to Handle:

The algorithm based on sorting and binary search handles duplicates correctly, as the difference between duplicates will be 0.

Array contains negative numbers

How to Handle:

The absolute difference calculation |nums[i] - nums[j]| correctly handles negative numbers.

Array contains very large numbers that might lead to integer overflow when calculating the difference

How to Handle:

Use long data type for intermediate calculations of differences to avoid overflow.

All numbers in the array are the same

How to Handle:

The binary search will converge to 0 quickly, as all differences will be 0.

p is equal to nums.length / 2

How to Handle:

The algorithm should handle the case where we need to create the maximum possible number of pairs efficiently.

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 constraints on the size of the input array `nums` and the value of `p`?
Can the elements in the `nums` array be negative, zero, or floating-point numbers?
If it's not possible to form `p` pairs, what should the function return?
Are duplicate numbers allowed in the `nums` array, and if so, how should they be handled?
Is it guaranteed that `p` will always be less than or equal to `nums.length / 2`?

Brute Force Solution

Approach

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

First, think about all the different ways you can pick pairs of numbers from the entire set of numbers.
Start by picking the first two numbers as a pair, then calculate the difference between them.
Next, pick the first number and the third number as a pair, calculate the difference, and so on, trying every possible pairing with the first number.
Then, move on to the second number and pair it with every number that comes after it, again calculating the differences.
Continue doing this until you've considered every possible pair of numbers.
Now, consider groups of multiple pairs. For each group of pairs, find the largest difference among all the pairs in that group.
Keep track of that largest difference for each possible group of pairs.
Finally, compare all the largest differences you've found from all the groups, and pick the smallest one. That's your answer.

Code Implementation

def minimize_maximum_difference_of_pairs_brute_force(numbers, number_of_pairs):
    number_of_elements = len(numbers)
    if number_of_elements < 2 * number_of_pairs:
        return -1

    minimum_maximum_difference = float('inf')

    # Iterate through all possible combinations of pairs.
    for i in range(1 << number_of_elements):
        pairs = []
        current_pair = []
        number_of_selected_elements = 0

        for j in range(number_of_elements):
            if (i >> j) & 1:
                number_of_selected_elements += 1
                current_pair.append(numbers[j])

                if len(current_pair) == 2:
                    pairs.append(abs(current_pair[0] - current_pair[1]))
                    current_pair = []

        # Ensure the correct number of pairs was selected.
        if len(pairs) == number_of_pairs and number_of_selected_elements == 2 * number_of_pairs:
            
            maximum_difference_in_current_pairs = max(pairs)

            #Keep track of minimum of the maximum differences
            minimum_maximum_difference = min(minimum_maximum_difference, maximum_difference_in_current_pairs)

    if minimum_maximum_difference == float('inf'):
        return -1
    return minimum_maximum_difference

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, arrange the numbers in order from smallest to largest. This makes it easier to pair them later.
Think of a range of possible answers. The smallest possible answer is zero (if two numbers are the same), and the largest is the difference between the biggest and smallest numbers.
Pick a number from that range as a 'guess' for the maximum difference we're allowed between pairs.
Go through the sorted numbers and try to create as many pairs as possible where the difference between the numbers in each pair is no more than our 'guess'.
If we can create enough pairs using our 'guess', it means the actual minimum possible maximum difference is probably equal to or smaller than our guess. If we can't create enough pairs, it means our guess was too small.
Adjust our 'guess' based on whether we could form enough pairs. If we made enough pairs, we try a smaller guess. If not, we try a larger guess.
Keep repeating the guessing and checking process, narrowing down the range of possible answers until we find the smallest possible maximum difference that allows us to create the required number of pairs.

Code Implementation

def minimize_the_maximum_difference_of_pairs(numbers, number_of_pairs):
    numbers.sort()

    def can_form_pairs(maximum_difference):
        pairs_formed = 0
        index = 0
        while index < len(numbers) - 1:
            if numbers[index + 1] - numbers[index] <= maximum_difference:
                pairs_formed += 1
                index += 2
            else:
                index += 1
        return pairs_formed >= number_of_pairs

    left_bound = 0
    right_bound = numbers[-1] - numbers[0]
    minimum_maximum_difference = right_bound

    # Binary search to find the minimum maximum difference
    while left_bound <= right_bound:
        middle = (left_bound + right_bound) // 2

        # Check if we can form the required number of pairs
        if can_form_pairs(middle):
            minimum_maximum_difference = middle

            # Try to minimize the maximum difference further
            right_bound = middle - 1
        else:

            # Need a larger difference to form enough pairs
            left_bound = middle + 1

    return minimum_maximum_difference

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 if p is 0, otherwise raise an IllegalArgumentException because it's impossible to form any pairs.
Input array with less than 2p elements (nums.length < 2p)	Return the smallest possible integer if p is 0, otherwise raise an IllegalArgumentException because not enough elements exist to form p pairs.
p is 0	Return 0 since no pairs are required, minimizing the maximum difference is trivially achieved.
Array contains duplicate values	The algorithm based on sorting and binary search handles duplicates correctly, as the difference between duplicates will be 0.
Array contains negative numbers	The absolute difference calculation \|nums[i] - nums[j]\| correctly handles negative numbers.
Array contains very large numbers that might lead to integer overflow when calculating the difference	Use long data type for intermediate calculations of differences to avoid overflow.
All numbers in the array are the same	The binary search will converge to 0 quickly, as all differences will be 0.
p is equal to nums.length / 2	The algorithm should handle the case where we need to create the maximum possible number of pairs efficiently.

Empty input array (nums is null or has length 0)

How to Handle:

Return 0 if p is 0, otherwise raise an IllegalArgumentException because it's impossible to form any pairs.

Input array with less than 2*p elements (nums.length < 2*p)

How to Handle:

Return the smallest possible integer if p is 0, otherwise raise an IllegalArgumentException because not enough elements exist to form p pairs.

p is 0

How to Handle:

Return 0 since no pairs are required, minimizing the maximum difference is trivially achieved.

Array contains duplicate values

How to Handle:

The algorithm based on sorting and binary search handles duplicates correctly, as the difference between duplicates will be 0.

Array contains negative numbers

How to Handle:

The absolute difference calculation |nums[i] - nums[j]| correctly handles negative numbers.

Array contains very large numbers that might lead to integer overflow when calculating the difference

How to Handle:

Use long data type for intermediate calculations of differences to avoid overflow.

All numbers in the array are the same

How to Handle:

The binary search will converge to 0 quickly, as all differences will be 0.

p is equal to nums.length / 2

How to Handle:

The algorithm should handle the case where we need to create the maximum possible number of pairs efficiently.

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