LeetCode – 3Sum

Problem:

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.

Note:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

Java Solution

This problem can be solved by using two pointers. Time complexity is O(n^2).

To avoid duplicate, we can take advantage of sorted arrays, i.e., move pointers by >1 to use same element only once.

public List<List<Integer>> threeSum(int[] nums) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
 
    if(nums == null || nums.length<3)
        return result;
 
    Arrays.sort(nums);
 
    for(int i=0; i<nums.length-2; i++){
        if(i==0 || nums[i] > nums[i-1]){
            int j=i+1;
            int k=nums.length-1;
 
            while(j<k){
                if(nums[i]+nums[j]+nums[k]==0){
                    List<Integer> l = new ArrayList<Integer>();
                    l.add(nums[i]);
                    l.add(nums[j]);
                    l.add(nums[k]);
                    result.add(l);
 
                    j++;
                    k--;
 
                    //handle duplicate here
                    while(j<k && nums[j]==nums[j-1])
                        j++;
                    while(j<k && nums[k]==nums[k+1])
                        k--;
 
                }else if(nums[i]+nums[j]+nums[k]<0){
                    j++;
                }else{
                    k--;
                }
            }
        }
 
    }
 
    return result;
}

Category >> Algorithms

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Serj Sintsov on 2014-4-12

Seems, I have a better solutions of this problem

boolean is3SumArray(int[] a) {
if (a.length < 3) return false;

Arrays.sort(a);

for (int i = 0; i < a.length-2; i++) {
int x = a[i];
int yI = i+1;
int zI = a.length-1;

while (yI 0) zI–;
else yI++;
}
}

return false;
}
doris717 on 2014-4-25

same complexity
H Chen on 2014-4-25

Your solution is wrong. You did not eliminate the duplicated combinations.
Yancey on 2014-5-29

Make some changes to solution1 and accepted!!

public class Solution {

public static ArrayList<ArrayList> threeSum(int[] num) {

//sort array

Arrays.sort(num);

ArrayList<ArrayList> result = new ArrayList<ArrayList>();

for(int i=0; i 0) break;

for(int j=i+1; j 0 && num[j] > 0) break;

for(int k=j+1; k<num.length; k++){

if(num[i] + num[j] + num[k] == 0) {

ArrayList each = new ArrayList();

each.add(num[i]);

each.add(num[j]);

each.add(num[k]);

result.add(each);

}

while (1+k<num.length && num[k]==num[k+1]) k++;

}

while (j+1<num.length && num[j]==num[j+1]) j++;

}

while (i+1<num.length && num[i]==num[i+1]) i++;

}

return result;

}

}
Ajay on 2014-6-24

why are you running only num.length-2 times.
Ajay on 2014-6-24

okay in last iteration we have only two element left so we should come out from the loop…
I got it !!!
Anirudh on 2014-6-29

if (i == 0 || num[i] > num[i – 1]) {

How does this execute for i = 0 case ? , wont you be trying to evaluate num[-1] ??
Kick Buttowski on 2014-8-4

my solution

int[] array = {-1, 0, 1, 2, -1, -4, 2,-1,-1};

int current = array[0];

for (int i = 0; i < array.length; i++) {

for (int j = i; j <= 2 + i && i+2 < array.length -1 ; j++) {

if (current + array[j] + array[j + 1] == 0) {

System.out.println(current + " " + array[j] + " " + array[j + 1]);

}

current = array[i];

}

}
Gaurav Sharma on 2014-12-8

If first condition is true then second condition will not be evaluated
Ranjit on 2015-1-8

int a[] = {2, 7, 11, 0, 9, 15, 7};

for(int i = 0; i<a.length; i++){

for(int j = i+1; j<a.length; j++){

for(int k = j+1; k<a.length; k++){

if(a[i]+a[j]+a[k] == 9){

System.out.println(i + " and "+ j + " and " + k);

}

}

}

}
Ranjit on 2015-1-8

int a[] = {2, 7, 11, 0, 9, 15, 7};

for(int i = 0; i<a.length; i++){
for(int j = i+1; j<a.length; j++){
for(int k = j+1; k<a.length; k++){
if(a[i]+a[j]+a[k] == 9){

System.out.println(i + " and "+ j + " and " + k);
}
}
}
}
PK on 2015-1-27

Pseudocode :
Iterate the array num[].
Int target = -num[i];
Now find twoSum for above target in array num using hashmap.
ryanlr on 2015-3-3

A great idea! Here is how to use hashmap to solve 2Sum problem: http://www.programcreek.com/2012/12/leetcode-solution-of-two-sum-in-java/
ryanlr on 2015-3-3

It looks working, but time is n^3.
Adil qureshi on 2015-6-17

A lot of code can be avoided by using a class Triplet {} and adding it to a set
Malay Shah on 2015-8-18

It will give time limit exceeded in O(n^2) solution…
Logic Luo on 2015-8-27

Hello, I have a problem. I don’t understand how avoid duplicate solutions using ”
if (i == 0 || num[i] > num[i – 1]) {“
Mohanasundaram Veeramuthu on 2015-9-7

Hi ryanlr,

Can you please correct my below prog that I wrote using PK’s algo & your 2Sum code using HashMap? I think I didn’t get the algorithm properly. Thank you.

public List<List> threeSum(int[] nums) {

List<List> result = new ArrayList<List>();
HashSet<List> hSet = new HashSet<List>();

if(nums.length < 3)
return result;

HashMap hMap = new HashMap();

for(int i=0; i<nums.length; i++) {

int target = -nums[i];
if(hMap.containsKey(nums[i])) {

List temp = new ArrayList();
temp.add(nums[i]);
temp.add(nums[hMap.get(nums[i])]);
temp.add(-target);

if(!hSet.contains(temp)) {

hSet.add(temp);
result.add(temp);
}
}
else
hMap.put(target – nums[i], i);
}
return result;
}
Omar Edgardo Lugo Sánchez on 2015-9-13

my solution complexity n*n

HashMap map = new HashMap();
int[] a = new int[] {4,5,6,7,8,-10};

// Arrays.sort(a);
for(int j=0;j<a.length;j++){
int suma = -1*a[j];
for(int i=j;i<a.length;i++){
if(i!=j){
int b = suma-a[i];
if(map.containsKey(a[i])==false){
map.put(b,a[i]);
}else{
Integer g = (Integer) map.get(a[i]);
System.out.println(a[i]+"+"+g+"+"+-1*suma+"=0");
}
}
}
map.clear();
}
Candis on 2015-10-12

Can someone plz explain why my code is getting timed out:

public static List<List> threeSum(int[] nums) {

Set<List> res = new HashSet<List>();

Arrays.sort(nums);

for(int i=1; i=0 && ki && j<nums.length){

int sumG = nums[k] + nums[j];

if(sumG + nums[i] == 0){

List list = new ArrayList();

list.add(nums[k]);

list.add(nums[i]);

list.add(nums[j]);

res.add(list);

k–; j++;

}

else if(sumG + nums[i] > 0){

k–;

}

else{

j++;

}

}

}

return new ArrayList(res);

}
AmourDeMai on 2015-10-28

The better solution may have a duplicate result. If input is [0, 0, 0, 0], output will be [[0,0,0], [0,0,0]]
To avoid it, before add to result, we should check if the temp ArrayList is already in the result.
Ankit on 2015-11-4

private static ArrayList<ArrayList> threeSum(int []S, int target) { ArrayList<ArrayList> list = new ArrayList<ArrayList>(); Arrays.sort(S); for(int i=0;i<S.length;i++) { for(int j=i+1;j<S.length;j++) { for(int k=j+1;k<S.length;k++) { if((S[i]+S[j]+S[k]) == target) { ArrayList result = new ArrayList(); result.add(S[i]); result.add(S[j]); result.add(S[k]); list.add(result); } } } } return list; }
Chetan on 2015-11-10

This solution still giving duplicate triplets. Following changes would make this solution complete I guess to avoid duplicates.

in case2, also add while loop for start++, same as in case 1

in case 3, also add while loop for end–, same as in case 1.

and finally at the end, inside for loop, add following while loop for avoiding duplicate negate values.

while(i < a.length-1 && a[i+1] == -negate) { i++; }
Gireesh on 2015-11-14

Here’s my accepted solution.
Reviews are welcome: https://github.com/mgireesh05/leetcode/blob/master/3sum/src/com/mgireesh/Solution.java

import java.util.Arrays; import java.util.LinkedList; import java.util.List; public class Solution { public List<List> threeSum(int[] nums) { List<List> answer = new LinkedList<List>(); if (nums.length < 3) { return answer; } Arrays.sort(nums); int find = 0; int prev_i = nums[0] - 1; for (int i = 0; i < nums.length - 2; i++) { if ((prev_i != nums[0] - 1) && (nums[i] == prev_i)) { continue; } prev_i = nums[i]; int prev_j = nums[0] - 1; for (int j = i + 1; j j; k--) { if ((prev_k != nums[0] - 1) && (find == prev_k)) { continue; } if (nums[k] == find) { List solSet = new LinkedList(); solSet.add(nums[i]); solSet.add(nums[j]); solSet.add(nums[k]); answer.add(solSet); prev_k = find; } else if (find > nums[k]) { break; } } } } return answer; } }
boyhou on 2015-11-21

The method below has any error? Or it just becauses that the complexity is n^3 ,so it cannot pass the leetcode.

public class Solution {

public List<List> threeSum(int[] nums) {

List<List> result = new ArrayList<List>();

List list = new ArrayList();

if(nums == null || nums.length == 0){

return result;

}

Arrays.sort(nums);

dfs(result, list, nums, 0, 0);

return result;

}

private void dfs(List<List> result, List list, int[] nums, int cur, int sum){

if(list.size() == 3 && sum == 0){

result.add(new ArrayList(list));

return;

}

if(list.size() > 3 || cur >= nums.length){

return;

}

for(int i=cur; icur && nums[i]==nums[i-1])

// continue;

list.add(nums[i]);

dfs(result, list, nums, i+1, sum+nums[i]);

list.remove(list.size()-1);

}

}

}
CP on 2016-4-2

public static boolean is3SumArray(int [] arr1, int sum1){
Arrays.sort(arr1);
for(int i=0; i<arr1.length-2; i++)
{
int left = i+1;
int right = arr1.length-1;
while(left<right)
{
int result = arr1[i] + arr1[left] + arr1[right];
if(result==sum1)
return true;
else if(resultsum1)
right–;
}
}
return false;
}
Gurpreet singh on 2016-7-10

Not the best or fasted but cleaner and easier to understand
public static List<List> threeSum(int[] nums) { List<List> result = new ArrayList<List>(); if(nums == null || nums.length == 0) { return result; }
Arrays.sort(nums); Map tripMap = new HashMap(); dumpArray(nums); for(int i = 0; i < nums.length - 2; i++){ int j = i + 1; int k = nums.length - 1; while ( j < k) { if(nums[i] + nums[j] + nums[k] == 0){ List trip = new ArrayList(Arrays.asList(new Integer[]{nums[i], nums[j] , nums[k]})); String tripKey = nums[i] + "" + nums[j] + "" + nums[k]; if(!tripMap.containsKey(tripKey)) { result.add(trip); tripMap.put(tripKey, true); } j++; k--; }else if(nums[i] + nums[j] + nums[k] < 0) { j++; } else { k--; } } } return result; }
Cherry Zhao on 2016-7-21

Thanks for the solution, it’s really helpful. This article http://bit.ly/29YmZI4 also has a in-depth analysis of 3sum question with variations.
Lasha K on 2016-10-19

if(i==0 || nums[i] > nums[i-1])

why do you need this condition? isn’t it always true for sorted array?

LeetCode – 3Sum

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