refactor(Array): optimize solution for 'how many numbers are smaller than the current number' problem

Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js

@@ -50,9 +50,28 @@
  */
 
 var smallerNumbersThanCurrent = function (nums) {
-  // Step 1: count the frequency of each number
+  // Step 1: Begin by initializing a [Frequency Map]()
   const freq = {};
-  for (let x of nums) freq[x] = (freq[x] || 0) + 1;
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
 
-  return nums.map((x) => freq[x] || 0);
+  // Step 2: Sort the numbers by ascending order
+  const sorted = Object.keys(freq).sort((a, b) => a - b);
+
+  // Step 3: Init a count of numbers smaller than the active number
+  let count = 0;
+
+  // Step 4: Init a map to track number of values smaller for each number
+  const smaller = {};
+
+  // Step 5: Iterate over the sorted list
+  for (let num of sorted) {
+    // Set count for active number
+    smaller[num] = count;
+
+    // Update the count by frequency
+    count += freq[num];
+  }
+
+  // Step 6: Use original list and find number of smaller values than it
+  return nums.map((n) => smaller[n]);
 };