Given an array nums
of n integers and an integer target
, find three integers in nums
such that the sum is closest to target
. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
My Solution:
This seems again like Leet 3Sum – I need to find a way to get close to a certain number using combinations of 3 nums of any list.
Since the target is not positive anymore I can’t eliminate certain cases, but still using sort will help me advance along certain operations and neglect other directions that will just decrease further.
#include<bits/stdc++.h>
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int diff = INT_MAX, res = 0;
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size()-2;i++)
{
for(int j=i+1,k=nums.size()-1;j<k;)
{
int sum = nums[i] + nums[j] + nums[k];
if(sum == target) return target;
if(diff>abs(sum-target))
{
res=sum;
diff=abs(sum-target);
}
(sum>target) ? k-- : j++;
}
}
return res;
}
};
Time Complexity:
O(nlog(n) +n^2) – O(nlog(n)) because I’m sorting the vectors + O(n*n) because we are traversing every vector against all other vectors and using two indexes(pointers/pivots) without this heuristic of avoiding unnecessary paths this would be an (On^3).
Space Complexity:
O(1) -Constant because I create a list of vectors, ahead of time
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