Solution 1 - Hashmap

Using a hash map, we can store the numbers in the array as keys and their indices as values. To check if a pair exists, we use the hashmap to quickly look up if any past value plus the current one equals $x$. If so, we output the current index and the one stored in the hashmap.

Implementation

Time Complexity: $\mathcal{O}(N \log N)$

Copy
n, x = map(int, input().split())
nums = list(map(int, input().split()))

m = {}
for i, a in enumerate(nums):
	if x - a in m:
		print(i + 1, m[x - a] + 1)
		break
	m[a] = i
else:
	print("IMPOSSIBLE")

Solution 2 - Two Pointers

We can sort the array and use a left pointer and right pointer. If the sum is greater than $x$, we decrement right. If the sum is less than $x$, we increment left.

Implementation

Time Complexity: $\mathcal{O}(N\log N)$

Copy
n, x = map(int, input().split())

nums = [(int(val), i) for i, val in enumerate(input().split())]
nums.sort()

l = 0
r = n - 1
while l < r:
	sum = nums[l][0] + nums[r][0]
	if sum == x: