Add solution to day 27

05- May/27- Stone Game III/27- Stone Game III (Ahmed Gamal).cpp

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+// Author: Ahmed Gamal
+
+// for this problem, we can use dynamic programming to solve it
+// the approach is similar to the approach of the problem 26- Stone Game II
+// the difference is that we will not use the m value, and we will check if that maximum score of the first player is larger than the sum of the rest of the elements or not
+
+class Solution {
+    static const int N = 5e4 + 5;
+    int memo[N][2];
+
+    int dp(int idx, bool turn, vector<int>& v) {
+        if(idx >= v.size()) {
+            return 0;
+        }
+
+        int& ret = memo[idx][turn];
+        if(~ret) {
+            return ret;
+        }
+
+        // if it's the first player's turn, we will try to maximize the score
+        // notice that we are initializing the ret with -2e9 instead of 0 because the maximum score of the first player can be negative
+        ret = turn ? -2e9 : 2e9;
+        int sum = 0;
+
+        // we will try to choose from 1 to 3 elements from the piles vector
+        for(int i = 0; i < 3; i++) {
+            if(idx + i >= v.size()) {
+                break;
+            }
+            sum += v[idx + i];
+            if(turn) {
+                ret = max(ret, sum + dp(idx + i + 1, false, v));
+            } else {
+                ret = min(ret, dp(idx + i + 1, true, v));
+            }
+        }
+
+        return ret;
+    }
+public:
+    string stoneGameIII(vector<int>& stoneValue) {
+        memset(memo, -1, sizeof memo);
+
+        // the maximum score of the first player
+        int ans = dp(0, true, stoneValue);
+
+        // the sum of the elements in the piles vector
+        int sum = accumulate(stoneValue.begin(), stoneValue.end(), 0);
+
+        return ans > sum - ans ? "Alice" : ans < sum - ans ? "Bob" : "Tie";
+    }
+};