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
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;
}