Given an array nums of n integers, are there elements a, b, c in nums such that a b c = 0? Find all unique triplets in the array which gives the sum of zero.
Note: The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
public static List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(nums == null || nums.length < 3)
return result;
int n = nums.length;
for(int i = 0; i < n - 2; i ) {
for(int j = i 1; j < n - 1; j ) {
for(int k = j 1; k < n; k ) {
if((nums[i] nums[j] nums[k]) == 0 ) {
addResult(result, new Integer[] {nums[i], nums[j], nums[k]});
}
}
}
}
return result;
}
private static void addResult(List<List<Integer>> result, Integer[] triplet) {
Set<Integer> threeSum = new HashSet<Integer>(Arrays.asList(triplet));
boolean alredayExists = false;
for(List<Integer> list: result) {
Set<Integer> res = new HashSet<Integer>(list);
if(res.equals(threeSum)) {
alredayExists = true;
break;
}
}
if(!alredayExists) {
result.add(Arrays.asList(triplet));
}
}Above implementation have Runtime complexity of O(n^3) and space complexity of O(1)
Runtime Complexity = O(n^3)
Space Complexity = O(1)
public static List<Integer[]> threeSum(int[] nums) {
List<Integer[]> result = new ArrayList<Integer[]>();
if(nums == null || nums.length < 3)
return result;
int n = nums.length;
Arrays.sort(nums);
for(int i = 0; i < n - 2; i ) {
int left = i 1;
int right = n - 1;
while(left < right) {
if( (nums[i] nums[left] nums[right]) > 0) {
right --;
} else if( (nums[i] nums[left] nums[right]) < 0) {
left ;
} else {
addResult(result, new Integer[] {nums[i], nums[left], nums[right]});
left ;
right--;
}
}
}
return result;
}
private static void addResult(List<Integer[]> result, Integer[] triplet) {
Set<Integer> threeSum = new HashSet<Integer>(Arrays.asList(triplet));
boolean alredayExists = false;
for(Integer[] list: result) {
Set<Integer> res = new HashSet<Integer>(Arrays.asList(list));
if(res.equals(threeSum)) {
alredayExists = true;
break;
}
}
if(!alredayExists) {
result.add(triplet);
}
}Above implementation have Runtime complexity of O(n^2) and space complexity of O(1)
Runtime Complexity = O(n^2)
Space Complexity = O(1)
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> response = new ArrayList<>();
if(nums == null || nums.length < 3)
return response;
Arrays.sort(nums);
int n = nums.length;
for(int i = 0; i < n-2; i ) {
// Handling duplicates : we should not consider same nums[i] again
if(i > 0 && nums[i] == nums[i-1]) {
continue;
}
int left = i 1;
int right = n-1;
while(left < right) {
int sum = nums[i] nums[left] nums[right];
if(sum == 0) {
response.add(Arrays.asList(nums[i], nums[left], nums[right]));
/** Handling Duplicates :
below 2 while loops, make sure we don't consider the same low and high numbers again
**/
while(left < right && nums[left] == nums[left 1])
left ;
while(left < right && nums[right] == nums[right-1])
right--;
left ;
right--;
} else if(sum > 0) {
right--;
} else {
left ;
}
}
}
return response;
}
}Above implementation have Runtime complexity of O(n^2) and space complexity of O(1)
Runtime Complexity = O(n^2)
Space Complexity = O(1)
Note that, in above implementation, we are not using addResult() method to handle duplicate triplets, rather we are handling duplicate scenarios, when we are adding a triplet to the final result. This optimizes the runtime of the algorithm and makes it fast.