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1799. Maximize Score After N Operations

Description

You are given nums, an array of positive integers of size 2 * n. You must perform n operations on this array.

In the ith operation (1-indexed), you will:

  • Choose two elements, x and y.
  • Receive a score of i * gcd(x, y).
  • Remove x and y from nums.

Return the maximum score you can receive after performing n operations.

The function gcd(x, y) is the greatest common divisor of x and y.

 

Example 1:

 Input: nums = [1,2] Output: 1 Explanation: The optimal choice of operations is: (1 * gcd(1, 2)) = 1

Example 2:

 Input: nums = [3,4,6,8] Output: 11 Explanation: The optimal choice of operations is: (1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11

Example 3:

 Input: nums = [1,2,3,4,5,6] Output: 14 Explanation: The optimal choice of operations is: (1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14

 

Constraints:

  • 1 <= n <= 7
  • nums.length == 2 * n
  • 1 <= nums[i] <= 106

Solutions

Solution 1: State Compression + Dynamic Programming

We can preprocess to get the greatest common divisor of any two numbers in the array nums, stored in the two-dimensional array $g$, where $g[i][j]$ represents the greatest common divisor of $nums[i]$ and $nums[j]$.

Then define $f[k]$ to represent the maximum score that can be obtained when the state after the current operation is $k$. Suppose $m$ is the number of elements in the array nums, then there are a total of $2^m$ states, that is, the range of $k$ is $[0, 2^m - 1]$.

Enumerate all states from small to large, for each state $k$, first determine whether the number of $1$s in the binary bits of this state $cnt$ is even, if so, perform the following operations:

Enumerate the positions where the binary bits in $k$ are 1, suppose they are $i$ and $j$, then the elements at positions $i$ and $j$ can perform one operation, and the score that can be obtained at this time is $\frac{cnt}{2} \times g[i][j]$, update the maximum value of $f[k]$.

The final answer is $f[2^m - 1]$.

The time complexity is $O(2^m \times m^2)$, and the space complexity is $O(2^m)$. Here, $m$ is the number of elements in the array nums.

  • class Solution { public int maxScore(int[] nums) { int m = nums.length; int[][] g = new int[m][m]; for (int i = 0; i < m; ++i) { for (int j = i + 1; j < m; ++j) { g[i][j] = gcd(nums[i], nums[j]); } } int[] f = new int[1 << m]; for (int k = 0; k < 1 << m; ++k) { int cnt = Integer.bitCount(k); if (cnt % 2 == 0) { for (int i = 0; i < m; ++i) { if (((k >> i) & 1) == 1) { for (int j = i + 1; j < m; ++j) { if (((k >> j) & 1) == 1) { f[k] = Math.max( f[k], f[k ^ (1 << i) ^ (1 << j)] + cnt / 2 * g[i][j]); } } } } } } return f[(1 << m) - 1]; } private int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } } 
  • class Solution { public: int maxScore(vector<int>& nums) { int m = nums.size(); int g[m][m]; for (int i = 0; i < m; ++i) { for (int j = i + 1; j < m; ++j) { g[i][j] = gcd(nums[i], nums[j]); } } int f[1 << m]; memset(f, 0, sizeof f); for (int k = 0; k < 1 << m; ++k) { int cnt = __builtin_popcount(k); if (cnt % 2 == 0) { for (int i = 0; i < m; ++i) { if (k >> i & 1) { for (int j = i + 1; j < m; ++j) { if (k >> j & 1) { f[k] = max(f[k], f[k ^ (1 << i) ^ (1 << j)] + cnt / 2 * g[i][j]); } } } } } } return f[(1 << m) - 1]; } }; 
  • class Solution: def maxScore(self, nums: List[int]) -> int: m = len(nums) f = [0] * (1 << m) g = [[0] * m for _ in range(m)] for i in range(m): for j in range(i + 1, m): g[i][j] = gcd(nums[i], nums[j]) for k in range(1 << m): if (cnt := k.bit_count()) % 2 == 0: for i in range(m): if k >> i & 1: for j in range(i + 1, m): if k >> j & 1: f[k] = max( f[k], f[k ^ (1 << i) ^ (1 << j)] + cnt // 2 * g[i][j], ) return f[-1] 
  • func maxScore(nums []int) int { m := len(nums) g := [14][14]int{} for i := 0; i < m; i++ { for j := i + 1; j < m; j++ { g[i][j] = gcd(nums[i], nums[j]) } } f := make([]int, 1<<m) for k := 0; k < 1<<m; k++ { cnt := bits.OnesCount(uint(k)) if cnt%2 == 0 { for i := 0; i < m; i++ { if k>>i&1 == 1 { for j := i + 1; j < m; j++ { if k>>j&1 == 1 { f[k] = max(f[k], f[k^(1<<i)^(1<<j)]+cnt/2*g[i][j]) } } } } } } return f[1<<m-1] } func gcd(a, b int) int { if b == 0 { return a } return gcd(b, a%b) } 
  • function maxScore(nums: number[]): number { const m = nums.length; const f: number[] = new Array(1 << m).fill(0); const g: number[][] = new Array(m).fill(0).map(() => new Array(m).fill(0)); for (let i = 0; i < m; ++i) { for (let j = i + 1; j < m; ++j) { g[i][j] = gcd(nums[i], nums[j]); } } for (let k = 0; k < 1 << m; ++k) { const cnt = bitCount(k); if (cnt % 2 === 0) { for (let i = 0; i < m; ++i) { if ((k >> i) & 1) { for (let j = i + 1; j < m; ++j) { if ((k >> j) & 1) { const t = f[k ^ (1 << i) ^ (1 << j)] + ~~(cnt / 2) * g[i][j]; f[k] = Math.max(f[k], t); } } } } } } return f[(1 << m) - 1]; } function gcd(a: number, b: number): number { return b ? gcd(b, a % b) : a; } function bitCount(i: number): number { i = i - ((i >>> 1) & 0x55555555); i = (i & 0x33333333) + ((i >>> 2) & 0x33333333); i = (i + (i >>> 4)) & 0x0f0f0f0f; i = i + (i >>> 8); i = i + (i >>> 16); return i & 0x3f; }

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