Problem Description

You are given two arrays of integers arr1 and arr2 that have equal lengths. Your task is to find the maximum value of the expression:

|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|

where you need to consider all possible pairs of indices i and j such that 0 <= i, j < arr1.length.

Breaking down the expression:

• |arr1[i] - arr1[j]| represents the absolute difference between elements at positions i and j in the first array
• |arr2[i] - arr2[j]| represents the absolute difference between elements at positions i and j in the second array
• |i - j| represents the absolute difference between the indices themselves

You need to evaluate this expression for all possible pairs of indices and return the maximum value found.

For example, if you have arr1 = [1, 2, 3] and arr2 = [4, 5, 6], you would evaluate the expression for all combinations like (i=0, j=0), (i=0, j=1), (i=0, j=2), (i=1, j=0), etc., and return the largest result.

Intuition

The key insight is recognizing that dealing with absolute values directly is complex, but we can transform them into simpler expressions by considering all possible sign combinations.

Let's denote x_i = arr1[i] and y_i = arr2[i]. Without loss of generality, we can assume i >= j (since if i < j, we can swap them and get the same result due to the absolute values).

When i >= j, the expression becomes: |x_i - x_j| + |y_i - y_j| + (i - j)

The absolute value |x_i - x_j| can be either (x_i - x_j) or -(x_i - x_j), depending on which value is larger. Similarly, |y_i - y_j| can be either (y_i - y_j) or -(y_i - y_j).

This gives us 4 possible combinations:

1. (x_i - x_j) + (y_i - y_j) + (i - j) = (x_i + y_i + i) - (x_j + y_j + j)
2. (x_i - x_j) - (y_i - y_j) + (i - j) = (x_i - y_i + i) - (x_j - y_j + j)
3. -(x_i - x_j) + (y_i - y_j) + (i - j) = (-x_i + y_i + i) - (-x_j + y_j + j)
4. -(x_i - x_j) - (y_i - y_j) + (i - j) = (-x_i - y_i + i) - (-x_j - y_j + j)

Notice the pattern: each expression is of the form (a*x_i + b*y_i + i) - (a*x_j + b*y_j + j) where a, b ∈ {-1, 1}.

To maximize this difference, we need to find the maximum and minimum values of a*x_i + b*y_i + i for each combination of a and b. The maximum difference will be max_value - min_value for each sign combination.

By iterating through all four sign combinations and tracking the maximum and minimum values of the transformed expression for each index, we can find the overall maximum in a single pass through the arrays.

Learn more about Math patterns.

Solution Approach

Based on our mathematical transformation, we need to evaluate a * x_i + b * y_i + i for all combinations where a, b ∈ {-1, 1}. This gives us 4 combinations: (1, 1), (1, -1), (-1, 1), and (-1, -1).

The implementation uses a clever way to iterate through these sign combinations. The dirs tuple (1, -1, -1, 1, 1) is used with pairwise to generate consecutive pairs: (1, -1), (-1, -1), (-1, 1), and (1, 1), which represent our (a, b) combinations.

Here's how the algorithm works:

1. Initialize variables: Set ans = -inf to track the maximum result across all sign combinations.

2. Iterate through sign combinations: For each pair (a, b) from the dirs tuple:

   • Initialize mx = -inf to track the maximum value of a * x + b * y + i
   • Initialize mi = inf to track the minimum value of a * x + b * y + i

3. Process each index: For each index i and corresponding values (x, y) from (arr1[i], arr2[i]):

   • Calculate the expression value: a * x + b * y + i
   • Update mx to be the maximum seen so far
   • Update mi to be the minimum seen so far
   • Update ans with mx - mi, which represents the maximum difference for the current sign combination

4. Return the result: After checking all four sign combinations, ans contains the maximum value of the original expression.

The algorithm efficiently computes the answer in O(n) time by:

• Making a single pass through the arrays for each sign combination (4 passes total)
• Maintaining running maximum and minimum values instead of comparing all pairs
• Using the mathematical transformation to avoid dealing with absolute values directly

The space complexity is O(1) as we only use a constant amount of extra variables regardless of input size.

Example Walkthrough

Let's walk through a small example with arr1 = [1, 3] and arr2 = [2, 5].

We need to find the maximum value of |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|.

Step 1: Understand what we're calculating

For each sign combination (a, b), we calculate a * arr1[i] + b * arr2[i] + i for each index i, then find the maximum difference between any two values.

Step 2: Process each sign combination

Let's trace through each combination:

Combination 1: (a=1, b=-1)

• For i=0: 1*1 + (-1)*2 + 0 = 1 - 2 + 0 = -1
• For i=1: 1*3 + (-1)*5 + 1 = 3 - 5 + 1 = -1
• Maximum = -1, Minimum = -1
• Difference = -1 - (-1) = 0

Combination 2: (a=-1, b=-1)

• For i=0: (-1)*1 + (-1)*2 + 0 = -1 - 2 + 0 = -3
• For i=1: (-1)*3 + (-1)*5 + 1 = -3 - 5 + 1 = -7
• Maximum = -3, Minimum = -7
• Difference = -3 - (-7) = 4

Combination 3: (a=-1, b=1)

• For i=0: (-1)*1 + 1*2 + 0 = -1 + 2 + 0 = 1
• For i=1: (-1)*3 + 1*5 + 1 = -3 + 5 + 1 = 3
• Maximum = 3, Minimum = 1
• Difference = 3 - 1 = 2

Combination 4: (a=1, b=1)

• For i=0: 1*1 + 1*2 + 0 = 1 + 2 + 0 = 3
• For i=1: 1*3 + 1*5 + 1 = 3 + 5 + 1 = 9
• Maximum = 9, Minimum = 3
• Difference = 9 - 3 = 6

Step 3: Find the maximum

The maximum difference across all combinations is 6.

Verification: Let's verify by checking the original expression for pairs (i=0, j=1) and (i=1, j=0):

• For (i=1, j=0): |3-1| + |5-2| + |1-0| = 2 + 3 + 1 = 6 ✓
• For (i=0, j=1): |1-3| + |2-5| + |0-1| = 2 + 3 + 1 = 6 ✓

The algorithm correctly identifies that the maximum value is 6, which occurs when comparing indices 0 and 1. The transformation handles all the absolute value cases automatically by considering different sign combinations.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The code iterates through 4 direction pairs (from pairwise(dirs) which generates 4 pairs). For each direction pair, it performs a single pass through both arrays simultaneously using enumerate(zip(arr1, arr2)), where n is the length of the arrays. Inside the inner loop, all operations (max, min, arithmetic operations) are O(1). Therefore, the total time complexity is O(4 * n) = O(n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• dirs tuple with 5 elements: O(1)
• Variables ans, mx, mi, a, b, i, x, y: O(1)
• The pairwise function generates pairs on-the-fly without storing all pairs at once
• The zip function also creates an iterator that doesn't store all pairs in memory

No additional data structures that scale with input size are used, resulting in O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling of Absolute Value Cases

Pitfall: A common mistake is trying to handle the absolute values directly or missing some of the sign combinations when transforming the expression. The expression |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j| can be expanded into 8 different cases based on the signs, but since i < j can be assumed without loss of generality (due to symmetry), we only need 4 cases.

Why it happens: Developers might try to enumerate all 8 cases or incorrectly reduce them, leading to wrong answers.

Solution: Recognize that for any pair (i, j), we can assume i < j without loss of generality. This reduces the expression to:

• (arr1[j] - arr1[i]) + (arr2[j] - arr2[i]) + (j - i) → coefficients (1, 1)
• (arr1[j] - arr1[i]) - (arr2[j] - arr2[i]) + (j - i) → coefficients (1, -1)
• -(arr1[j] - arr1[i]) + (arr2[j] - arr2[i]) + (j - i) → coefficients (-1, 1)
• -(arr1[j] - arr1[i]) - (arr2[j] - arr2[i]) + (j - i) → coefficients (-1, -1)

2. Brute Force O(n²) Approach Leading to Time Limit Exceeded

Pitfall: The naive approach of checking all pairs directly:

max_val = 0
for i in range(len(arr1)):
    for j in range(len(arr1)):
        val = abs(arr1[i] - arr1[j]) + abs(arr2[i] - arr2[j]) + abs(i - j)
        max_val = max(max_val, val)
return max_val

Why it happens: The problem statement naturally suggests comparing all pairs, which leads to an O(n²) solution.

Solution: Transform the problem mathematically to recognize that for each sign combination, we're looking for the maximum difference between expressions of the form a*arr1[i] + b*arr2[i] + i. This allows us to compute the result in a single pass for each combination, reducing complexity to O(n).

3. Forgetting to Initialize Max/Min Values Properly

Pitfall: Using incorrect initial values like 0 for maximum or minimum tracking:

max_value = 0  # Wrong! Should be float('-inf')
min_value = 0  # Wrong! Should be float('inf')

Why it happens: Arrays might contain negative values, and the expression could evaluate to negative numbers during intermediate steps.

Solution: Always initialize:

• max_value = float('-inf') for tracking maximums
• min_value = float('inf') for tracking minimums

4. Misunderstanding the Index Component

Pitfall: Forgetting to include the index i in the expression calculation or treating it as j - i instead of just i or j:

# Wrong: forgetting the index
current_expression = coeff_arr1 * value1 + coeff_arr2 * value2

# Wrong: using difference when we should use the index directly
current_expression = coeff_arr1 * value1 + coeff_arr2 * value2 + (j - i)

Why it happens: The mathematical transformation shows that we need a*arr1[j] + b*arr2[j] + j - (a*arr1[i] + b*arr2[i] + i), so each term needs its own index added.

Solution: Include the index in the expression calculation:

current_expression = coeff_arr1 * value1 + coeff_arr2 * value2 + index