LeetCode 2708: Maximum Strength of a Group Solution

Master LeetCode problem 2708 (Maximum Strength of a Group), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2708. Maximum Strength of a Group

Medium

Array Dynamic Programming Backtracking Greedy Bit Manipulation Sorting Enumeration

Problem Explanation

Explanation

To find the maximum strength of a group, we need to consider all possible groups of students and calculate their strength. We can achieve this by iterating through all possible subsets of the input array nums and calculating the product of elements in each subset. The maximum of these products will be the maximum strength of the group.

Algorithm

Initialize a variable maxStrength to store the maximum strength found.
Use a bitmask to iterate through all possible subsets of nums.
For each subset represented by the current bitmask, calculate the product of elements in that subset.
Update maxStrength if the calculated product is greater than the current maximum.
Return maxStrength as the maximum strength of the group.

Time Complexity

The time complexity of this approach is O(2^n * n), where n is the number of elements in the input array nums.

Space Complexity

The space complexity is O(1) as we are not using any extra space proportional to the input size.

Solution Code

class Solution {
    public int maxGroupStrength(int[] nums) {
        int n = nums.length;
        int maxStrength = Integer.MIN_VALUE;
        
        for (int mask = 1; mask < (1 << n); mask++) {
            int product = 1;
            for (int i = 0; i < n; i++) {
                if ((mask & (1 << i)) != 0) {
                    product *= nums[i];
                }
            }
            maxStrength = Math.max(maxStrength, product);
        }
        
        return maxStrength;
    }
}

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