Minimum Sum of Values by Dividing Array

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ArraysDynamic ProgrammingBit Manipulation

You are given two arrays nums and andValues of length n and m respectively.

The value of an array is equal to the last element of that array.

You have to divide nums into m disjoint contiguous subarrays such that for the i^th subarray [l_i, r_i], the bitwise AND of the subarray elements is equal to andValues[i], in other words, nums[l_i] & nums[l_i + 1] & ... & nums[r_i] == andValues[i] for all 1 <= i <= m, where & represents the bitwise AND operator.

Return the minimum possible sum of the values of the m subarrays nums is divided into. If it is not possible to divide nums into m subarrays satisfying these conditions, return -1.

Example 1:

Input: nums = [1,4,3,3,2], andValues = [0,3,3,2]

Output: 12

Explanation:

The only possible way to divide nums is:

[1,4] as 1 & 4 == 0.
[3] as the bitwise AND of a single element subarray is that element itself.
[3] as the bitwise AND of a single element subarray is that element itself.
[2] as the bitwise AND of a single element subarray is that element itself.

The sum of the values for these subarrays is 4 + 3 + 3 + 2 = 12.

Example 2:

Input: nums = [2,3,5,7,7,7,5], andValues = [0,7,5]

Output: 17

Explanation:

There are three ways to divide nums:

[[2,3,5],[7,7,7],[5]] with the sum of the values 5 + 7 + 5 == 17.
[[2,3,5,7],[7,7],[5]] with the sum of the values 7 + 7 + 5 == 19.
[[2,3,5,7,7],[7],[5]] with the sum of the values 7 + 7 + 5 == 19.

The minimum possible sum of the values is 17.

Example 3:

Input: nums = [1,2,3,4], andValues = [2]

Output: -1

Explanation:

The bitwise AND of the entire array nums is 0. As there is no possible way to divide nums into a single subarray to have the bitwise AND of elements 2, return -1.

Constraints:

1 <= n == nums.length <= 10⁴
1 <= m == andValues.length <= min(n, 10)
1 <= nums[i] < 10⁵
0 <= andValues[j] < 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 within the array? Can they be negative, zero, or floating-point numbers?
What are the allowed sizes of the two subarrays after division? Are there any restrictions on how many elements must be in each?
If it's impossible to divide the array in a way that results in a minimum sum, what should the function return?
Is the goal to minimize the *sum* of the subarray values, or to minimize the *difference* between the subarray sums (i.e., find subarrays whose sums are as close as possible)?
Are there any size limits on the input array?

Brute Force Solution

Approach

The brute force approach is all about trying absolutely everything. We'll explore every single possible way to split up the list of numbers into different groups, and then figure out which split gives us the smallest sum.

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

First, let's consider the scenario where the first group has only the very first number.
Next, let's consider the scenario where the first group has the first two numbers.
We'll continue increasing the size of the first group, one number at a time, until it contains all the numbers.
For each of these possible sizes of the first group, we'll calculate its 'value' (based on the problem's rules).
Then, for whatever is left over after making the first group, we repeat the same process of creating smaller subgroups, calculating each group's value.
We keep doing this until all the numbers have been assigned to a group.
Whenever we complete a possible division of the original list into groups, we add up each of the group 'values' to get the final sum for that division.
Finally, we compare all these final sums, and the smallest one is our answer.

Code Implementation

def minimum_sum_of_values_brute_force(numbers):
    def calculate_group_value(group):
        if not group:
            return 0
        return sum(group)

    def calculate_minimum_sum(current_numbers):
        # If the list is empty, the sum is zero.
        if not current_numbers:
            return 0

        minimum_total_value = float('inf')
        
        # Iterate through all possible sizes of the first group
        for first_group_size in range(1, len(current_numbers) + 1):
            first_group = current_numbers[:first_group_size]
            
            # Recursively calculate for the rest of the array
            remaining_numbers = current_numbers[first_group_size:]

            first_group_value = calculate_group_value(first_group)

            remaining_value = calculate_minimum_sum(remaining_numbers)
            
            total_value = first_group_value + remaining_value

            # Update the minimum value if necessary
            minimum_total_value = min(minimum_total_value, total_value)
            
        return minimum_total_value

    return calculate_minimum_sum(numbers)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves exploring all possible ways to divide the array into subarrays. Each element can either belong to the current subarray being formed or start a new subarray. This binary choice for each of the n elements results in 2^n possible combinations. Therefore, the algorithm effectively explores all possible subsets, leading to an exponential time complexity. The cost is driven by the recursive exploration of all possible partitions of the input array.

Space Complexity

O(N) – The brute force approach uses recursion to explore all possible splits of the array. The maximum depth of the recursion is N, where N is the number of elements in the array, corresponding to the scenario where each element forms its own group. Each recursive call creates a new stack frame, which consumes memory. Therefore, the space complexity is proportional to the depth of the recursion, which is O(N).

Optimal Solution

Approach

The goal is to divide a list of numbers into smaller groups, and find the smallest total we can get after calculating a 'score' for each group. The trick is to build up the solution little by little, remembering the best result we've seen so far for each point in the list. This helps us avoid redoing calculations and speeds things up significantly.

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

Imagine you're walking through the list of numbers, one at a time.
For each number, consider all the possible groups you can create that end with that number.
Calculate the 'score' for each of these groups.
Remember the best (smallest) total 'score' you've found so far to reach the end of that group.
When you reach the end of the list, the smallest 'score' you've remembered is your answer.

Code Implementation

def minimum_sum_of_values(values):
    number_of_values = len(values)

    # Initialize the DP table with a large default value.
    minimum_sums = [float('inf')] * (number_of_values + 1)
    minimum_sums[0] = 0

    for i in range(1, number_of_values + 1):
        for j in range(1, i + 1):
            # Consider all possible groups ending at index i.

            group = values[i - j:i]
            group_score = calculate_group_score(group)

            # Find best minimum sum so far.
            minimum_sums[i] = min(minimum_sums[i],
                                  minimum_sums[i - j] + group_score)

    # Return minimum sum of the entire array.
    return minimum_sums[number_of_values]

def calculate_group_score(group):
    if not group:
        return 0
    return sum(group)

def min_sum_optimized(values):
    number_of_values = len(values)
    dp = [float('inf')] * (number_of_values + 1)
    dp[0] = 0

    for i in range(1, number_of_values + 1):
        for j in range(1, i + 1):
            sub_array = values[i-j:i]
            
            # Calculate group score for sub-array
            group_score = sum(sub_array)

            # Update minimum sum by adding group score to previous min
            dp[i] = min(dp[i], dp[i-j] + group_score)

    return dp[number_of_values]

def calculate_group_score_optimized(group):
    if not group:
        return 0

    return sum(group)

def minimum_sum(values):
    number_of_values = len(values)

    # Initialize DP table to store min sums for prefixes.
    minimum_sums = [float('inf')] * (number_of_values + 1)
    minimum_sums[0] = 0

    for i in range(1, number_of_values + 1):
        for j in range(1, i + 1):
            # Iterate through all possible group starting points.

            group = values[i - j:i]
            group_score = sum(group)

            # Update minimum sum with the current group's score.
            minimum_sums[i] = min(minimum_sums[i],
                                  minimum_sums[i - j] + group_score)

    # The final element holds min sum for entire array.
    return minimum_sums[number_of_values]

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array of n numbers. For each element at index i, it considers all possible subarrays ending at i. This involves calculating the 'score' for each subarray from index j to i, where j ranges from 0 to i. Therefore, for each element, there's a loop that iterates up to i times. The total number of operations can be approximated by the sum of integers from 1 to n, which is n*(n+1)/2. Simplifying this, we arrive at a time complexity of O(n²).

Space Complexity

O(N) – The explanation mentions remembering the best (smallest) total 'score' found so far to reach the end of each group. This implies the use of an auxiliary data structure, likely an array or list, to store these intermediate minimum scores. Since we're storing one score for each point in the list of numbers (size N), the space required grows linearly with the input size N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or undefined input array	Return null or throw an IllegalArgumentException, depending on problem specifications.
Empty array	Return a default value (e.g., Infinity or a large number) indicating no valid sum exists, or throw an exception depending on problem requirements.
Array with only one element	Return a default value (e.g., Infinity or a large number) or throw an exception as no division is possible.
Array with very large numbers, leading to potential integer overflow	Use long or consider using BigInteger to prevent overflow during intermediate calculations if needed.
Array contains negative numbers	Ensure the algorithm correctly handles negative numbers by considering absolute values or adapting the logic.
Array contains zero(s)	Handle the case where a subarray sum could be zero, which may affect the division result and the overall minimum sum.
Array with all the same values	Ensure the algorithm doesn't get stuck or produce incorrect results when all values are identical.
Array size exceeding memory constraints	If the array is extremely large, consider using an online algorithm or a divide-and-conquer approach to manage memory efficiently.

Null or undefined input array

How to Handle:

Return null or throw an IllegalArgumentException, depending on problem specifications.

Empty array

How to Handle:

Return a default value (e.g., Infinity or a large number) indicating no valid sum exists, or throw an exception depending on problem requirements.

Array with only one element

How to Handle:

Return a default value (e.g., Infinity or a large number) or throw an exception as no division is possible.

Array with very large numbers, leading to potential integer overflow

How to Handle:

Use long or consider using BigInteger to prevent overflow during intermediate calculations if needed.

Array contains negative numbers

How to Handle:

Ensure the algorithm correctly handles negative numbers by considering absolute values or adapting the logic.

Array contains zero(s)

How to Handle:

Handle the case where a subarray sum could be zero, which may affect the division result and the overall minimum sum.

Array with all the same values

How to Handle:

Ensure the algorithm doesn't get stuck or produce incorrect results when all values are identical.

Array size exceeding memory constraints

How to Handle:

If the array is extremely large, consider using an online algorithm or a divide-and-conquer approach to manage memory efficiently.