Two Sum Leetcode Program Solution in C++ Java & Python
Given an array of integers numbers 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.
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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.
Time Complexity: O(n)
Space Complexity: O(n)
Two Sum Leetcode Program Solution in C++
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> numToIndex;
for (int i = 0; i < nums.size(); ++i) {
if (numToIndex.count(target - nums[i]))
return {numToIndex[target - nums[i]], i};
numToIndex[nums[i]] = i;
}
throw;
}
};
Two Sum Leetcode Program Solution in Java
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> numToIndex = new HashMap<>();
for (int i = 0; i < nums.length; ++i) {
if (numToIndex.containsKey(target - nums[i]))
return new int[] {numToIndex.get(target - nums[i]), i};
numToIndex.put(nums[i], i);
}
throw new IllegalArgumentException();
}
}
Two Sum Leetcode Program Solution in Python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
numToIndex = {}
for i, num in enumerate(nums):
if target - num in numToIndex:
return numToIndex[target - num], i
numToIndex[num] = i
