some more sorting problems

Author: beck chen

Leetcode Visuals/Sorting and Searching.drawio

@@ -0,0 +1,23 @@
+/**
+ * @param {number[][]} intervals
+ * @return {number[][]}
+ */
+var merge = function (intervals) {
+    intervals.sort((interval1, interval2) => interval1[0] - interval2[0]);
+
+    const mergedCollection = [intervals[0]];
+    for (let intervalIndex = 1; intervalIndex < intervals.length; intervalIndex++) {
+        const intervalObj = intervals[intervalIndex];
+        const mergedObj = mergedCollection[mergedCollection.length - 1];
+
+        // no overlap
+        if (intervalObj[0] > mergedObj[1]) {
+            // simply append to mergedCollection
+            mergedCollection.push(intervalObj);
+        } else {
+            mergedObj[1] = Math.max(mergedObj[1], intervalObj[1]);
+        }
+    }
+
+    return mergedCollection;
+};

leetcode practice/sorting and searching/kthLargestElementInAnArray.js

@@ -25,29 +25,29 @@
  * @param {Interval[]} intervals
  * @return {Interval[]}
  */
-var merge = function(intervals) {
-    if (!intervals || intervals.length === 0){ return [];}
-    
+var merge = function (intervals) {
+    if (!intervals || intervals.length === 0) { return []; }
+
     // sort intervals forwards, as in one with small start comes first
     // ex. [[1,3],[8,10],[2,6]] -> [[1,3],[2,6],[8,10]] 
-    intervals.sort((elem1, elem2) => elem1.start-elem2.start);
+    intervals.sort((elem1, elem2) => elem1.start - elem2.start);
     // put the biggest one, [8,10] into results first
     let results = [intervals[0]];
-    
+
     // loop through the rest
-    for(let ii=1;ii<intervals.length;ii++){
+    for (let ii = 1; ii < intervals.length; ii++) {
         let elem = intervals[ii];
-        let prevElem = results[results.length-1];
+        let prevElem = results[results.length - 1];
         // this is to handle case of no overlap
         // in this case, we'll want to keep prevElem in our results since it's in order but also add elem to results (like when we see [2,6],[8,10]) results should contain both of these)
-        if (elem.start > prevElem.end){
+        if (elem.start > prevElem.end) {
             results.push(elem);
-        } 
+        }
         // this is where we handle the overlapping case. we want to keep the prevElem, but update its "end" to the end of the current one
         else {
-            results[results.length-1].end = Math.max(elem.end, prevElem.end);
+            results[results.length - 1].end = Math.max(elem.end, prevElem.end);
         }
     }
-    
+
     return results;
 };