Minimize the Difference Between Target and Chosen Elements

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ArraysDynamic Programming

You are given an m x n integer matrix mat and an integer target.

Choose one integer from each row in the matrix such that the absolute difference between target and the sum of the chosen elements is minimized.

Return the minimum absolute difference.

The absolute difference between two numbers a and b is the absolute value of a - b.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], target = 13
Output: 0
Explanation: One possible choice is to:
- Choose 1 from the first row.
- Choose 5 from the second row.
- Choose 7 from the third row.
The sum of the chosen elements is 13, which equals the target, so the absolute difference is 0.

Example 2:

Input: mat = [[1],[2],[3]], target = 100
Output: 94
Explanation: The best possible choice is to:
- Choose 1 from the first row.
- Choose 2 from the second row.
- Choose 3 from the third row.
The sum of the chosen elements is 6, and the absolute difference is 94.

Example 3:

Input: mat = [[1,2,9,8,7]], target = 6
Output: 1
Explanation: The best choice is to choose 7 from the first row.
The absolute difference is 1.

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 70
1 <= mat[i][j] <= 70
1 <= target <= 800

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 is the range of values for the elements within the matrices and the target value? Are negative numbers allowed?
Can any of the input matrices be empty, or have rows with varying lengths?
If multiple combinations of elements minimize the difference, should I return any one of them or is there a specific selection criterion (e.g., lexicographically smallest)?
What should I return if it's impossible to choose one element from each matrix such that the difference with the target is minimized (e.g., if the input matrices are all empty)?
Are we optimizing primarily for correctness, or are there specific memory constraints I should be aware of?

Brute Force Solution

Approach

The brute force approach to this problem is all about exploring every single possible combination. We will select one number from each of the given lists in every way imaginable. After each selection, we'll calculate the difference between the sum of the selected numbers and the target, and then find the smallest difference across all combinations.

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

Start by picking one number from the first list.
For that chosen number, consider every possible number from the second list.
For each pair of numbers, consider every possible number from the third list.
Continue this process until you've picked one number from each list.
Calculate the sum of the numbers you picked from all the lists.
Find the difference between that sum and the target number.
Remember this difference. If it's smaller than any difference you found before, remember that difference instead.
Repeat steps 1-7 for every single possible combination of numbers from the lists.
Finally, after you've tried all the combinations, return the smallest difference you found.

Code Implementation

def minimize_the_difference_brute_force(lists_of_numbers, target_number):
    minimum_difference = float('inf')

    def find_minimum_difference(current_index, current_sum):
        nonlocal minimum_difference

        # If we've gone through all lists, calculate the difference and update the minimum.
        if current_index == len(lists_of_numbers):

            difference = abs(current_sum - target_number)
            minimum_difference = min(minimum_difference, difference)
            return

        # Iterate through the current list.
        for number in lists_of_numbers[current_index]:

            # Recursively call the function with the updated sum and next list index.
            find_minimum_difference(current_index + 1, current_sum + number)

    # Start the recursion with the first list and an initial sum of 0.
    find_minimum_difference(0, 0)

    return minimum_difference

Big(O) Analysis

Time Complexity

O(m^n) – The provided brute force approach explores all possible combinations of choosing one element from each list. Assuming there are 'n' lists and each list has 'm' elements, the algorithm iterates through 'm' choices for the first list, then 'm' choices for the second list, and so on, for all 'n' lists. Therefore, the total number of combinations is m * m * ... * m (n times), which is m^n. Consequently, the time complexity is O(m^n).

Space Complexity

O(1) – The brute force approach described iterates through the lists without storing intermediate results in any significant data structure. It only maintains variables to keep track of the current combination being considered (e.g., the current index in each list) and to store the minimum difference found so far. The space required for these variables remains constant regardless of the size or number of input lists, therefore the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to pick one number from each row of a grid such that their sum is as close as possible to a given target. We can solve this by building up the sums incrementally, keeping track of the best possible sums we can achieve as we go from row to row.

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

Consider the first row. Note down all the possible sums you can achieve by just picking one number from this row. These become your initial sums.
Now, look at the second row. For each sum you recorded in the previous step, try adding each number from the second row to it. This generates new sums. Keep track of these.
Repeat this process for each row: take the sums you've generated so far, and add each number from the current row to them. Record the results.
After processing all rows, you will have a collection of possible sums. Find the sum that is closest to the target value. That's the best possible sum you can achieve.
The key is that you are progressively building sums, row by row, always working towards the closest result without exhaustively trying every single combination of numbers from the grid.

Code Implementation

def minimize_the_difference(
    grid_of_numbers, target_value):
    possible_sums = {0}

    for row_of_numbers in grid_of_numbers:
        new_sums = set()

        # Build new sums by adding each row value
        # to existing possible sums.
        for current_sum in possible_sums:
            for number in row_of_numbers:
                new_sums.add(current_sum + number)

        possible_sums = new_sums

    # Find the closest sum to the target
    # value from all possible sums.
    minimum_difference = float('inf')
    closest_sum = None

    for final_sum in possible_sums:
        difference = abs(final_sum - target_value)
        if difference < minimum_difference:
            minimum_difference = difference
            closest_sum = final_sum

    return minimum_difference

Big(O) Analysis

Time Complexity

O(m * n * k) – Let m be the number of rows in the grid, n be the maximum number of columns in any row, and k be the number of distinct possible sums after processing each row. For each row, we iterate through all possible previous sums (k) and for each of these sums, we iterate through the elements of the current row (maximum n). Thus, processing each row has a time complexity of O(n * k). Since we have m rows, we perform this process m times, leading to a total time complexity of O(m * n * k). Since we are trying to minimize the difference with the target, k can potentially grow proportionally with the values in the grid.

Space Complexity

O(M * target) – The auxiliary space is dominated by the storage of possible sums at each step. In the worst case, after processing each row, we might have sums ranging from a minimum value to a maximum value that could be as large as the target itself. If M is the number of columns, then each number of the first row can be added to the next row, and so on. This leads to storing up to target unique sums after each row is processed, leading to O(M * target) space needed to keep track of these sums, where M is the number of columns, and target is the given target value, with the sums updated at each stage to be closer to the target value as possible.

Edge Cases

Case	How to Handle
Empty matrix (null or zero rows/columns)	Return 0, indicating a difference of zero is achievable with an empty selection.
Matrix with only one row	Return the absolute difference between the single element in the row and the target.
Target is significantly smaller than all possible sums	The algorithm should converge to the smallest possible sum, minimizing the difference with the target.
Target is significantly larger than all possible sums	The algorithm should converge to the largest possible sum, minimizing the difference with the target.
Large matrix dimensions (potential memory issues with DP)	Consider using a space-optimized DP approach or pruning intermediate results to avoid excessive memory usage.
Matrix contains negative numbers	The algorithm should correctly handle negative numbers when calculating the minimum difference.
Matrix contains very large numbers (potential integer overflow)	Use appropriate data types (e.g., long) to prevent integer overflow during summation or absolute difference calculations.
All elements in each row are identical	The optimal solution would be the sum of any element from each row; handle this edge case to improve performance.

Empty matrix (null or zero rows/columns)

How to Handle:

Return 0, indicating a difference of zero is achievable with an empty selection.

Matrix with only one row

How to Handle:

Return the absolute difference between the single element in the row and the target.

Target is significantly smaller than all possible sums

How to Handle:

The algorithm should converge to the smallest possible sum, minimizing the difference with the target.

Target is significantly larger than all possible sums

How to Handle:

The algorithm should converge to the largest possible sum, minimizing the difference with the target.

Large matrix dimensions (potential memory issues with DP)

How to Handle:

Consider using a space-optimized DP approach or pruning intermediate results to avoid excessive memory usage.

Matrix contains negative numbers

How to Handle:

The algorithm should correctly handle negative numbers when calculating the minimum difference.

Matrix contains very large numbers (potential integer overflow)

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow during summation or absolute difference calculations.

All elements in each row are identical

How to Handle:

The optimal solution would be the sum of any element from each row; handle this edge case to improve performance.