Maximum Product of Three Numbers - Practice Coding Problems

Maximum Product of Three Numbers - Problem

Given an integer array nums, find three numbers whose product is maximum and return the maximum product.

Note: The solution is guaranteed to fit in a 32-bit integer.

Input & Output

Example 1 — Basic Case

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

› Output: 6

💡 Note: Only one triplet possible: 1 × 2 × 3 = 6

Example 2 — Negative Numbers

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

› Output: 24

💡 Note: Maximum product from three largest: 2 × 3 × 4 = 24

Example 3 — Mixed Signs

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

› Output: -6

💡 Note: All negative numbers: (-1) × (-2) × (-3) = -6

Constraints

3 ≤ nums.length ≤ 10⁴
-1000 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Given integer array with positive and negative values

Find Extremes

Identify largest and smallest values efficiently

Compare Products

Calculate and compare key combinations

Key Takeaway

🎯 Key Insight: Two negative numbers multiply to positive, potentially creating larger products than three positives

Asked in

a Amazon 25 M Microsoft 18 Apple 12

The key insight is that maximum product comes from either three largest positive numbers or two smallest negative numbers times the largest positive. Best approach is single pass tracking with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n³)	O(1)	Check all possible combinations of three numbers
Sort First Approach	O(n log n)	O(1)	Sort the array then consider key combinations for maximum product
Single Pass Optimal	O(n)	O(1)	Find the required elements in one pass without sorting

Brute Force — Algorithm Steps

Step 1: Use three nested loops to generate all triplets
Step 2: Calculate product for each triplet
Step 3: Keep track of maximum product

Visualization

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Step-by-Step Walkthrough

Generate Triplets

Use three nested loops to create all combinations

Calculate Products

Multiply the three numbers in each triplet

Track Maximum

Keep the highest product found

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int maxProduct = INT_MIN;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            for (int k = j + 1; k < numsSize; k++) {
                int product = nums[i] * nums[j] * nums[k];
                if (product > maxProduct) {
                    maxProduct = product;
                }
            }
        }
    }
    
    return maxProduct;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int count = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops iterate through all possible combinations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track maximum

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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