Approach

Scan left to right while storing previously seen values in a hashmap:

• At index j, the needed partner value is target - a[j].
• If that value is already in the hashmap at index i, output i + 1 and j + 1.
• Otherwise, store a[j] with its index and continue.

Because each lookup and insert is ~O(1)~ on average, total time complexity is ~O(n)~.

Solution (Python)

n, target = (int(x) for x in input().split())
a = [int(x) for x in input().split()]

seen: dict[int, int] = {}

for i in range(n):
    need = target - a[i]
    if need in seen:
        print(seen[need] + 1, i + 1)
        break
    seen[a[i]] = i