Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
HAPPY CASE
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Example 2:
Input: nums = [0,1,1]
Output: []
EDGE CASE
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
For Array problems, we want to consider the following approaches:
⚠️ Common Mistakes
[nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. You may notice that there a solution where j == k is possible. i, j and k are indexes and not the elements of the array. You cannot use same element more than once in the triplet, but you can definitely use same number on different indexes. Remember you've to return triplet of elements and not triplet of indexes unlike the two sum problem where you've to return pair of indexes.Plan the solution with appropriate visualizations and pseudocode.
General Idea: use the two pointers pattern to make sure the result contains unique triplets
First, sort the array.
Have two pointers l and r where l starts from i + 1 and r in reverse. ie len(nums)-1. Here i is the index of n.
As the array is sorted, if n + nums[l] + nums[r] > 0 then decreare r or else decrease l.
If equal to zero, append to ans.
n == nums[i-1] explanation:
if 2 numbers are same, skip the number.
eg [-1,0,1,2,-1,-4]
After sorting, [-4,-1,-1,0,1,2]
Here -1 is repeated. So for i = 2, n = -1, nums[i-1] == n. Thus move to next step (or else it will create duplicates)
Implement the code to solve the algorithm.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort() # will make spoting of duplicates easy
triplets = []
length = len(nums)
for i in range(length-2): # ignore last two
# check if element is a duplicate. the first cannot be a duplicate
if i > 0 and nums[i] == nums[i-1]:
# skip handling an element if it's similar to the one before it
# because it is sorted, we effectively skip duplicates
continue
# TWO SUM for a sorted array
# 1. find elements that will add up to 0
# 2. check inner elements
left = i + 1
right = length - 1
while left < right:
# will be used to check if the sum is equal to 0
total = nums[i] + nums[left] + nums[right]
# if total is less than 0 we try to increase it's value
if total < 0:
left += 1 # moving left to a lerger value
# if total is more than 0 we try to decrease it's value
elif total > 0:
right -= 1 # moving right to a smaller value
# 1. add list of elements to triplets
# 2. check inner elements
else:
# add elements to triplets
triplets.append([nums[i], nums[left], nums[right]])
# check inner elements
# 1. skip similar elements
# 2. move to inner elements
# skip:
# no need to continue with an element with the same value as l/r
# Skip all similar to the current left and right so that,
# when we are moving to the next element, we dont move to an element with the same value
while left < right and nums[left] == nums[left+1]:
left += 1
while left < right and nums[right] == nums[right-1]:
right -= 1
# move to inner elements
left += 1
right -= 1
return tripletsclass Solution {
public List<List<Integer>> threeSum(int[] arr) {
Arrays.sort(arr);
List<List<Integer>> ls = new ArrayList<>();
for (int i = 0; i<arr.length-2; i++) {
if (i == 0 || (i>0 && arr[i-1]!=arr[i])) { //check for duplicates to avoid copy we've used arr[i-1]!=arr[i] instead of arr[i+1]!=arr[i] because we must take the duplicate value in condition i.e. in example 1 [-1,-1,2] is also and output so if we do arr[i+1]!=arr[i] then we'll skip to the next -1 and this output will not come
int start = i+1;
int end = arr.length-1;
int sum = 0-arr[i];
while (start<end) {
if (arr[start] + arr[end] == sum) {
ls.add(Arrays.asList(arr[i], arr[start], arr[end]));
while (start<end && arr[start] == arr[start+1]) start++;//avoid all the same values
while (start<end && arr[end] == arr[end-1]) end--;//avoid all the same values
start++;
end--;
} else if (arr[start] + arr[end] < sum) {
start++;
} else end--;
}
}
}
return ls;
}
}Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Given n elements,
