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Interview Question 3Sum

Last update: 12/28/2016 9:37:00 AM

By Fatih KABAKCI

The problem is to figure out all unique triplets in a given array which sum of any 3 elements in that array is zero. O(n2) is accepted.

Problem #15 - 3Sum

Description: Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Difficulty: Medium

Solution in Java

When the problem is seen, the laziest solution would be brute force which is O(n3) algorithm complexity. Designing three loop for each element gives O(n3) complexity as the following.

Brute Force - O(n3)

This method is considered Time Limit Exceeded on LeetCode platform. However it might be still good idea to understand problem from beginning, basic point.

public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> l = new ArrayList<List<Integer>>();

		for (int i = 0; i < nums.length; i++) {
			int a = nums[i];
			for (int j = i + 1; j < nums.length; j++) {
				int b = nums[j];
				for (int k = j + 1; k < nums.length; k++) {
					int c = nums[k];
					if (a + b + c == 0) {
						List<Integer> sl = new ArrayList<Integer>();						
						sl.add(a);
						sl.add(b);
						sl.add(c);
						Collections.sort(sl);
						l.add(sl);
					}
				}
			}
		}
		l = new ArrayList<List<Integer>>(new LinkedHashSet<List<Integer>>(l));
		return l;
    }

However, the solution which is above, would not be acceptable well enough. Because, the entire code looks all possibilities even though some of them are unnecessary to look for. So, what we can do is to try to diminish complexity.

Better Solution - O(n2)

We might start to sort the input array first. Then we can get 3 elements from first, second and last one. In regard to result of sum of them, which might be zero, less than zero or greater than zero, we move loop variables to up or down. Here might be better solution as below.

For example,



As seen above, we can code the solution with O(n2) complexity in Java as below.

package com.fatihkabakci.Medium._3Sum;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Set;

/**
 * 
 * @author fkabakci
 * 
 * Problem Description: Given an array S of n integers, are there elements a, b, c in S such that 
 * a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
 * 
 *  For example, given array S = [-1, 0, 1, 2, -1, -4],
 *
 *	A solution set is:
 *	[
 *	  [-1, 0, 1],
 *	  [-1, -1, 2]
 *	]
 * 
 * Solution:
 * 
 * 1. Sort the input array.
 * 2. Start with getting first, second and last element.
 * 3. Add up them.
 * 4. If sum equals to zero,
 * 		 then add them into the list, and skip to third one, and previous one element of the last one.
 * 	  If sum is less than zero,
 * 	     then close to higher one. (Skip to third one by increasing loop variable j)
 * 	  If sum is greater than zero,
 * 		 then close to lower one. (Skip to previous one of the last one by decreasing loop variable k)
 * 5. Lastly, get rid of duplicated elements by using Collection Set.
 */

public class _3Sum {
	public List<List<Integer>> threeSum(int[] nums) {
		if (nums.length < 3)
			return new ArrayList<List<Integer>>();
		
		Arrays.sort(nums);
		List<List<Integer>> list = new ArrayList<List<Integer>>();
		
		int i = 0, j, k;
		while (i < nums.length - 1) {
			j = i + 1;
			k = nums.length - 1;
			while (j < k) {
				int sum = nums[i] + nums[j] + nums[k];
				if (sum == 0) {
					List<Integer> tsum = new ArrayList<Integer>();
					tsum.add(nums[i]);
					tsum.add(nums[j]);
					tsum.add(nums[k]);
					list.add(tsum);
					j++;
					k--;
				} else if (sum < 0)
					j++;
				else
					k--;
			}
			i++;
		}
		Set<List<Integer>> set = new LinkedHashSet<List<Integer>>(list);
		return new ArrayList<List<Integer>>(set);
	}
}

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