Add solution #985

Author: JoshCrozier

solutions/0985-sum-of-even-numbers-after-queries.js

@@ -27,3 +27,20 @@ var sumEvenAfterQueries = function(A, queries) {
     return [...sums, A.reduce((sum, n) => sum + (n % 2 === 0 ? n : 0), 0)];
   }, []);
 };
+
+// Significantly faster, less elegant approach:
+/**
+ * @param {number[]} A
+ * @param {number[][]} queries
+ * @return {number[]}
+ */
+var sumEvenAfterQueries = function(A, queries) {
+  return queries.reduce((sums, query) => {
+    const [v, i] = query;
+    const prev = A[i];
+    const lastSum = sums.length ? sums[sums.length - 1] : A.reduce((sum, n) => sum + (n % 2 === 0 ? n : 0), 0);
+    A[i] += v;
+    sums.push(lastSum - (prev % 2 === 0 ? prev : 0) + (A[i] % 2 === 0 ? A[i] : 0));
+    return sums;
+  }, []);
+};