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Problem Statement
Given an integer array nums of length n 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 1:
Input: nums = [-1,2,1,-4], target = 1 Output: 2 Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
Example 2:
Input: nums = [0,0,0], target = 1 Output: 0
Constraints:
3 <= nums.length <= 1000-1000 <= nums[i] <= 1000-104 <= target <= 104
- Time: O(n^2)O(n2)
- Space: O(|\texttt{ans}|)O(∣ans∣)
3 Sum Closest Program Problem Solution In C++
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int ans = nums[0] + nums[1] + nums[2];
sort(begin(nums), end(nums));
for (int i = 0; i + 2 < nums.size(); ++i) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
// choose nums[i] as the first num in the triplet,
// and search the remaining nums in [i + 1, n - 1]
int l = i + 1;
int r = nums.size() - 1;
while (l < r) {
const int sum = nums[i] + nums[l] + nums[r];
if (sum == target)
return sum;
if (abs(sum - target) < abs(ans - target))
ans = sum;
if (sum < target)
++l;
else
--r;
}
}
return ans;
}
};
3 Sum Closest Program Problem Solution in JAVA
class Solution {
public int threeSumClosest(int[] nums, int target) {
int ans = nums[0] + nums[1] + nums[2];
Arrays.sort(nums);
for (int i = 0; i + 2 < nums.length; ++i) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
// choose nums[i] as the first num in the triplet,
// and search the remaining nums in [i + 1, n - 1]
int l = i + 1;
int r = nums.length - 1;
while (l < r) {
final int sum = nums[i] + nums[l] + nums[r];
if (sum == target)
return sum;
if (Math.abs(sum - target) < Math.abs(ans - target))
ans = sum;
if (sum < target)
++l;
else
--r;
}
}
return ans;
}
}
3 Sum Closest Program Problem Solution In Python
class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
ans = nums[0] + nums[1] + nums[2]
nums.sort()
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
l = i + 1
r = len(nums) - 1
while l < r:
sum = nums[i] + nums[l] + nums[r]
if sum == target:
return sum
if abs(sum - target) < abs(ans - target):
ans = sum
if sum < target:
l += 1
else:
r -= 1
return ans
