Maximum Product of Three Numbers

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

Tap to expand

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

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

Try every possible triplet of numbers in the array and keep track of the maximum product found. This ensures we don't miss any combination.

Single Pass Optimal

⏱️ Time: O(n) Space: O(1)

Instead of sorting, find the three largest and two smallest elements in a single pass. Then compute the maximum product from the same two combinations.

Sort First Approach

⏱️ Time: O(n log n) Space: O(1)

Sort the array and then check the two most promising combinations: either the three largest numbers, or the two smallest (most negative) times the largest positive number.

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

Tap to expand

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

35.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler