LeetCode #2708 — MEDIUM

Maximum Strength of a Group

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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, ... , i_k is defined as nums[i₀] * nums[i₁] * nums[i₂] * ... * nums[i_k].

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

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Backtracking · Greedy · Bit Manipulation

Example 1

[3,-1,-5,2,5,-9]

Example 2

[-4,-5,-4]

Related Problems

Maximum Strength of K Disjoint Subarrays (maximum-strength-of-k-disjoint-subarrays)

Step 02

Core Insight

What unlocks the optimal approach

Try to generate all pairs of subsets and check which group provides maximal strength.
It can also be solved in O(NlogN) by sorting the array and using all positive integers.
Use negative integers only in pairs such that their product becomes positive.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2708: Maximum Strength of a Group
class Solution {
    public long maxStrength(int[] nums) {
        long ans = (long) -1e14;
        int n = nums.length;
        for (int i = 1; i < 1 << n; ++i) {
            long t = 1;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    t *= nums[j];
                }
            }
            ans = Math.max(ans, t);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2708: Maximum Strength of a Group
func maxStrength(nums []int) int64 {
	ans := int64(-1e14)
	for i := 1; i < 1<<len(nums); i++ {
		var t int64 = 1
		for j, x := range nums {
			if i>>j&1 == 1 {
				t *= int64(x)
			}
		}
		ans = max(ans, t)
	}
	return ans
}

# Accepted solution for LeetCode #2708: Maximum Strength of a Group
class Solution:
    def maxStrength(self, nums: List[int]) -> int:
        ans = -inf
        for i in range(1, 1 << len(nums)):
            t = 1
            for j, x in enumerate(nums):
                if i >> j & 1:
                    t *= x
            ans = max(ans, t)
        return ans

// Accepted solution for LeetCode #2708: Maximum Strength of a Group
/**
 * [2708] Maximum Strength of a Group
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn max_strength(nums: Vec<i32>) -> i64 {
        if nums.len() == 1 {
            return nums[0] as i64;
        }

        let mut nums = nums;
        nums.sort();

        let mut now = 0;
        let mut result = 1;
        let mut found = false;

        while now < nums.len() {
            if nums[now] > 0 {
                break;
            }

            if now + 1 < nums.len() && nums[now] * nums[now + 1] > 0 {
                let r = nums[now] * nums[now + 1];
                result *= r as i64;
                found = true;
                now += 2;
            } else {
                now += 1;
            }
        }

        while now < nums.len() {
            result *= nums[now] as i64;
            found = true;

            now += 1;
        }

        match found {
            true => result,
            false => 0,
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2708() {
        assert_eq!(20, Solution::max_strength(vec![-4, -5, -4]));
        assert_eq!(1350, Solution::max_strength(vec![3, -1, -5, 2, 5, -9]));
    }
}

// Accepted solution for LeetCode #2708: Maximum Strength of a Group
function maxStrength(nums: number[]): number {
    let ans = -Infinity;
    const n = nums.length;
    for (let i = 1; i < 1 << n; ++i) {
        let t = 1;
        for (let j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                t *= nums[j];
            }
        }
        ans = Math.max(ans, t);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.