New Problem Solution - "Sum of Even Numbers After Queries"

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|985|[Sum of Even Numbers After Queries](https://leetcode.com/problems/sum-of-even-numbers-after-queries/) | [C++](./algorithms/cpp/sumOfEvenNumbersAfterQueries/SumOfEvenNumbersAfterQueries.cpp)|Easy|
 |984|[String Without AAA or BBB](https://leetcode.com/problems/string-without-aaa-or-bbb/) | [C++](./algorithms/cpp/stringWithoutAAAOrBBB/StringWithoutAAAOrBBB.cpp)|Easy|
 |983|[Minimum Cost For Tickets](https://leetcode.com/problems/minimum-cost-for-tickets/) | [C++](./algorithms/cpp/minimumCostForTickets/MinimumCostForTickets.cpp)|Medium|
 |981|[Time Based Key-Value Store](https://leetcode.com/problems/time-based-key-value-store/) | [C++](./algorithms/cpp/timeBasedKeyValueStore/TimeBasedKeyValueStore.cpp)|Medium|

algorithms/cpp/sumOfEvenNumbersAfterQueries/SumOfEvenNumbersAfterQueries.cpp

@@ -0,0 +1,68 @@
+// Source : https://leetcode.com/problems/sum-of-even-numbers-after-queries/
+// Author : Hao Chen
+// Date   : 2019-02-05
+
+/***************************************************************************************************** 
+ *
+ * We have an array A of integers, and an array queries of queries.
+ * 
+ * For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the 
+ * answer to the i-th query is the sum of the even values of A.
+ * 
+ * (Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the 
+ * array A.)
+ * 
+ * Return the answer to all queries.  Your answer array should have answer[i] as the answer to the 
+ * i-th query.
+ * 
+ * Example 1:
+ * 
+ * Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
+ * Output: [8,6,2,4]
+ * Explanation: 
+ * At the beginning, the array is [1,2,3,4].
+ * After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
+ * After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
+ * After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
+ * After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
+ * 
+ * Note:
+ * 
+ * 	1 <= A.length <= 10000
+ * 	-10000 <= A[i] <= 10000
+ * 	1 <= queries.length <= 10000
+ * 	-10000 <= queries[i][0] <= 10000
+ * 	0 <= queries[i][1] < A.length
+ * 
+ ******************************************************************************************************/
+class Solution {
+public:
+    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
+        int sum = 0;
+        for(int i=0; i<A.size(); i++) {
+            if (A[i] % 2 == 0) sum += A[i];
+        }
+        vector<int> result;
+        for(auto query : queries) {
+            int i = query[1];
+            int x = A[i] + query[0];
+            
+            if (A[i] % 2 == 0 && x %2 == 0) {
+                sum += ( -A[i] + x ); 
+                A[i] = x;
+            } else if (A[i] % 2 != 0 && x %2 == 0) {
+                sum += x;
+                A[i] = x;
+            } else if (A[i] % 2 == 0 && x %2 != 0) {
+                sum -= A[i];
+                A[i] = x;
+            } else if (A[i] % 2 != 0 && x %2 != 0){ 
+                A[i] = x;
+            }
+            result.push_back(sum);
+        }
+        
+        if (result.size()<=0) result.push_back(0);
+        return result;
+    }
+};