Refactor hasPairWithSum using Set for O(n) lookup

Author: Ify-O

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,59 @@
 /**
  * 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)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * The original solution used nested loops, resulting in O(n²) time.
+ * This version uses a Set to store previously seen numbers,
+ * allowing constant-time lookups.
+ *
+ * @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) {
+  const seen = new Set();
+
+  for (const number of numbers) {
+    const complement = target - number;
+
+    if (seen.has(complement)) {
+      return true;
+    }
+
+    seen.add(number);
+  }
+
+  return false;
+}/**
+ * Find if there is a pair of numbers that sum to a given target value.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * The original solution used nested loops, resulting in O(n²) time.
+ * This version uses a Set to store previously seen numbers,
+ * allowing constant-time lookups.
  *
  * @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 number of numbers) {
+    const complement = target - number;
+
+    if (seen.has(complement)) {
+      return true;
     }
+
+    seen.add(number);
   }
+
   return false;
-}
+}