Minimum Number of Operations to Make Arrays Similar

Hard

Walmart

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Topics:

ArraysGreedy Algorithms

You are given two positive integer arrays nums and target, of the same length.

In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and:

set nums[i] = nums[i] + 2 and
set nums[j] = nums[j] - 2.

Two arrays are considered to be similar if the frequency of each element is the same.

Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target.

Example 1:

Input: nums = [8,12,6], target = [2,14,10]
Output: 2
Explanation: It is possible to make nums similar to target in two operations:
- Choose i = 0 and j = 2, nums = [10,12,4].
- Choose i = 1 and j = 2, nums = [10,14,2].
It can be shown that 2 is the minimum number of operations needed.

Example 2:

Input: nums = [1,2,5], target = [4,1,3]
Output: 1
Explanation: We can make nums similar to target in one operation:
- Choose i = 1 and j = 2, nums = [1,4,3].

Example 3:

Input: nums = [1,1,1,1,1], target = [1,1,1,1,1]
Output: 0
Explanation: The array nums is already similiar to target.

Constraints:

n == nums.length == target.length
1 <= n <= 10⁵
1 <= nums[i], target[i] <= 10⁶
It is possible to make nums similar to target.

Minimum Number of Operations to Make Arrays Similar

Hard

Walmart

0 views

Topics:

ArraysGreedy Algorithms

You are given two positive integer arrays nums and target, of the same length.

In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and:

set nums[i] = nums[i] + 2 and
set nums[j] = nums[j] - 2.

Two arrays are considered to be similar if the frequency of each element is the same.

Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target.

Example 1:

Input: nums = [8,12,6], target = [2,14,10]
Output: 2
Explanation: It is possible to make nums similar to target in two operations:
- Choose i = 0 and j = 2, nums = [10,12,4].
- Choose i = 1 and j = 2, nums = [10,14,2].
It can be shown that 2 is the minimum number of operations needed.

Example 2:

Input: nums = [1,2,5], target = [4,1,3]
Output: 1
Explanation: We can make nums similar to target in one operation:
- Choose i = 1 and j = 2, nums = [1,4,3].

Example 3:

Input: nums = [1,1,1,1,1], target = [1,1,1,1,1]
Output: 0
Explanation: The array nums is already similiar to target.

Constraints:

n == nums.length == target.length
1 <= n <= 10⁵
1 <= nums[i], target[i] <= 10⁶
It is possible to make nums similar to target.

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 and values within the arrays? Are negative numbers allowed?
If it's impossible to make the arrays similar, what value should I return?
Are duplicate numbers allowed within each array, and if so, how should they be handled?
Can you provide a more precise definition of 'similar'? Does it mean the arrays must be identical after applying the operations, or is a looser form of similarity acceptable?
If there are multiple sequences of operations that achieve the minimum, is any one of them acceptable, or is there a specific criteria for choosing between them?

Brute Force Solution

Approach

The brute force method for making arrays similar involves exploring all possible ways to transform one array into the other. This means we consider every combination of moves and alterations to see which one requires the fewest actions. It's like trying every key on a keyring until you find the one that unlocks the door.

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

First, look at each possible pair of numbers, one from the first set and one from the second set.
For each pair, calculate how many operations it would take to make the number from the first set match the number from the second set.
Consider every possible pairing between numbers from the two sets.
For each complete pairing, sum the number of operations needed for each pair to get the total number of operations for that pairing.
Keep track of the pairing that requires the smallest total number of operations.
After checking every possible pairing, report the smallest total number of operations found.

Code Implementation

def min_operations_brute_force(array1, array2):
    # Check if arrays have different lengths
    if len(array1) != len(array2):
        return -1

    minimum_operations = float('inf')

    import itertools

    # Iterate through all possible permutations to try every pairing
    for permutation in itertools.permutations(array2):
        current_operations = 0

        # Calculate the cost for the current permutation
        for index in range(len(array1)):
            current_operations += abs(array1[index] - permutation[index])

        # Keep track of the minimum number of operations required
        minimum_operations = min(minimum_operations, current_operations)

    return minimum_operations

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Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores all possible pairings between elements of two arrays of size n. Generating all possible pairings (permutations) between the two sets involves considering n! (n factorial) different arrangements. For each pairing, we iterate through all n pairs to compute the operations required to make the numbers match. Therefore, the time complexity is driven by generating all permutations, resulting in a time complexity of O(n!).

Space Complexity

O(N!) – The brute force method, as described, implicitly explores all possible pairings between the two input arrays. To keep track of the current pairing being evaluated and to avoid re-evaluating the same pairing, it requires storing the current permutation. In the worst case, we need to generate all possible permutations of array indices, which has a space requirement proportional to the number of permutations. Since there can be N! different pairings, the implicit call stack in the recursive implementation to generate each pairing will require O(N!) space where N is the number of elements in the arrays.

Optimal Solution

Approach

The problem asks to find the minimum number of operations to make two arrays similar. The trick is to realize that only odd numbers can be turned into other odd numbers, and even numbers into other even numbers. This lets us separate and compare only the relevant numbers.

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

First, separate the numbers in each group into two new groups: one containing only odd numbers and one containing only even numbers.
Next, sort each of these groups of odd and even numbers independently.
After sorting, compare the odd numbers in one group to the odd numbers in the other, and likewise for the even numbers.
For each pair of numbers at the corresponding position, find the difference between them. Add up all the positive differences that result from comparing corresponding positions within the odd number groups or the even number groups.
Since each operation effectively moves two numbers at once, divide the final sum of positive differences by two to get the minimum number of operations.

Code Implementation

def minimum_operations_to_make_arrays_similar(numbers_a, numbers_b): 
    odd_numbers_a = []
    even_numbers_a = []
    odd_numbers_b = []
    even_numbers_b = []

    for number in numbers_a:
        if number % 2 == 0:
            even_numbers_a.append(number)
        else:
            odd_numbers_a.append(number)

    for number in numbers_b:
        if number % 2 == 0:
            even_numbers_b.append(number)
        else:
            odd_numbers_b.append(number)

    odd_numbers_a.sort()
    even_numbers_a.sort()
    odd_numbers_b.sort()
    even_numbers_b.sort()

    positive_difference_sum = 0

    # Sum differences between odd numbers.
    for i in range(len(odd_numbers_a)): 
        difference = odd_numbers_a[i] - odd_numbers_b[i]
        if difference > 0:
            positive_difference_sum += difference

    # Sum differences between even numbers. 
    for i in range(len(even_numbers_a)): 
        difference = even_numbers_a[i] - even_numbers_b[i]
        if difference > 0:
            positive_difference_sum += difference

    # Each operation changes two numbers.
    return positive_difference_sum // 2

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log n) – Separating odd and even numbers takes O(n) time. Sorting the odd and even number subarrays, each of which could be of size n/2 in the worst case, takes O((n/2) log (n/2)) for each. Since we sort two such subarrays, the combined sorting time is 2 * O((n/2) log (n/2)), which simplifies to O(n log n). Comparing and summing the differences takes O(n) time. Therefore, the dominant factor is the sorting step, resulting in an overall time complexity of O(n log n).

Space Complexity

O(N) – The algorithm creates four new lists: two for storing odd numbers (one for each input array) and two for storing even numbers (one for each input array). In the worst case, all numbers in the input arrays could be either odd or even. Thus the size of each of these lists could be at most N, where N is the size of the input arrays. This results in an auxiliary space complexity of O(N) because the space used grows linearly with the size of the input.

Edge Cases

Case	How to Handle
Empty or null input arrays	Return 0 if either array is null or empty, as no operations are needed to make them similar.
Arrays with different lengths	Return -1 immediately if the arrays have different lengths because they cannot be made similar.
Arrays already similar	Return 0 if the sorted odd and even numbers are identical, as no operations are needed.
Large integer values in arrays causing overflow	Use long long for intermediate calculations to prevent integer overflow during difference calculation and division.
Arrays with only even or only odd numbers	The algorithm should correctly process arrays containing only even or only odd numbers by ensuring correct separation.
Arrays with a very large number of elements.	The sorting algorithm must be efficient (O(n log n)) to avoid timeout.
Arrays where the sum of odd numbers differs from the sum of odd numbers of the other array.	If the sums of the odd or even numbers are different then return -1 since the arrays cannot be made similar.
Input arrays contain duplicate odd or even numbers.	The algorithm correctly handles duplicates as it sorts and compares the lists of odd and even numbers.