Minimum Operations to Make Array Equal II

Medium

Asked by:

4 views

Topics:

ArraysGreedy Algorithms

You are given two integer arrays nums1 and nums2 of equal length n and an integer k. You can perform the following operation on nums1:

Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k. In other words, nums1[i] = nums1[i] + k and nums1[j] = nums1[j] - k.

nums1 is said to be equal to nums2 if for all indices i such that 0 <= i < n, nums1[i] == nums2[i].

Return the minimum number of operations required to make nums1 equal to nums2. If it is impossible to make them equal, return -1.

Example 1:

Input: nums1 = [4,3,1,4], nums2 = [1,3,7,1], k = 3
Output: 2
Explanation: In 2 operations, we can transform nums1 to nums2.
1^st operation: i = 2, j = 0. After applying the operation, nums1 = [1,3,4,4].
2^nd operation: i = 2, j = 3. After applying the operation, nums1 = [1,3,7,1].
One can prove that it is impossible to make arrays equal in fewer operations.

Example 2:

Input: nums1 = [3,8,5,2], nums2 = [2,4,1,6], k = 1
Output: -1
Explanation: It can be proved that it is impossible to make the two arrays equal.

Constraints:

n == nums1.length == nums2.length
2 <= n <= 10⁵
0 <= nums1[i], nums2[j] <= 10⁹
0 <= k <= 10⁵

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 values in `nums1`, `nums2`, and the value of `k`? Can they be negative, zero, or very large?
What should I return if `nums1` and `nums2` cannot be made equal with the given operation?
Is `k` guaranteed to be a positive integer?
Can I assume that the lengths of `nums1` and `nums2` will always be equal, and will never be empty or null?
If there are multiple ways to achieve the minimum number of operations, is any valid solution acceptable?

Brute Force Solution

Approach

The brute force approach for this problem involves trying out every possible combination of adjustments to the numbers. We want to find the fewest moves needed to make the numbers in the first list the same as the numbers in the second list, given we can only change them by a specific amount in each move.

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

First, think about all the possible values the numbers in the first list could eventually have. Since we can only adjust by a certain amount, these values are limited.
Next, try setting all numbers that need to be increased to one possible target value and all numbers that need to be decreased to another possible target value.
Calculate how many moves it would take to change the first list's numbers to match the second list's numbers for each possible pairing of target values.
Keep track of the minimum number of moves found so far.
Repeat steps 2 and 3 for all possible combinations of target values.
Finally, after checking all the combinations, return the minimum number of moves you recorded.

Code Implementation

def min_operations_brute_force(first_array, second_array, difference):
    min_operations = float('inf')

    def arrays_are_equal(array1, array2):
        if len(array1) != len(array2):
            return False
        for index in range(len(array1)):
            if array1[index] != array2[index]:
                return False
        return True

    def find_min_operations(current_first_array, operations_count):
        nonlocal min_operations

        if arrays_are_equal(current_first_array, second_array):
            min_operations = min(min_operations, operations_count)
            return

        if operations_count >= min_operations:
            return

        for index in range(len(first_array)):
            # Try adding multiples of difference to the current element
            for multiplier in range(1, 3):
                new_array = current_first_array[:]
                new_array[index] += difference * multiplier

                find_min_operations(new_array, operations_count + multiplier)

            # Try subtracting multiples of difference from the current element
            for multiplier in range(1, 3):
                new_array = current_first_array[:]
                new_array[index] -= difference * multiplier

                # Check all possible subtractions
                find_min_operations(new_array, operations_count + multiplier)

    # Initiate recursive search starting from the original first array
    find_min_operations(first_array[:], 0)

    if min_operations == float('inf'):
        return -1
    else:
        return min_operations

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through all possible pairs of target values for increasing and decreasing elements. In the worst case, each element in the input array of size n could potentially be a target value for both increasing and decreasing. Therefore, we might consider all pairs of elements, resulting in a nested loop structure. The outer loop iterates through n possible target values for increasing, and the inner loop iterates through n possible target values for decreasing. For each pair of target values, it iterates through the array again to calculate the number of moves required. This results in roughly n * n * n operations. Simplification yields O(n^2).

Space Complexity

O(1) – The provided brute force approach, as described, does not explicitly use any auxiliary data structures that scale with the input size N (where N is the number of elements in the arrays). It primarily involves iterating and comparing values, potentially storing a constant number of variables to track the minimum number of moves. Therefore, the auxiliary space used is constant regardless of the input size. Thus, the space complexity is O(1).

Optimal Solution

Approach

The goal is to equalize two number lists with a fixed change value. The optimal solution focuses on the differences between corresponding numbers and ensures these differences balance out using the allowed change value.

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

First, calculate the difference between each pair of corresponding numbers from the two lists.
Separate the differences into two groups: positive differences and negative differences.
Recognize that to equalize the lists, every positive difference needs a corresponding negative difference to cancel it out. In other words, we're transferring values from elements that are 'too high' to elements that are 'too low'.
Verify that the total of all positive differences exactly equals the absolute value of the total of all negative differences. If they aren't equal, it's impossible to equalize the lists.
Focus on the positive differences. For each positive difference, check if it can be achieved by adding the fixed change value a whole number of times. If not, it's also impossible to equalize.
If all checks pass, determine how many times you need to apply the fixed change value to each positive difference to reach the required amount. Sum up these counts to find the minimum number of operations needed.

Code Implementation

def min_operations(nums1, nums2, change_limit):
    total_difference = 0
    for i in range(len(nums1)):
        total_difference += nums1[i] - nums2[i]

    # If total difference isn't divisible, arrays can't be equal
    if total_difference % change_limit != 0:
        return -1

    increase_sum = 0
    decrease_sum = 0

    for i in range(len(nums1)):
        difference = nums1[i] - nums2[i]

        # Categorize differences to calc increases and decreases
        if difference > 0:
            increase_sum += difference
        elif difference < 0:
            decrease_sum -= difference

    #Increases must equal decreases for arrays to be equal
    if increase_sum != decrease_sum:
        return -1

    # The number of moves is the sum of increases divided by limit
    return increase_sum // change_limit

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the input arrays of size n to calculate the differences (step 1). It then iterates through these differences to separate positive and negative values and to calculate their sums (steps 2 and 4). A final iteration is used to calculate the number of operations for the positive differences (step 6). Since each step involves a single loop through the array, the time complexity is dominated by these linear operations, resulting in O(n).

Space Complexity

O(N) – The algorithm calculates the differences between corresponding numbers from the two input lists, implying the creation of an auxiliary array or list to store these differences. If the input lists have N elements each, this difference array will also have a maximum size of N. The positive and negative differences are also conceptually grouped, contributing to further space usage dependent on N. Therefore, the auxiliary space used scales linearly with the input size N.

Edge Cases

Case	How to Handle
Empty or null input arrays	Return 0 if both are empty, or -1 if only one is empty, or if either is null.
Arrays of different lengths	Return -1 immediately as the problem states the arrays must be of equal length.
k is zero	If k is zero, check if nums1 and nums2 are equal; return 0 if they are, -1 otherwise.
Large input arrays (performance)	Ensure the solution uses linear time complexity to handle large arrays efficiently (e.g., avoid nested loops).
Integer overflow when calculating differences	Use long data type to store the differences between elements to prevent potential overflows.
The sum of differences is not divisible by k	Return -1 if the absolute value of the sum of (nums1[i]-nums2[i]) is not divisible by k, indicating an unsolvable state.
Positive differences do not equal negative differences	Ensure the sum of increases equals the sum of decreases; return -1 if they do not match since it's impossible to balance.
Negative numbers in input arrays	The solution should correctly handle negative numbers without modification as the problem statement does not restrict the sign of numbers.

Empty or null input arrays

How to Handle:

Return 0 if both are empty, or -1 if only one is empty, or if either is null.

Arrays of different lengths

How to Handle:

Return -1 immediately as the problem states the arrays must be of equal length.

k is zero

How to Handle:

If k is zero, check if nums1 and nums2 are equal; return 0 if they are, -1 otherwise.

Large input arrays (performance)

How to Handle:

Ensure the solution uses linear time complexity to handle large arrays efficiently (e.g., avoid nested loops).

Integer overflow when calculating differences

How to Handle:

Use long data type to store the differences between elements to prevent potential overflows.

The sum of differences is not divisible by k

How to Handle:

Return -1 if the absolute value of the sum of (nums1[i]-nums2[i]) is not divisible by k, indicating an unsolvable state.

Positive differences do not equal negative differences

How to Handle:

Ensure the sum of increases equals the sum of decreases; return -1 if they do not match since it's impossible to balance.

Negative numbers in input arrays

How to Handle:

The solution should correctly handle negative numbers without modification as the problem statement does not restrict the sign of numbers.