Question

1799. Maximize Score After N Operations

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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:
  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]

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];
  }
};

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;
}