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2708. Maximum Strength of a Group

Description

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

Solutions

  • class Solution {
    public long maxStrength(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        if (n == 1) {
            return nums[0];
        }
        if (nums[1] == 0 && nums[n - 1] == 0) {
            return 0;
        }
        long ans = 1;
        int i = 0;
        while (i < n) {
            if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
                ans *= nums[i] * nums[i + 1];
                i += 2;
            } else if (nums[i] <= 0) {
                i += 1;
            } else {
                ans *= nums[i];
                i += 1;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    long long maxStrength(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        if (n == 1) {
            return nums[0];
        }
        if (nums[1] == 0 && nums[n - 1] == 0) {
            return 0;
        }
        long long ans = 1;
        int i = 0;
        while (i < n) {
            if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
                ans *= nums[i] * nums[i + 1];
                i += 2;
            } else if (nums[i] <= 0) {
                i += 1;
            } else {
                ans *= nums[i];
                i += 1;
            }
        }
        return ans;
    }
};

  • class Solution:
    def maxStrength(self, nums: List[int]) -> int:
        nums.sort()
        n = len(nums)
        if n == 1:
            return nums[0]
        if nums[1] == nums[-1] == 0:
            return 0
        ans, i = 1, 0
        while i < n:
            if nums[i] < 0 and i + 1 < n and nums[i + 1] < 0:
                ans *= nums[i] * nums[i + 1]
                i += 2
            elif nums[i] <= 0:
                i += 1
            else:
                ans *= nums[i]
                i += 1
        return ans


  • func maxStrength(nums []int) int64 {
	sort.Ints(nums)
	n := len(nums)
	if n == 1 {
		return int64(nums[0])
	}
	if nums[1] == 0 && nums[n-1] == 0 {
		return 0
	}
	ans := int64(1)
	for i := 0; i < n; i++ {
		if nums[i] < 0 && i+1 < n && nums[i+1] < 0 {
			ans *= int64(nums[i] * nums[i+1])
			i++
		} else if nums[i] > 0 {
			ans *= int64(nums[i])
		}
	}
	return ans
}

  • function maxStrength(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    if (n === 1) {
        return nums[0];
    }
    if (nums[1] === 0 && nums[n - 1] === 0) {
        return 0;
    }
    let ans = 1;
    for (let i = 0; i < n; ++i) {
        if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
            ans *= nums[i] * nums[i + 1];
            ++i;
        } else if (nums[i] > 0) {
            ans *= nums[i];
        }
    }
    return ans;
}

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