Add solution #1010

Author: JoshCrozier

README.md

@@ -145,6 +145,7 @@
 985|[Sum of Even Numbers After Queries](./0985-sum-of-even-numbers-after-queries.js)|Easy|
 989|[Add to Array-Form of Integer](./0989-add-to-array-form-of-integer.js)|Easy|
 1009|[Complement of Base 10 Integer](./1009-complement-of-base-10-integer.js)|Easy|
+1010|[Pairs of Songs With Total Durations Divisible by 60](./1010-pairs-of-songs-with-total-durations-divisible-by-60.js)|Medium|
 1037|[Valid Boomerang](./1037-valid-boomerang.js)|Easy|
 1103|[Distribute Candies to People](./1103-distribute-candies-to-people.js)|Easy|
 1108|[Defanging an IP Address](./1108-defanging-an-ip-address.js)|Easy|

solutions/1010-pairs-of-songs-with-total-durations-divisible-by-60.js

@@ -0,0 +1,24 @@
+/**
+ * 1010. Pairs of Songs With Total Durations Divisible by 60
+ * https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/
+ * Difficulty: Medium
+ *
+ * You are given a list of songs where the ith song has a duration of time[i] seconds.
+ *
+ * Return the number of pairs of songs for which their total duration in seconds is
+ * divisible by 60. Formally, we want the number of indices i, j such that i < j with
+ * (time[i] + time[j]) % 60 == 0.
+ */
+
+/**
+ * @param {number[]} time
+ * @return {number}
+ */
+var numPairsDivisibleBy60 = function(time) {
+  const mods = new Array(60).fill(0);
+  return time.reduce((sum, t) => {
+    sum += mods[(60 - t % 60) % 60];
+    mods[t % 60]++;
+    return sum;
+  }, 0);
+};