# Question

Formatted question description: https://leetcode.ca/all/1.html

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?

# Algorithm

In general, increasing time complexity comes at the cost of using more space, creating a trade-off. However, in this particular problem, I aim to achieve linear time complexity by only traversing the array once and using a HashMap to establish a mapping between each number and its coordinate position.

Since the search efficiency of a HashMap is constant, while traversing the array, we can subtract the current number from the target to find the other number needed. We can then look up this other number directly in the HashMap. Note that the number found may not be the immediate preceding number. For example, if the target is 4 and we have traversed a 2, the other number needed may not be the previous number, which is also 2.

The implementation steps for this approach are as follows: first traverse the array to create the HashMap mapping, and then traverse it again to search for the target numbers and record their indices.

# Code

• import java.util.HashMap;

class Two_Sum {

public static void main(String[] args) {

Two_Sum out = new Two_Sum();
Solution s = out.new Solution();

int[] a = {2, 7, 11, 15};
int target = 9;

int[] result = s.twoSum(a, target);

for (int each : result) {
System.out.println(each);
}
}

// time: O(N)
// space: O(N)
public class Solution {

public int[] twoSum(int[] nums, int target) {

// set up hashmap: remaining to original. thread-unsafe but for here is ok
HashMap<Integer, Integer> hm = new HashMap<>();

for (int i = 0; i < nums.length; i++) {

// if key in hm, result found.
// so that only one pass
if (hm.containsKey(nums[i])) {
int otherIndex = hm.get(nums[i]);

return new int[]{Math.min(i, otherIndex) + 1, Math.max(i, otherIndex) + 1};
}
// put new record into hashmap
else {
hm.put(target - nums[i], i);
}
}

return null;
}

}
}

############

class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> m = new HashMap<>();
for (int i = 0;; ++i) {
int x = nums[i];
int y = target - x;
if (m.containsKey(y)) {
return new int[] {m.get(y), i};
}
m.put(x, i);
}
}
}

• // OJ: https://leetcode.com/problems/two-sum/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
vector<int> twoSum(vector<int>& A, int target) {
vector<vector<int>> v;
int N = A.size(), L = 0, R = N - 1;
for (int i = 0; i < N; ++i) v.push_back({ A[i], i });
sort(begin(v), end(v));
while (L < R) {
int sum = v[L][0] + v[R][0];
if (sum == target) return { v[L][1], v[R][1] };
if (sum < target) ++L;
else --R;
}
return {};
}
};

• class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
m = {}
for i, v in enumerate(nums):
x = target - v
if x in m:
return [m[x], i]
m[v] = i

############

class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
d = {}
for i, num in enumerate(nums):
if target - num in d:
return [d[target - num], i]
d[num] = i
# no special case handling because it's assumed that it has only one solution


• func twoSum(nums []int, target int) []int {
m := map[int]int{}
for i := 0; ; i++ {
x := nums[i]
y := target - x
if j, ok := m[y]; ok {
return []int{j, i}
}
m[x] = i
}
}

• /**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function (nums, target) {
const m = new Map();
for (let i = 0; ; ++i) {
const x = nums[i];
const y = target - x;
if (m.has(y)) {
return [m.get(y), i];
}
m.set(x, i);
}
};


• class Solution {

/**
* @param Integer[] $nums * @param Integer$target
* @return Integer[]
*/
function twoSum($nums,$target) {
$len = count($nums);
for ($i = 0;$i < $len;$i++) {
for ($j = 1 +$i; $j <$len; $j++) { if ($nums[$i] +$nums[$j] ==$target) {
return [$i,$j];
}
}
}
}
}

• # @param {Integer[]} nums
# @param {Integer} target
# @return {Integer[]}
def two_sum(nums, target)
nums.each_with_index do |x, idx|
if nums.include? target - x
return [idx, nums.index(target - x)] if nums.index(target - x) != idx
end
next
end
end


• import scala.collection.mutable

object Solution {
def twoSum(nums: Array[Int], target: Int): Array[Int] = {
var map = new mutable.HashMap[Int, Int]()
for (i <- 0 to nums.length) {
if (map.contains(target - nums(i))) {
return Array(map(target - nums(i)), i)
} else {
map += (nums(i) -> i)
}
}
Array(0, 0)
}
}


• public class Solution {
public int[] TwoSum(int[] nums, int target) {
var m = new Dictionary<int, int>();
for (int i = 0, j; ; ++i) {
int x = nums[i];
int y = target - x;
if (m.TryGetValue(y, out j)) {
return new [] {j, i};
}
if (!m.ContainsKey(x)) {
}
}
}
}

• #[
Author: @joe733
]#

import std/enumerate

proc twoSum(nums: seq[int], target: int): seq[int] =
var
bal: int
tdx: int
for idx, val in enumerate(nums):
bal = target - val
if bal in nums:
tdx = nums.find(bal)
if idx != tdx:
return @[idx, tdx]


• use std::collections::HashMap;

pub fn soluation(nums: Vec<i32>, target: i32) -> Vec<i32> {
let mut map = HashMap::new();
for (i, item) in nums.iter().enumerate() {
if map.contains_key(item) {
return vec![i as i32, map[item]];
} else {
let x = target - nums[i];
map.insert(x, i as i32);
}
}
unreachable!()
}


• class Solution {
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
var m = [Int: Int]()
var i = 0
while true {
let x = nums[i]
let y = target - nums[i]
if let j = m[target - nums[i]] {
return [j, i]
}
m[nums[i]] = i
i += 1
}
}
}