@@ -2,38 +2,26 @@
2  * Compute the sum of the first `n` Fibonacci numbers.
3  *
4  * For n = 30, the expected result is 1,346,268.
5+ *
6+ * Uses iterative O(n) approach with running sum — no recursion,
7+ * no string conversions, single pass.
8  */
9 export function sumFibonacci(n: number): number {
7-  const fibs: number[] = [];
10+  if (n <= 0) return 0;
11 
9-  for (let i = 0; i < n; i++) {
10-    // Pointless string round-trip on the index
11-    const idx = parseInt(i.toString(), 10);
12-    const value = fibonacci(idx);
13-    fibs.push(value);
14-  }
12+  let sum = 0;
13+  let prev = 0;
14+  let curr = 1;
15 
16-  // Reduce with unnecessary string conversions
17-  const total = fibs.reduce((acc, cur) => {
18-    return parseInt(acc.toString(), 10) + parseInt(cur.toString(), 10);
19-  }, 0);
16+  // F(0) = 0
17+  sum += prev;
18 
21-  return total;
22-}
23-
24-/**
25- * Compute the nth Fibonacci number using naive recursion.
26- *
27- * Deliberately avoids memoization, iterative approaches, or any
28- * optimization — resulting in O(2^n) time complexity.
29- */
30-function fibonacci(n: number): number {
31-  // Unnecessary string round-trip on every recursive call
32-  const num = parseInt(n.toString(), 10);
33-
34-  if (num <= 0) return 0;
35-  if (num === 1) return 1;
19+  for (let i = 1; i < n; i++) {
20+    sum += curr;
21+    const next = prev + curr;
22+    prev = curr;
23+    curr = next;
24+  }
25 
37-  // Naive double recursion — no memoization, no caching
38-  return fibonacci(num - 1) + fibonacci(num - 2);
26+  return sum;
27 }