@@ -1,21 +1,43 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Inefficiency: the original used a loop inside a loop to check every
+ * possible pair of numbers.
+ *
+ * Originally:
+ * Time Complexity: O(n²) - for each number, the inner loop scans the
+ *   remaining numbers, so about n * (n - 1) / 2 pair checks.
+ * 
+ * Space Complexity: O(1) - only the loop counters i and j are used.
+ * 
+ * Optimal Time Complexity: O(n) - every number must be read at least once,
+ *   and a Set gives O(1) average lookups, so one pass is enough
+ *
+ * After Refactor (reduced from quadratic to linear time by remembering the
+ * numbers already seen in a Set instead of re-scanning the list):
+ * 
+ * Time Complexity: O(n) - one loop, with O(1) average has() and add().
+ * 
+ * Space Complexity: O(n) - visitedNumbers can hold up to n numbers.
+ * 
+ * Optimal Time Complexity: O(n) - every number must be read at least once,
+ *   so we can't do better than linear.
  *
  * @param {Array<number>} numbers - Array of numbers to search through
  * @param {number} target - Target sum to find
  * @returns {boolean} True if pair exists, false otherwise
  */
 export function hasPairWithSum(numbers, target) {
-  for (let i = 0; i < numbers.length; i++) {
-    for (let j = i + 1; j < numbers.length; j++) {
-      if (numbers[i] + numbers[j] === target) {
-        return true;
-      }
+  const visitedNumbers = new Set();
+
+  for (let num of numbers) {
+    const partnerNum = target - num;
+
+    if (visitedNumbers.has(partnerNum)) {
+      return true;
     }
+    visitedNumbers.add(num);
   }
+
   return false;
-}
+}