Easy
Given an array of integers nums
and an integer target
, return indices of the two numbers such that they add up to target
.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
Follow-up: Can you come up with an algorithm that is less than O(n2)
time complexity?
(define (two-sum-iter nums target hash index)
(cond
((null? nums) '())
((hash-has-key? hash (car nums) ) (list (hash-ref hash (car nums) ) index ) )
(else (two-sum-iter (cdr nums) target (hash-set hash (- target (car nums ) ) index ) (+ index 1) ))
)
)
(define/contract (two-sum nums target)
(-> (listof exact-integer?) exact-integer? (listof exact-integer?))
(two-sum-iter nums target (make-immutable-hash) 0)
)