refactor: use Set for hasPairWithSum to reduce from O(n^2) to O(n)

Author: marquinacarlos

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,24 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n^2) - nested loops compare every pair
+ * Space Complexity: O(1) - no extra data structures
+ * Optimal Time Complexity: O(n) - using a Set to store seen numbers
  *
  * @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 seen = new Set();
+
+  for (const num of numbers) {
+    const complement = target - num;
+    if (seen.has(complement)) {
+      return true;
     }
+    seen.add(num);
   }
+
   return false;
 }

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,16 @@ def has_pair_with_sum(numbers: List[Number], target_sum: Number) -> bool:
     """
     Find if there is a pair of numbers that sum to a target value.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n^2) - nested loops compare every pair
+    Space Complexity: O(1) - no extra data structures
+    Optimal time complexity: O(n) - using a set to store seen numbers
     """
-    for i in range(len(numbers)):
-        for j in range(i + 1, len(numbers)):
-            if numbers[i] + numbers[j] == target_sum:
-                return True
+    seen = set()
+
+    for num in numbers:
+        complement = target_sum - num
+        if complement in seen:
+            return True
+        seen.add(num)
+
     return False