Two Sum – LeetCode

1. Two Sum

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
• Only one valid answer exists.

Solution

Approach 1: Brute Force

class Solution {
    public int[] twoSum(int[] nums, int target) {
        for(int i = 0;i < nums.length; i++){
            for(int j = i+1;j < nums.length; j++){
                if(nums[i]+nums[j] == target)
                    return new int[]{i,j};
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

Time complexity – O(n²) Space complexity – O(1)

Approach 2: Using HashMap to store difference with index

class Solution {
	public int[] twoSum1(int[] nums, int target) {
        HashMap<Integer,Integer> preComputed = new HashMap<>();
        for(int i = 0;i< nums.length;i++){
            if(preComputed.containsKey(target-nums[i])){
                return new int[]{preComputed.get(target-nums[i]), i};
            }
            preComputed.put(nums[i],i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

Time complexity – O(n) Space complexity – O(n)