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Sep 1, 2023
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Add 16. 3Sum Closest #322

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16. 3Sum Closest
cheehwatang Sep 1, 2023
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Added 16. 3Sum Closest
cheehwatang Sep 1, 2023
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67 changes: 55 additions & 12 deletions README.md
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<th>Solution</th>
<th>Topics</th>
</tr>
<tr>
<td align="center">September 1st</td>
<td>16. <a href="https://leetcode.com/problems/3sum-closest/">3Sum Closest</a></td>
<td align="center">$\text{\color{Dandelion}Medium}$</td>
<td align="center">
<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/16.%203Sum%20Closest/ThreeSumClosest.java">Sorting & Two Pointers</a>
</td>
<td align="center">
<a href="#array">Array</a>,
<a href="#sorting">Sorting</a>,
<a href="#two-pointers">Two Pointers</a>
</td>
</tr>
<tr>
<td align="center">August 31st</td>
<td>15. <a href="https://leetcode.com/problems/3sum/">3Sum</a></td>
Expand Down Expand Up @@ -76,18 +89,6 @@
<a href="#string">String</a>
</td>
</tr>
<tr>
<td align="center">August 27th</td>
<td>3. <a href="https://leetcode.com/problems/longest-substring-without-repeating-characters/">Longest Substring Without Repeating Characters</a></td>
<td align="center">$\text{\color{Dandelion}Medium}$</td>
<td align="center">
<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/3.%20Longest%20Substring%20Without%20Repeating%20Characters/LongestSubstringWithoutRepeatingCharacters.java">Sliding Window</a>
</td>
<td align="center">
<a href="#sliding-window">Sliding Window</a>,
<a href="#string">String</a>
</td>
</tr>
</table>
</br>
<hr>
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</td>
<td></td>
</tr>
<tr>
<td align="center">16</td>
<td><a href="https://leetcode.com/problems/3sum-closest/">3Sum Closest</a></td>
<td align="center">
<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/16.%203Sum%20Closest/ThreeSumClosest.java">Java</a>
</td>
<td align="center">$\text{\color{Dandelion}Medium}$</td>
<td align="center">
<a href="#array">Array</a>,
<a href="#sorting">Sorting</a>,
<a href="#two-pointers">Two Pointers</a>
</td>
<td></td>
</tr>
<tr>
<td align="center">26</td>
<td><a href="https://leetcode.com/problems/remove-duplicates-from-sorted-array/">Remove Duplicates from Sorted Array</a></td>
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</td>
<td></td>
</tr>
<tr>
<td align="center">16</td>
<td><a href="https://leetcode.com/problems/3sum-closest/">3Sum Closest</a></td>
<td align="center">
<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/16.%203Sum%20Closest/ThreeSumClosest.java">Java</a>
</td>
<td align="center">$\text{\color{Dandelion}Medium}$</td>
<td align="center">
<a href="#array">Array</a>,
<a href="#sorting">Sorting</a>,
<a href="#two-pointers">Two Pointers</a>
</td>
<td></td>
</tr>
<tr>
<td align="center">49</td>
<td><a href="https://leetcode.com/problems/group-anagrams/">Group Anagrams</a></td>
Expand Down Expand Up @@ -9189,6 +9218,20 @@
</td>
<td></td>
</tr>
<tr>
<td align="center">16</td>
<td><a href="https://leetcode.com/problems/3sum-closest/">3Sum Closest</a></td>
<td align="center">
<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/16.%203Sum%20Closest/ThreeSumClosest.java">Java</a>
</td>
<td align="center">$\text{\color{Dandelion}Medium}$</td>
<td align="center">
<a href="#array">Array</a>,
<a href="#sorting">Sorting</a>,
<a href="#two-pointers">Two Pointers</a>
</td>
<td></td>
</tr>
<tr>
<td align="center">19</td>
<td><a href="https://leetcode.com/problems/remove-nth-node-from-end-of-list/">Remove Nth Node From End of List</a></td>
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93 changes: 93 additions & 0 deletions solutions/16. 3Sum Closest/ThreeSumClosest.java
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package com.cheehwatang.leetcode;

import java.util.Arrays;

// Time Complexity : O(n^2),
// where 'n' is the length of 'nums'.
// We traverse 'nums' to check each 'nums[i]', with each element we use 2 pointers,
// thus it is a nested loop, with O(n^2) time complexity.
// The Arrays.sort() function has a time complexity of O(n logn).
//
// Space Complexity : O(1),
// as the auxiliary space used is independent of the input.

public class ThreeSumClosest {

// Approach:
// Use sorting and two pointers to search for all the triplets and check for the closest sum to 'target'.
// For each 'nums[i]' in 'nums', we can check the sum with nums[j] and nums[k] using two pointers,
// 'left' and 'right' to traverse the remaining numbers.
// In order to use two pointers in this manner, we have to first sort the array in ascending order.
// This allows use to check if nums[i] + nums[j] + nums[k] is greater or less than 0.
// If the sum is less than 'target', then we shift the 'left' pointer to the right.
// If the sum is greater than 'target', then we shift the 'right' pointer to the left.
// If we found the sum to be equal to 'target', then we return 'target'.
//
// To find the closest number to 'target', we use a variable to keep track and update when found a closer number.

public int threeSumClosest(int[] nums, int target) {
// Sort the array, and use 2 variables to keep track of the minimum difference and the sum of the triplets.
Arrays.sort(nums);

int minDiff = Integer.MAX_VALUE;
int threeSum = 0;
for (int i = 0; i < nums.length - 2; i++) {
// If nums[i] is identical with the previous number skip the number to shorten the time.
if (i > 0 && nums[i] == nums[i - 1]) continue;

// The following 'highSum' and 'lowSum' is additional check to shorten the time.
// Otherwise, the rest of the code works on their own.
//
// The 'highSum' is to check if the largest possible sum for nums[i] is less than the target.
// Since we have sorted the 'nums', we know all the remaining numbers will be smaller than the 'highSum'.
// Thus, we can skip the remaining numbers if 'highSum' is lesser than target.
int highSum = nums[i] + nums[nums.length - 1] + nums[nums.length - 2];
if (highSum < target) {
if (target - highSum < minDiff) {
threeSum = highSum;
minDiff = target - highSum;
}
continue;
}
// The 'lowSum' is to check if the smallest possible sum for nums[i] is greater than the target.
// Since we have sorted the 'nums', we know all the remaining numbers will be greater than the 'lowSum'.
// If the 'lowSum' is greater than the target,
// the remaining numbers, including the subsequent i-th numbers will be greater than the 'lowSum'.
// Thus, we are sure that 'lowSum' is the closest triplet to 'target', if found.
int lowSum = nums[i] + nums[i + 1] + nums[i + 2];
if (lowSum > target) {
if (lowSum - target < minDiff) {
return lowSum;
}
}

// After the initial checks, use two pointers, 'left' and 'right'.
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[left] + nums[right] + nums[i];
int difference = Math.abs(sum - target);
// If the triplet's 'minDiff' is lesser, then record it as the closest 'threeSum' for now.
if (difference < minDiff) {
minDiff = difference;
threeSum = sum;
}
// Otherwise, decrease the 'right' if sum > target,
if (sum > target) {
right--;
while (left < right && nums[right] == nums[right + 1]) right--;
}
// and increase 'left' if sum < target.
else if (sum < target) {
left++;
while (left < right && nums[left] == nums[left - 1]) left++;
}
// If found sum == target, then it is the closest possible.
else {
return sum;
}
}
}
return threeSum;
}
}