refactor 435

fishercoder1534 · fishercoder1534 · commit f53161a79e18 · 2021-10-13T11:19:55.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_435.java b/src/main/java/com/fishercoder/solutions/_435.java
@@ -11,6 +11,9 @@ public static class Solution1 {
          * and https://discuss.leetcode.com/topic/65594/java-least-is-most
          * Sort the intervals by their end time, if equal, then sort by their start time.
          * Then merge based on ending time.
+         *
+         * This comment is inspiring as of why sorting by end time could guarantee the correct answer: https://leetcode.com/problems/non-overlapping-intervals/discuss/91713/Java:-Least-is-Most/96271
+         * In my words: if the earliest finished interval is not included, it's more likely that others (with later finish time) will overlap with others.
          */
         public int eraseOverlapIntervals(int[][] intervals) {
             Arrays.sort(intervals, (a, b) -> a[1] != b[1] ? a[1] - b[1] : a[0] - b[0]);
