Refactor hasPairWithSum using Set

Author: Zobeir-Rigi

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,40 @@
 /**
  * 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(n²)
+ * Space Complexity: O(1)
+ * Optimal Time Complexity: O(n)
  *
- * @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
+ * Original complexity:
+ * The original solution uses nested loops. For each number,
+ * it checks every remaining number in the array looking for
+ * a pair whose sum equals the target. Since each number can
+ * be compared with many other numbers, the time complexity
+ * is O(n²).
+ *
+ * Refactor:
+ * Use a Set to store numbers that have already been seen.
+ * For each number, calculate the value needed to reach the
+ * target and check whether it exists in the Set.
+ * Set lookups are O(1), so we only need one pass through
+ * the array.
+ *
+ * Refactored Complexity:
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
  */
 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 needed = target - num;
+
+    if (seen.has(needed)) {
+      return true;
     }
+
+    seen.add(num);
   }
+
   return false;
-}
+}