Equal Sum Arrays With Minimum Number of Operations

Medium

American Express

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Topics:

ArraysGreedy Algorithms

You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.

In one operation, you can change any integer's value in any of the arrays to any value between 1 and 6, inclusive.

Return the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. Return -1 if it is not possible to make the sum of the two arrays equal.

Example 1:

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].

Example 2:

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.

Example 3:

Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed. 
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].

Constraints:

1 <= nums1.length, nums2.length <= 10⁵
1 <= nums1[i], nums2[i] <= 6

Try coding it in LeetCode

Equal Sum Arrays With Minimum Number of Operations

Medium

American Express

2 views

Topics:

ArraysGreedy Algorithms

You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.

In one operation, you can change any integer's value in any of the arrays to any value between 1 and 6, inclusive.

Return the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. Return -1 if it is not possible to make the sum of the two arrays equal.

Example 1:

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].

Example 2:

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.

Example 3:

Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed. 
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].

Constraints:

1 <= nums1.length, nums2.length <= 10⁵
1 <= nums1[i], nums2[i] <= 6

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 `nums1` and `nums2`? What is the maximum possible length of each array?
What is the range of values within the input arrays `nums1` and `nums2` before any operations are performed? Can I assume they are positive integers as indicated in the problem description?
If it's impossible to make the sums equal, the problem statement says to return -1. Are there any other conditions under which -1 should be returned?
Are there any restrictions on the order in which I choose elements from the arrays to modify, or can I choose them arbitrarily?
In the case where multiple sequences of operations achieve the minimum number of operations, is any one of those sequences acceptable?

Brute Force Solution

Approach

The brute force approach to solving this problem involves trying out every possible combination of moves we can make. We explore all ways to make the two arrays have equal sums, keeping track of how many moves each way takes.

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

First, we calculate the initial sums of both number collections.
If one collection's total is already bigger than the other, we try changing a number in the bigger collection to a smaller number, or a number in the smaller collection to a bigger number.
We continue doing this, one change at a time, until the totals are equal. We remember how many changes we made.
We also explore different change combinations: maybe swapping two pairs of numbers, or swapping three, and so on. We keep track of the changes each time.
We try every single combination of number swaps until we have tried them all, or until we find a solution that takes fewer moves than all the other solutions we've found so far.
Finally, after trying literally everything, we pick the combination of changes that made the totals equal with the fewest number of individual changes.

Code Implementation

def equal_sum_arrays_brute_force(first_array, second_array):
    minimum_operations = float('inf')

    def calculate_array_sum(array):
        array_sum = 0
        for number in array:
            array_sum += number
        return array_sum

    def find_equal_sum(current_first_array, current_second_array, operations_count):
        nonlocal minimum_operations

        first_array_sum = calculate_array_sum(current_first_array)
        second_array_sum = calculate_array_sum(current_second_array)

        if first_array_sum == second_array_sum:
            minimum_operations = min(minimum_operations, operations_count)
            return

        if operations_count >= minimum_operations:
            return

        # Check all possible moves to reduce larger array sum
        if first_array_sum > second_array_sum:
            for first_index in range(len(current_first_array)):
                for change_value in range(1, current_first_array[first_index]):
                    temp_first_array = current_first_array[:]
                    temp_first_array[first_index] = current_first_array[first_index] - change_value
                    find_equal_sum(temp_first_array, current_second_array, operations_count + 1)

        elif second_array_sum > first_array_sum:
            for second_index in range(len(current_second_array)):
                for change_value in range(1, 7 - current_second_array[second_index]):
                    temp_second_array = current_second_array[:]
                    temp_second_array[second_index] = current_second_array[second_index] + change_value
                    find_equal_sum(current_first_array, temp_second_array, operations_count + 1)

    find_equal_sum(first_array, second_array, 0)

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

Big(O) Analysis

Time Complexity

O(6^n) – The described brute force approach explores every possible combination of moves. In the worst case, we might need to consider changing each element in both arrays to every possible value between 1 and 6. If the combined size of the two arrays is n, each element has up to 6 possible values. This leads to a search space that grows exponentially with the input size, specifically on the order of 6 raised to the power of n. Thus, the time complexity is O(6^n), although in practicality the constant 6 may be different depending on the actual number constraints. The complexity becomes apparent in the enumeration of combinations needed, making it inefficient for larger inputs.

Space Complexity

O(N^K) – The brute force approach explores every possible combination of moves, where N is the length of the longer input array and K is the maximum number of operations explored in a single branch of the search. The recursion depth can go up to K, and at each level, we are storing possible combinations or states of the arrays which can be up to N in size. Therefore, the recursion stack and storing these combinations contribute to a space complexity of O(N^K), where N represents the maximum length of the input arrays and K represents the depth of recursion which is correlated to the number of operations.

Optimal Solution

Approach

The goal is to make the sums of two sets of numbers equal by changing numbers in each set. We want to do this with the fewest possible changes. The best approach involves understanding which set needs to increase or decrease its sum and then strategically making the most impactful changes to the numbers in each set.

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

First, calculate the sum of the numbers in each of the two sets.
Determine which set has the smaller sum and which set has the larger sum.
If the sums are already equal, then we don't need to make any changes, so the answer is zero.
Otherwise, we want to increase the numbers in the set with the smaller sum and decrease the numbers in the set with the larger sum. We need to do this in the most efficient way.
To increase the smaller sum as quickly as possible, we want to change the smallest numbers in the smaller set to the largest possible value (which is 6).
To decrease the larger sum as quickly as possible, we want to change the largest numbers in the larger set to the smallest possible value (which is 1).
We prioritize these two changes (increasing the smaller sum and decreasing the larger sum) to reduce the difference between the two sums as quickly as we can.
Keep track of how many changes we make along the way.
If we can't make the sums equal, it means there's no solution.
The total number of changes we make is the answer.

Code Implementation

def min_operations(first_array, second_array):
    first_array_sum = sum(first_array)
    second_array_sum = sum(second_array)

    if first_array_sum == second_array_sum:
        return 0

    if first_array_sum < second_array_sum:
        smaller_array = first_array
        larger_array = second_array
    else:
        smaller_array = second_array
        larger_array = first_array

    smaller_array.sort()
    larger_array.sort(reverse=True)

    smaller_array_index = 0
    larger_array_index = 0
    operations_count = 0

    # Need to equalize sums
    while first_array_sum != second_array_sum:
        # If no more operations can be done, impossible to equalize
        if smaller_array_index == len(smaller_array) and larger_array_index == len(larger_array):
            return -1

        # Prioritize increasing the smaller array
        if (smaller_array_index < len(smaller_array) and 
            larger_array_index < len(larger_array) and
            larger_array[larger_array_index] - 1 > 6 - smaller_array[smaller_array_index]):

            difference = larger_array[larger_array_index] - 1
            second_array_sum -= difference
            first_array_sum += difference

            larger_array_index += 1

        elif smaller_array_index < len(smaller_array):
            difference = 6 - smaller_array[smaller_array_index]
            second_array_sum -= difference
            first_array_sum += difference

            smaller_array_index += 1

        else:
            difference = larger_array[larger_array_index] - 1
            second_array_sum -= difference
            first_array_sum += difference
            larger_array_index += 1

        operations_count += 1

        # If it took too many steps, it's impossible
        if operations_count > (len(first_array) + len(second_array)):
            return -1

    return operations_count

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm calculates sums of two arrays, which takes O(n) time where n is the length of the longer array. The core of the algorithm involves sorting one or both arrays to efficiently find the optimal numbers to change. Sorting the arrays takes O(n log n) time using efficient sorting algorithms like merge sort or heap sort. After sorting, the algorithm iterates through the sorted arrays performing constant time operations to reduce the difference. Thus the dominant operation is sorting, resulting in an overall time complexity of O(n log n).

Space Complexity

O(1) – The described algorithm calculates sums, compares values, and modifies array elements in place. It primarily uses a fixed number of integer variables to store sums, differences, and a count of operations. No auxiliary data structures like lists or hash maps are created that scale with the input size. Therefore, the space complexity is constant, independent of the sizes of the input arrays.

Edge Cases

Case	How to Handle
One or both input arrays are null or empty.	Return -1 if either array is null or empty because it's impossible to make the sums equal.
The sum of nums1 is significantly smaller than the sum of nums2, and no number of operations can equalize them.	Return -1 if max possible sum of nums1 is still less than the min possible sum of nums2 or vice versa.
The lengths of the input arrays are very different, potentially leading to inefficiency if one array is extremely large.	Optimize by always iterating through the smaller array, or using a counting sort-like structure for large arrays.
All elements in one array are 1 and all elements in the other array are 6.	This represents a worst-case scenario and tests the efficiency of the operation counting logic.
Arrays are already equal in sum.	Return 0 immediately as no operations are needed.
Integer overflow when calculating sums for very large arrays and element values.	Use long data type to store the sums to prevent integer overflow during calculations.
Input array contains a large number of duplicates.	The counting sort or frequency map approach handles duplicates efficiently without needing special considerations.
The difference between the sums is a prime number greater than 6	Ensure the algorithm correctly iterates and checks all possible changes to reach the required sum difference, even with non-trivial prime differences.