New Problem Solution -"1770. Maximum Score from Performing Multiplication Operations"

haoel · haoel · commit b6ae9dec0f98 · 2021-04-01T10:10:05.000+08:00
diff --git a/README.md b/README.md
@@ -37,6 +37,7 @@ LeetCode
 |1774|[Closest Dessert Cost](https://leetcode.com/problems/closest-dessert-cost/) | [C++](./algorithms/cpp/closestDessertCost/ClosestDessertCost.cpp)|Medium|
 |1773|[Count Items Matching a Rule](https://leetcode.com/problems/count-items-matching-a-rule/) | [C++](./algorithms/cpp/countItemsMatchingARule/CountItemsMatchingARule.cpp)|Easy|
 |1771|[Maximize Palindrome Length From Subsequences](https://leetcode.com/problems/maximize-palindrome-length-from-subsequences/) | [C++](./algorithms/cpp/maximizePalindromeLengthFromSubsequences/MaximizePalindromeLengthFromSubsequences.cpp)|Hard|
+|1770|[Maximum Score from Performing Multiplication Operations](https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/) | [C++](./algorithms/cpp/maximumScoreFromPerformingMultiplicationOperations/MaximumScoreFromPerformingMultiplicationOperations.cpp)|Medium|
 |1769|[Minimum Number of Operations to Move All Balls to Each Box](https://leetcode.com/problems/minimum-number-of-operations-to-move-all-balls-to-each-box/) | [C++](./algorithms/cpp/minimumNumberOfOperationsToMoveAllBallsToEachBox/MinimumNumberOfOperationsToMoveAllBallsToEachBox.cpp)|Medium|
 |1768|[Merge Strings Alternately](https://leetcode.com/problems/merge-strings-alternately/) | [C++](./algorithms/cpp/mergeStringsAlternately/MergeStringsAlternately.cpp)|Easy|
 |1765|[Map of Highest Peak](https://leetcode.com/problems/map-of-highest-peak/) | [C++](./algorithms/cpp/mapOfHighestPeak/MapOfHighestPeak.cpp)|Medium|
diff --git a/algorithms/cpp/maximumScoreFromPerformingMultiplicationOperations/MaximumScoreFromPerformingMultiplicationOperations.cpp b/algorithms/cpp/maximumScoreFromPerformingMultiplicationOperations/MaximumScoreFromPerformingMultiplicationOperations.cpp
@@ -0,0 +1,81 @@
+// Source : https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/
+// Author : Hao Chen
+// Date   : 2021-04-01
+
+/***************************************************************************************************** 
+ *
+ * You are given two integer arrays nums and multipliers of size n and m respectively, where n >= m. 
+ * The arrays are 1-indexed.
+ * 
+ * You begin with a score of 0. You want to perform exactly m operations. On the i^th operation 
+ * (1-indexed), you will:
+ * 
+ * 	Choose one integer x from either the start or the end of the array nums.
+ * 	Add multipliers[i] * x to your score.
+ * 	Remove x from the array nums.
+ * 
+ * Return the maximum score after performing m operations.
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [1,2,3], multipliers = [3,2,1]
+ * Output: 14
+ * Explanation: An optimal solution is as follows:
+ * - Choose from the end, [1,2,3], adding 3 * 3 = 9 to the score.
+ * - Choose from the end, [1,2], adding 2 * 2 = 4 to the score.
+ * - Choose from the end, [1], adding 1 * 1 = 1 to the score.
+ * The total score is 9 + 4 + 1 = 14.
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]
+ * Output: 102
+ * Explanation: An optimal solution is as follows:
+ * - Choose from the start, [-5,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score.
+ * - Choose from the start, [-3,-3,-2,7,1], adding -3 * -5 = 15 to the score.
+ * - Choose from the start, [-3,-2,7,1], adding -3 * 3 = -9 to the score.
+ * - Choose from the end, [-2,7,1], adding 1 * 4 = 4 to the score.
+ * - Choose from the end, [-2,7], adding 7 * 6 = 42 to the score. 
+ * The total score is 50 + 15 - 9 + 4 + 42 = 102.
+ * 
+ * Constraints:
+ * 
+ * 	n == nums.length
+ * 	m == multipliers.length
+ * 	1 <= m <= 10^3
+ * 	m <= n <= 10^5 
+ * 	-1000 <= nums[i], multipliers[i] <= 1000
+ ******************************************************************************************************/
+
+const int MAX_SIZE = 1000;
+class Solution {
+private:
+    int cache[MAX_SIZE][MAX_SIZE]; // num of left picked, num of right picked.
+    int m, n;
+public:
+    int maximumScore(vector<int>& nums, vector<int>& multipliers) {
+        memset(cache, -1, sizeof(cache));
+        n = nums.size();
+        m = multipliers.size();
+        return maximumScoreDFS(nums, 0, n-1, multipliers, 0 );
+    }
+    
+    int maximumScoreDFS(vector<int>& nums, int left, int right, 
+                        vector<int>& multipliers, int midx) {
+        
+        if(midx >= m )  return 0;
+        
+        int nLeft = left; // num of left nums[] picked.
+        int nRight = (n-1)-right; // num of right nums[] picked.
+        if (cache[nLeft][nRight]!=-1) return cache[nLeft][nRight];
+        
+        int pickLeft = maximumScoreDFS(nums, left+1, right, multipliers, midx+1) +
+            multipliers[midx] * nums[left];
+        
+        int pickRight = maximumScoreDFS(nums, left, right-1, multipliers, midx+1) +
+            multipliers[midx] * nums[right];
+        
+        cache[nLeft][nRight] = max(pickLeft, pickRight);
+        return cache[nLeft][nRight];
+    }
+};
