From f1731158ce040849d3de05c626af1a1e569b2261 Mon Sep 17 00:00:00 2001
From: ahmedgamal2212 <ahmedgemy2212@gmail.com>
Date: Sun, 14 May 2023 08:46:33 +0300
Subject: [PATCH] Add solution to day 14

---
 ...Score After N Operations (Ahmed Gamal).cpp | 72 +++++++++++++++++++
 1 file changed, 72 insertions(+)
 create mode 100644 05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Gamal).cpp

diff --git a/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Gamal).cpp b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Gamal).cpp
new file mode 100644
index 000000000..4a7c5af8a
--- /dev/null
+++ b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Gamal).cpp	
@@ -0,0 +1,72 @@
+// Author: Ahmed Gamal
+
+// for this problem we will use dp on subsets (memoization) (using bitmask)
+// the idea is to try all possible pairs of numbers and try to add them to the final answer
+// we will use dp[msk] to be the maximum score we can get from the remaining numbers in the array if we already used the numbers in the mask
+// we will try to add two numbers to the answer and continue building the rest of the answer
+// to get the number of the operation we are in, we will use the number of ones in the mask
+// every operation we will take two numbers and mark them as used in the mask, so the number of ones in the mask will be the number of operations we did * 2
+// to get the current operation we are in, we will divide the number of ones in the mask by 2 after adding 2 to it (the two elements we are going to use in the current operation)
+
+class Solution {
+    // N: the maximum number of elements in the array
+    // memo: the dp array
+    static const int N = 14;
+    int memo[1 << N | 1], n;
+    vector<int> v;
+
+    // dp(msk): the maximum score we can get from the remaining numbers in the array if we already used the numbers in the mask (msk
+    int dp(int msk) {
+        // if we already used all the numbers in the array, we will return 0
+        if(__builtin_popcount(msk) == n) {
+            return 0;
+        }
+
+        // if we already calculated the answer, we will return it
+        int& ret = memo[msk];
+        if(~ret) {
+            return ret;
+        }
+
+        // op: the current operation we are in
+        int op = (__builtin_popcount(msk) + 2) / 2;
+
+        ret = 0;
+
+        // we will try to add two numbers to the answer and continue building the rest of the answer
+        for(int i = 0; i < n; i++) {
+            // if we already used the number i, we will skip it
+            if(msk & (1 << i))
+                continue;
+
+            // cur: the mask of the current element we are going to use
+            int cur = 1 << i;
+
+            // we will try to add the number i to the answer and continue building the rest of the answer
+            for(int j = i + 1; j < n; j++) {
+
+                // if we already used the number j, we will skip it
+                if(msk & (1 << j))
+                    continue;
+
+                // we will try to add the number j to the answer and continue building the rest of the answer
+                ret = max(ret, op * __gcd(v[i], v[j]) + dp(msk | cur | (1 << j)));
+            }
+        }
+
+        return ret;
+    }
+public:
+    int maxScore(vector<int>& nums) {
+        // n: the number of elements in the array
+        // v: the array of numbers
+        n = int(nums.size());
+        v = nums;
+
+        // initialize the dp array with -1
+        memset(memo, -1, sizeof memo);
+
+        // we will start building the answer from the first operation
+        return dp(0);
+    }
+};
\ No newline at end of file