Python3 Program to Find all triplets with zero sum

06 Jan, 2022
Given an array of distinct elements. The task is to find triplets in the array whose sum is zero.
Examples : 
Input : arr[] = {0, -1, 2, -3, 1}
Output : (0 -1 1), (2 -3 1)

Explanation : The triplets with zero sum are
0 + -1 + 1 = 0 and 2 + -3 + 1 = 0  

Input : arr[] = {1, -2, 1, 0, 5}
Output : 1 -2  1
Explanation : The triplets with zero sum is
1 + -2 + 1 = 0   
Method 1: This is a simple method that takes O(n3) time to arrive at the result.
Approach: The naive approach runs three loops and check one by one that sum of three elements is zero or not. If the sum of three elements is zero then print elements otherwise print not found.
Algorithm: Run three nested loops with loop counter i, j, kThe first loops will run from 0 to n-3 and second loop from i+1 to n-2 and the third loop from j+1 to n-1. The loop counter represents the three elements of the triplet.Check if the sum of elements at i’th, j’th, k’th is equal to zero or not. If yes print the sum else continue.
Run three nested loops with loop counter i, j, kThe first loops will run from 0 to n-3 and second loop from i+1 to n-2 and the third loop from j+1 to n-1. The loop counter represents the three elements of the triplet.Check if the sum of elements at i’th, j’th, k’th is equal to zero or not. If yes print the sum else continue.
Run three nested loops with loop counter i, j, k
The first loops will run from 0 to n-3 and second loop from i+1 to n-2 and the third loop from j+1 to n-1. The loop counter represents the three elements of the triplet.
Check if the sum of elements at i’th, j’th, k’th is equal to zero or not. If yes print the sum else continue.
Below is the implementation of the above approach: 
Python3
# A simple Python 3 program # to find three elements whose # sum is equal to zero  # Prints all triplets in # arr[] with 0 sumdef findTriplets(arr, n):      found = False    for i in range(0, n-2):              for j in range(i+1, n-1):                      for k in range(j+1, n):                              if (arr[i] + arr[j] + arr[k] == 0):                    print(arr[i], arr[j], arr[k])                    found = True                        # If no triplet with 0 sum     # found in array    if (found == False):        print(" not exist ")  # Driver codearr = [0, -1, 2, -3, 1]n = len(arr)findTriplets(arr, n)  # This code is contributed by Smitha Dinesh Semwal    
0 -1 1
2 -3 1
Complexity Analysis: 
Time Complexity: O(n3). As three nested loops are required, so the time complexity is O(n3).
Auxiliary Space: O(1). Since no extra space is required, so the space complexity is constant.
 Method 2: The second method uses the process of Hashing to arrive at the result and is solved at a lesser time of O(n2). 
Approach: This involves traversing through the array. For every element arr[i], find a pair with sum “-arr[i]”. This problem reduces to pair sum and can be solved in O(n) time using hashing.
Algorithm: 
Create a hashmap to store a key-value pair.Run a nested loop with two loops, the outer loop from 0 to n-2 and the inner loop from i+1 to n-1Check if the sum of ith and jth element multiplied with -1 is present in the hashmap or notIf the element is present in the hashmap, print the triplet else insert the j’th element in the hashmap.
Create a hashmap to store a key-value pair.
Run a nested loop with two loops, the outer loop from 0 to n-2 and the inner loop from i+1 to n-1
Check if the sum of ith and jth element multiplied with -1 is present in the hashmap or not
If the element is present in the hashmap, print the triplet else insert the j’th element in the hashmap.
Below is the implementation of the above approach: 
Python3
# Python3 program to find triplets # in a given array whose sum is zero   # function to print triplets with 0 sum def findTriplets(arr, n):    found = False    for i in range(n - 1):          # Find all pairs with sum         # equals to "-arr[i]"         s = set()        for j in range(i + 1, n):            x = -(arr[i] + arr[j])            if x in s:                print(x, arr[i], arr[j])                found = True            else:                s.add(arr[j])    if found == False:        print("No Triplet Found")  # Driver Codearr = [0, -1, 2, -3, 1]n = len(arr)findTriplets(arr, n)  # This code is contributed by Shrikant13
-1 0 1
-3 2 1
Complexity Analysis: 
Time Complexity: O(n2). Since two nested loops are required, so the time complexity is O(n2).
Auxiliary Space: O(n). Since a hashmap is required, so the space complexity is linear.
 Method 3: This method uses Sorting to arrive at the correct result and is solved in O(n2) time.  
Approach: The above method requires extra space. The idea is based on method 2 of this post. For every element check that there is a pair whose sum is equal to the negative value of that element.
Algorithm: 
Sort the array in ascending order.Traverse the array from start to end.For every index i, create two variables l = i + 1 and r = n – 1Run a loop until l is less than r if the sum of array[i], array[l] and array[r] is equal to zero then print the triplet and break the loopIf the sum is less than zero then increment the value of l, by increasing the value of l the sum will increase as the array is sorted, so array[l+1] > array [l]If the sum is greater than zero then decrement the value of r, by increasing the value of l the sum will decrease as the array is sorted, so array[r-1] < array [r].
Sort the array in ascending order.
Traverse the array from start to end.
For every index i, create two variables l = i + 1 and r = n – 1
Run a loop until l is less than r if the sum of array[i], array[l] and array[r] is equal to zero then print the triplet and break the loop
If the sum is less than zero then increment the value of l, by increasing the value of l the sum will increase as the array is sorted, so array[l+1] > array [l]
If the sum is greater than zero then decrement the value of r, by increasing the value of l the sum will decrease as the array is sorted, so array[r-1] < array [r].
Below is the implementation of the above approach: 
Python3
# python program to find triplets in a given# array whose sum is zero  # function to print triplets with 0 sumdef findTriplets(arr, n):      found = False      # sort array elements    arr.sort()      for i in range(0, n-1):              # initialize left and right        l = i + 1        r = n - 1        x = arr[i]        while (l < r):                      if (x + arr[l] + arr[r] == 0):                # print elements if it's sum is zero                print(x, arr[l], arr[r])                l+=1                r-=1                found = True                            # If sum of three elements is less            # than zero then increment in left            elif (x + arr[l] + arr[r] < 0):                l+=1              # if sum is greater than zero than            # decrement in right side            else:                r-=1              if (found == False):        print(" No Triplet Found")    # Driven sourcearr = [0, -1, 2, -3, 1]n = len(arr)findTriplets(arr, n)  # This code is contributed by Smitha Dinesh Semwal
-3 1 2
-1 0 1
Complexity Analysis: 
Time Complexity : O(n2). Only two nested loops are required, so the time complexity is O(n2).
Auxiliary Space : O(1), no extra space is required, so the time complexity is constant.
Please refer complete article on Find all triplets with zero sum for more details!
Facebook
Google
two-pointer-algorithm
Arrays
Hash
Python
Searching
Sorting
Google
Facebook
two-pointer-algorithm
Arrays
Searching
Hash
Sorting
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Maximum and minimum of an array using minimum number of comparisons
Top 50 Array Coding Problems for Interviews
Multidimensional Arrays in Java
Stack Data Structure (Introduction and Program)
Linear Search
Given an array A[] and a number x, check for pair in A[] with sum as x (aka Two Sum)
What is Hashing | A Complete Tutorial
Internal Working of HashMap in Java
Hashing | Set 1 (Introduction)
Count pairs with given sum