Question 1
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] Output: 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?=> Coding
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> twoSum;
for(int i=0;i<nums.size();i++){
for(int j=i+1;j<nums.size();j++){
if((nums[i]+nums[j])==target){
twoSum.push_back(i);
twoSum.push_back(j);
}
}
}
return twoSum;
}
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