source: https://leetcode.com/problems/two-sum
C/C++ Solution to LeetCode problem 1. Two Sum.
Problem
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.
Examples
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
Two solutions using hashmap (in reality the second one is an optimization of the first one).
- Add all values to a hash table as
key
and value a vector of the indexes where the value is located. - Iterate through the hash table, we need to find a value
x = target - current_key
.- if the
x
key exist, then we get the index for that value and for thecurrent_key
. - if not, we move the the next key.
- if the
The second solution performs the search of x
, if not found then the current_key
it is added to the hash table.
Solution 1:
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class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, vector<int>> hasht;
for(int i=0; i<nums.size(); i+=1)
hasht[nums[i]].push_back(i);
for(auto const& [key, value]: hasht) {
int need = target - key;
auto a = hasht.find(need);
if(a != hasht.end()) {
if (key == a->first) {
if (value.size() == 1)
continue;
return {value[0], value[1]};
}
else
return {value[0], a->second[0]};
}
}
return {};
}
};
Solution 2:
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class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> hasht;
for(int i = 0; i < nums.size(); i++){
if(hasht.find(target - nums[i]) == hasht.end())
hasht[nums[i]] = i;
else
return {hasht[target - nums[i]], i};
}
return {};
}
};