1. Two Sum
Published in
1 min readJul 19, 2024
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?
Solution:
Approach 1 [Naive Approach]:
// Two sum
class Solution {
public int[] twoSum(int[] nums, int target) {
int a = nums.length;
int arr[] = new int[2];
for(int i=0;i<a;i++)
{
for(int j=i+1;j<a;j++)
{
if((nums[i]+nums[j])==target){
arr[0]=i;
arr[1]=j;
break;
}
}
}
return arr;
}
}
Approach 2 [Optimized Approach]:
import java.util.HashMap;
import java.util.Map;
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
}
}