README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2700-2799/2708.Maximum%20Strength%20of%20a%20Group/README_EN.md	1502	Biweekly Contest 105 Q3	Greedy Bit Manipulation Array Dynamic Programming Backtracking Enumeration Sorting

2708. Maximum Strength of a Group

中文文档

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

Solution 1: Binary Enumeration

The problem is actually to find the maximum product of all subsets. Since the length of the array does not exceed $13$, we can consider using the method of binary enumeration.

We enumerate all subsets in the range of $[1, 2^n)$, and for each subset, we calculate its product, and finally return the maximum value.

The time complexity is $O(2^n \times n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

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

Java

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;
    }
}

C++

class Solution {
public:
    long long maxStrength(vector<int>& nums) {
        long long ans = -1e14;
        int n = nums.size();
        for (int i = 1; i < 1 << n; ++i) {
            long long t = 1;
            for (int j = 0; j < n; ++j) {
                if (i >> j & 1) {
                    t *= nums[j];
                }
            }
            ans = max(ans, t);
        }
        return ans;
    }
};

Go

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
}

TypeScript

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;
}

Solution 2: Sorting + Greedy

First, we can sort the array. Based on the characteristics of the array, we can draw the following conclusions:

• If there is only one element in the array, then the maximum strength value is this element.
• If there are two or more elements in the array, and $nums[1] = nums[n - 1] = 0$, then the maximum strength value is $0$.
• Otherwise, we traverse the array from small to large. If the current element is less than $0$ and the next element is also less than $0$, then we multiply these two elements and accumulate the product into the answer. Otherwise, if the current element is less than or equal to $0$, we skip it directly. If the current element is greater than $0$, we multiply this element into the answer. Finally, we return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}