LeetCode 2708 - Maximum Strength of a Group

The problem asks us to select a non-empty subset of integers from a given array nums such that the product of the numbers in the subset is maximized. Each number represents a student’s exam score, and the subset represents a group of students.

leetcodemediumarraydynamic-programmingbacktrackinggreedybit-manipulationsortingenumeration

LeetCode Problem 2708

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Backtracking, Greedy, Bit Manipulation, Sorting, Enumeration

Solution

Problem Understanding

The problem asks us to select a non-empty subset of integers from a given array nums such that the product of the numbers in the subset is maximized. Each number represents a student’s exam score, and the subset represents a group of students. The "strength" of a group is the product of its scores. The array nums may contain positive numbers, negative numbers, and zeros, and the array length is small, from 1 to 13.

The expected output is a single integer representing the maximum achievable strength for any non-empty subset of nums.

Constraints are important: the small size of the array (<= 13) means brute-force exploration of all subsets is technically feasible, because there are at most $2^{13} - 1 = 8191$ non-empty subsets. The score values being small (-9 to 9) indicates that products will not overflow typical integer types in Python or Go.

Key edge cases include arrays with: all negative numbers, arrays with zeros, single-element arrays, and arrays where excluding a negative number is optimal to maximize the product. We must also handle subsets of length 1, since the group cannot be empty.

Approaches

The brute-force approach is to enumerate all possible non-empty subsets of nums, compute the product of each subset, and return the maximum. This works because the number of subsets is small for n <= 13. However, enumerating all subsets requires exponential time, which is inefficient for larger arrays.

The optimal approach leverages the properties of numbers in multiplication. Specifically, to maximize a product:

  1. All positive numbers should be included.
  2. Negative numbers should be included in pairs, because the product of two negatives is positive.
  3. The largest negative number (closest to zero) may need to be excluded if the total count of negatives is odd.
  4. Zeros should be ignored unless the array contains only zeros or a single negative number, where zero may be preferable.

By separating positive and negative numbers, counting them, and carefully selecting which negatives to include, we can compute the maximum product without generating all subsets.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n * n) O(n) Enumerates all subsets, computes product for each, feasible for n ≤ 13
Optimal O(n log n) O(n) Separate positive and negative numbers, sort negatives, multiply positives and largest even count of negatives

Algorithm Walkthrough

  1. Separate nums into positives, negatives, and count zeros.
  2. If all numbers are zeros, return 0 because no positive product exists.
  3. If all numbers are negative and there are no positives, exclude the largest negative to maximize the product.
  4. Sort negative numbers by absolute value in ascending order. If the count of negatives is odd, exclude the one closest to zero (largest negative).
  5. Multiply all remaining negatives and all positive numbers.
  6. If the product is empty (i.e., all numbers were zeros or one negative), return the maximum number in nums as the group strength.
  7. Return the final product.

Why it works: This algorithm guarantees that we always maximize the product because multiplying all positives increases the product, and pairing negatives ensures negative contributions become positive. By selectively excluding one negative if the count is odd, we avoid reducing the product.

Python Solution

from typing import List
import math

class Solution:
    def maxStrength(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]
        
        positives = [x for x in nums if x > 0]
        negatives = [x for x in nums if x < 0]
        
        if len(negatives) % 2 != 0:
            negatives.sort()
            negatives.pop()  # remove the largest negative (closest to zero)
        
        product = 1
        for x in positives + negatives:
            product *= x
        
        if product == 1 and not positives and negatives == []:
            # All zeros
            return 0
        return product

Implementation Walkthrough: The code first handles the single-element array. It separates positives and negatives, sorts negatives if needed to drop the least impactful one, and multiplies the rest. The final check ensures that if the product is 1 but no numbers contributed, we return 0.

Go Solution

import "sort"

func maxStrength(nums []int) int64 {
    if len(nums) == 1 {
        return int64(nums[0])
    }
    
    positives := []int{}
    negatives := []int{}
    
    for _, num := range nums {
        if num > 0 {
            positives = append(positives, num)
        } else if num < 0 {
            negatives = append(negatives, num)
        }
    }
    
    if len(negatives) % 2 != 0 {
        sort.Ints(negatives)
        negatives = negatives[:len(negatives)-1] // remove largest negative
    }
    
    product := int64(1)
    if len(positives) + len(negatives) == 0 {
        // only zeros
        return 0
    }
    
    for _, num := range positives {
        product *= int64(num)
    }
    for _, num := range negatives {
        product *= int64(num)
    }
    
    return product
}

Go-Specific Notes: Explicit int64 is used to avoid overflow. Slices are used to separate positives and negatives, and the largest negative is removed by slicing. Go does not require extra checks for empty slices beyond zeros handling.

Worked Examples

Example 1: nums = [3,-1,-5,2,5,-9]

  1. Positives: [3,2,5]
  2. Negatives: [-1,-5,-9] → sorted [-9,-5,-1] → count odd → remove -1 → remaining [-9,-5]
  3. Multiply: 3*2*5*(-9)*(-5) = 30*45 = 1350

Example 2: nums = [-4,-5,-4]

  1. Positives: []
  2. Negatives: [-4,-5,-4] → sorted [-5,-4,-4] → count odd → remove -4 → remaining [-5,-4]
  3. Multiply: -5*-4 = 20

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting negatives dominates, n ≤ 13
Space O(n) Storing positives and negatives separately

The algorithm is efficient due to small input size. Sorting ensures correct selection of negatives, and multiplication is linear in remaining numbers.

Test Cases

# Provided examples
assert Solution().maxStrength([3,-1,-5,2,5,-9]) == 1350  # mixed positives and negatives
assert Solution().maxStrength([-4,-5,-4]) == 20          # all negatives
# Single element
assert Solution().maxStrength([0]) == 0                  # single zero
assert Solution().maxStrength([5]) == 5                  # single positive
assert Solution().maxStrength([-7]) == -7               # single negative
# Zeros and negatives
assert Solution().maxStrength([0,-1,-2]) == 2           # ignore zero, remove -1
assert Solution().maxStrength([0,0,0]) == 0             # all zeros
# Mixed
assert Solution().maxStrength([1,-2,3,-4,0]) == 24      # remove zero implicitly
Test Why
[3,-1,-5,2,5,-9] mixed positives and negatives
[-4,-5,-4] all negatives, odd count
[0] single zero
[5] single positive
[-7] single negative
[0,-1,-2] zeros and negatives
[0,0,0] all zeros
[1,-2,3,-4,0] mixed with zero

Edge Cases

  1. Single-element array: The group cannot be empty, so the only number must be returned. Handles positive, negative, and zero.
  2. All zeros: Any subset product is zero. The implementation explicitly checks for empty products and returns 0.
  3. Odd count of negatives: To maximize product, the least negative (closest to zero) must be removed. Sorting ensures the correct negative is removed.

This approach robustly handles mixed cases, zeros, and small arrays, guaranteeing the maximum strength group.