Maximum Product Difference Between Two Pairs

Maximum Product Difference Between Two Pairs - Problem

Array Sorting Easy

The product difference between two pairs (a, b) and (c, d) is defined as (a * b) - (c * d).

For example, the product difference between (5, 6) and (2, 7) is (5 * 6) - (2 * 7) = 30 - 14 = 16.

Given an integer array nums, choose four distinct indices w, x, y, and z such that the product difference between pairs (nums[w], nums[x]) and (nums[y], nums[z]) is maximized.

Return the maximum such product difference.

Input & Output

Example 1 — Basic Case

$ Input: nums = [5,6,2,7,1,4]

› Output: 40

💡 Note: Choose indices (1,3) for pair (6,7) and (2,4) for pair (2,1). Product difference is (6×7) - (2×1) = 42 - 2 = 40.

Example 2 — Small Array

$ Input: nums = [4,2,5,9]

› Output: 37

💡 Note: Choose indices (2,3) for pair (5,9) and (0,1) for pair (4,2). Product difference is (5×9) - (4×2) = 45 - 8 = 37.

Example 3 — Negative Numbers

$ Input: nums = [-1,-2,-3,-4]

› Output: 10

💡 Note: Choose indices (2,3) for pair (-3,-4) and (0,1) for pair (-1,-2). Product difference is ((-3)×(-4)) - ((-1)×(-2)) = 12 - 2 = 10.

Constraints

4 ≤ nums.length ≤ 10⁴
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

a Amazon 25 M Microsoft 18 G Google 12

The key insight is that maximum product difference comes from multiplying the two largest numbers and subtracting the product of the two smallest numbers. Best approach is the one-pass solution that tracks extremes while scanning. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n⁴) Space: O(1)

Generate all possible combinations of 4 indices and calculate the product difference for each. Keep track of the maximum difference found.

One Pass (Optimal)

⏱️ Time: O(n) Space: O(1)

Track the two largest and two smallest elements while scanning the array once. This avoids the need to sort the entire array.

Sort First

⏱️ Time: O(n log n) Space: O(1)

Sort the array first, then pick the two largest elements for the first pair and two smallest for the second pair to maximize the product difference.

Brute Force — Algorithm Steps

Generate all combinations of 4 indices
Calculate product difference for each combination
Return the maximum difference

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Try all possible sets of 4 distinct indices

Calculate Differences

For each combination, compute product difference

Find Maximum

Track and return the maximum difference found

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int maxDiff = INT_MIN;
    
    for (int w = 0; w < numsSize; w++) {
        for (int x = 0; x < numsSize; x++) {
            if (x == w) continue;
            for (int y = 0; y < numsSize; y++) {
                if (y == w || y == x) continue;
                for (int z = 0; z < numsSize; z++) {
                    if (z == w || z == x || z == y) continue;
                    int diff = nums[w] * nums[x] - nums[y] * nums[z];
                    if (diff > maxDiff) {
                        maxDiff = diff;
                    }
                }
            }
        }
    }
    
    return maxDiff;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n⁴)

Four nested loops to generate all combinations of 4 indices

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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