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-rw-r--r--max_score_n_operations.cpp59
1 files changed, 0 insertions, 59 deletions
diff --git a/max_score_n_operations.cpp b/max_score_n_operations.cpp
deleted file mode 100644
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--- a/max_score_n_operations.cpp
+++ /dev/null
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-// 1799. Maximize Score After N Operations
-#include "leetcode.h"
-
-/*
- * given 'nums', an array of positive integers of size '2 * n'. you must perform
- * 'n' operations on this array. in the i'th operation (1-index), 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.
- */
-
-class Solution {
-public:
- int maxScore(vector<int> nums) {
- m = nums.size();
- n = m / 2;
- return dp(nums, 1, 0);
- }
-
-private:
- int gcd_memo[14][14] = {};
- int memo[1 << 4] = {};
- int m, n;
- int dp(vector<int> &nums, int op, int mask) {
- if (op > n)
- return 0;
- if (memo[mask])
- return memo[mask];
- for (int i = 0; i < m; ++i) {
- if ((mask >> i) & 0x1)
- continue;
- for (int j = i + 1; j < m; ++j) {
- if ((mask >> j) & 0x1)
- continue;
- int new_mask = (0x1 << i) | (0x1 << j) | mask;
- int score = op * gcd(nums, i, j) + dp(nums, op + 1, new_mask);
- memo[mask] = max(memo[mask], score);
- }
- }
- return memo[mask];
- }
- int gcd(vector<int> &nums, int i, int j) {
- if (gcd_memo[i][j])
- return gcd_memo[i][j];
- return gcd_memo[i][j] = __gcd(nums[i], nums[j]);
- }
-};
-
-int main() {
- Solution obj;
- vector<int> n1 = {1, 2};
- vector<int> n2 = {3, 4, 6, 8};
- vector<int> n3 = {1, 2, 3, 4, 5, 6};
- printf("%d\n", obj.maxScore(n1)); // expect: 1
- printf("%d\n", obj.maxScore(n2)); // expect: 11
- printf("%d\n", obj.maxScore(n3)); // expect: 14
-}