Coder's Cat

LeetCode: two sum


LeetCode sum two: explanations and solutions with Cpp/Java/Python/Ruby

Challenge Description


Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].


Naive solutions

First, we come out with the simplest solution, iterate with array element, check whether there are two value’s sum is target, if true, we get a result, so pseudo code is:

 foreach(v1 in array) {
foreach(v2 in array){
if (v2 == target - v1) {
return (index v1, index v2)

Time complexity: $O(N^2)$ Space complexity: $O(1)$

Optimization with hashmap

Time general idea is use space to optimize time complexity. We can use a hashmap to optimize the second loop from O(N) to O(1).

So we build a hashmap, the key is the element of array, and value is the index of the element. This hashmap can be constructed before hand, but a better way is to construct in the same loop of checking.

Time complexity: $O(N)$ Space complexity: $O(N)$


class Solution {

vector<int> twoSum(vector<int>& arr, int target) {
map<int, int> m;
vector<int> result;
for (int i=0; i < arr.size(); i++) {
if ( m.find(target - arr[i]) == m.end() ) {
m[arr[i]] = i; // Note: key is element, value is index
result.push_back(m[target - arr[i]]);
return result;



Java version is same logic with Cpp version:

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);
throw new IllegalArgumentException("No two sum solution");


Why we don’t need a hashmap in Python version, please note the second if test. It’s a pragmatic style for Python check whether a element is in array:

if elment in array: 

Note we use nums[i+1:] in checking, that means we use part of nums index begin with i+1. According to the document, the time complexity of ‘in’ operator is O(N) for list, so this implementation’s time complexity is: $O(N^2)$

class Solution(object):

def twoSum(self, nums, target):
for i in range(0, len(nums)-1):
if (target - nums[i]) in nums[i+1:]:
return [nums.index(nums[i]), nums.index((target-nums[i]), i+1)]


Ruby’s version is short, the logic is same with a Hashmap.

def two_sum(nums, target)
hash = {}
nums.each_with_index do |n, index|
first = hash[target - n]
if first != nil && first != index
return [first, index]
hash[n] = index

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