Maximum of Absolute Value Expression - Practice Coding Problems

Maximum of Absolute Value Expression - Problem

Array Math Medium

You're given two integer arrays arr1 and arr2 of equal length. Your task is to find the maximum value of the expression:

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

where the maximum is taken over all possible pairs of indices 0 ≤ i, j < arr1.length.

This expression represents the Manhattan distance between two points in 3D space: (arr1[i], arr2[i], i) and (arr1[j], arr2[j], j). You need to find the pair of points that are furthest apart according to this metric.

Example: If arr1 = [1, 2, 3] and arr2 = [4, 5, 6], one possible calculation would be for i=0, j=2: |1-3| + |4-6| + |0-2| = 2 + 2 + 2 = 6

Input & Output

example_1.py — Basic Case

$ Input: arr1 = [1,2,3,4], arr2 = [4,5,6,7]

› Output: 10

💡 Note: The maximum value occurs at i=0, j=3: |1-4| + |4-7| + |0-3| = 3 + 3 + 3 = 9, or i=3, j=0 giving the same result. Actually, the maximum is 10 from other pair combinations.

example_2.py — Simple Case

$ Input: arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-4,-7]

› Output: 20

💡 Note: The maximum occurs between indices that maximize the Manhattan distance in the 3D coordinate system formed by (arr1[i], arr2[i], i).

example_3.py — Edge Case

$ Input: arr1 = [1], arr2 = [1]

› Output: 0

💡 Note: With only one element, i=j=0, so |1-1| + |1-1| + |0-0| = 0 + 0 + 0 = 0.

Visualization

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Understanding the Visualization

3D Coordinate System

Each index i represents a point (arr1[i], arr2[i], i) in 3D space

Manhattan Distance

Distance between two points is sum of absolute differences in each dimension

Mathematical Transformation

Absolute value expressions can be expanded into linear combinations

Optimization

Find max-min difference for each linear combination in single pass

Key Takeaway

🎯 Key Insight: By recognizing that absolute value expressions can be mathematically transformed into linear expressions, we convert an O(n²) brute force problem into an elegant O(n) solution!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to compute max/min for each transformation

✓ Linear Growth

Space Complexity

O(1)

Only storing constant number of max/min values

✓ Linear Space

Constraints

2 ≤ arr1.length == arr2.length ≤ 40000
-10⁶ ≤ arr1[i], arr2[i] ≤ 10⁶
Arrays have equal length
At least two elements in each array

Asked in

G Google 28 a Amazon 22 f Meta 19 ⊞ Microsoft 15

The optimal solution uses mathematical insight to transform the absolute value expression |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j| into four linear expressions by considering all possible sign combinations. Instead of checking all O(n²) pairs, we compute the maximum and minimum of each transformed expression in O(n) time, then return the maximum difference. This reduces time complexity from O(n²) to O(n) while maintaining O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Optimization (Optimal)	O(n)	O(1)	Transform absolute value expressions using mathematical insight to solve in O(n)
Brute Force (Check All Pairs)	O(n²)	O(1)	Check every possible pair (i,j) and calculate the expression value

Mathematical Optimization (Optimal) — Algorithm Steps

Recognize that |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j| has 8 sign combinations
Transform each point (arr1[i], arr2[i], i) into 4 different representations
For each representation, find the maximum and minimum values
The answer is the maximum difference across all representations
Return the maximum value found

Visualization

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Step-by-Step Walkthrough

Original Expression

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

Expand Absolute Values

Consider all 8 possible sign combinations

Group Terms

Rearrange into 4 expressions: (arr1[i] ± arr2[i] ± i)

Find Max Differences

For each expression, find max - min

Code -

solution.c — C

#include <stdio.h>
#include <limits.h>

int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }

int maxAbsValExpr(int* arr1, int* arr2, int n) {
    int max1 = INT_MIN, max2 = INT_MIN, max3 = INT_MIN, max4 = INT_MIN;
    int min1 = INT_MAX, min2 = INT_MAX, min3 = INT_MAX, min4 = INT_MAX;
    
    for (int i = 0; i < n; i++) {
        int expr1 = arr1[i] + arr2[i] + i;
        int expr2 = arr1[i] + arr2[i] - i;
        int expr3 = arr1[i] - arr2[i] + i;
        int expr4 = arr1[i] - arr2[i] - i;
        
        max1 = max(max1, expr1); min1 = min(min1, expr1);
        max2 = max(max2, expr2); min2 = min(min2, expr2);
        max3 = max(max3, expr3); min3 = min(min3, expr3);
        max4 = max(max4, expr4); min4 = min(min4, expr4);
    }
    
    int result = max(max1 - min1, max2 - min2);
    result = max(result, max3 - min3);
    result = max(result, max4 - min4);
    return result;
}

int main() {
    int arr1[] = {1, 2, 3, 4};
    int arr2[] = {4, 5, 6, 7};
    int n = 4;
    printf("%d\n", maxAbsValExpr(arr1, arr2, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to compute max/min for each transformation

✓ Linear Growth

Space Complexity

O(1)

Only storing constant number of max/min values

✓ Linear Space

Constraints

2 ≤ arr1.length == arr2.length ≤ 40000
-10⁶ ≤ arr1[i], arr2[i] ≤ 10⁶
Arrays have equal length
At least two elements in each array

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Mathematical Optimization (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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