1. Two Sum
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] Explanation: 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
- Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than O(n2)
time complexity?
Python solution using HashMap
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
new_hash_map = {}
for i, n in enumerate(nums):
difference = target - n
if difference in new_hash_map:
return(i, new_hash_map[difference])
else:
new_hash_map[n] = i
Time complexity O(N)
Java Solution
class Solution {
public boolean containsDuplicate(int[] nums) {
Set<Integer> numbers = new HashSet<Integer>();
for (int num : nums) {
if (numbers.contains(num)) return true;
numbers.add(num);
}
return false;
}
}
Time complexity O(N)
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