2708. Maximum Strength of a Group

Problem

You are given a 0-indexed integer array nums representing the score of students in an exam. The teacher would like to form one non-empty group of students with maximal strength, where the strength of a group of students of indices i0, i1, i2, … , ik is defined as nums[i0] * nums[i1] * nums[i2] * ... * nums[ik​].

Return the maximum strength of a group the teacher can create.

Example 1:

Input: nums = [3,-1,-5,2,5,-9]
Output: 1350
Explanation: One way to form a group of maximal strength is to group the students at indices [0,2,3,4,5]. Their strength is 3 * (-5) * 2 * 5 * (-9) = 1350, which we can show is optimal.

Example 2:

Input: nums = [-4,-5,-4]
Output: 20
Explanation: Group the students at indices [0, 1] . Then, we’ll have a resulting strength of 20. We cannot achieve greater strength.

Constraints:

• 1 <= nums.length <= 13
• -9 <= nums[i] <= 9

Solution (Java)

import java.util.*;

class Solution {
    public long maxStrength(int[] nums) {
        int n = nums.length;
        List<Long> negs = new ArrayList<>();
        List<Long> pos = new ArrayList<>();
        for (int num : nums) {
            if (num < 0)
                negs.add((long) num);
            if (num > 0)
                pos.add((long) num);
        }
        long prod = 1;
        long x = Collections.frequency(Arrays.asList(nums), 0);
        Collections.sort(negs);
        if (negs.size() <= 1 && pos.size() == 0)
            return Arrays.stream(nums).max().getAsInt();
        if (negs.size() % 2 == 0) {
            for (long num : negs)
                prod = prod * num;
            for (long num : pos)
                prod = prod * num;
            return prod;
        } else {
            for (int i = 0; i < negs.size() - 1; ++i)
                prod = prod * negs.get(i);
            for (long num : pos)
                prod = prod * num;
            return prod;
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).