two-sum is a algorithm
Intro
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.
Solution
S1 - Brute Force Algorithm
Python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]
JavaScript
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function (nums, target) {
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] == target) {
return [i, j];
}
}
}
};
Java
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};
}
}
}
// return null if no result
return null;
}
}
S2 - Brute Force Algorithm
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
hashmap = {}
for i in range(len(nums)):
hashmap[nums[i]] = i
for i in range(len(nums)):
cmpt = target - nums[i] # complement
if cmpt in hashmap and hashmap[cmpt] != i:
return [i, hashmap[cmpt]]