All unique triplets that sum up to a given value

Last Updated : 25 May, 2021

Given an array and a sum value, find all possible unique triplets in that array whose sum is equal to the given sum value. If no such triplets can be formed from the array, then print “No triplets can be formed”, else print all the unique triplets. For example, if the given array is {12, 3, 6, 1, 6, 9} and the given sum is 24, then the unique triplets are (3, 9, 12) and (6, 6, 12) whose sum is 24.

Examples:

Input : array = {12, 3, 6, 1, 6, 9} sum = 24
Output : [[3, 9, 12], [6, 6, 12]]

Input : array = {-2, 0, 1, 1, 2} sum = 0
Output : [[-2, 0, 2], [-2, 1, 1]]

Input : array = {-2, 0, 1, 1, 2} sum = 10
Output : No triplets can be formed

Method 1
In a previous post, Find a triplet that sum to a given value we have discussed whether the triplets can be formed from the array or not.
Here we need to print all unique set of triplets that sum up to a given value

Sort the input array.
Find three indexes from the array i, j and k where A[i]+A[j]+A[k] = given sum value.
Fix the first element as A[i] and iterate i from 0 to array size – 2.
For each iteration of i, take j to be index of the first element in the remaining elements and k to be the index of the last element.
Check for the triplet combination A[i]+A[j]+A[k] = given sum value.
If triplet is obtained (ie., A[i]+A[j]+A[k] = given sum value)
1. Add all the triplet in a TreeSet with “:” separated value to get the unique triplets.
2. Increment the second value index
3. Decrement the third value index.
4. Repeat step 4 & 5 till j < k
Else if, A[i]+A[j]+A[k] < given sum value, increment the second value index
Else if, A[i]+A[j]+A[k] > given sum value, decrement the third value index

Below is the implementation of the above idea:

C++

// C++ program to find unique triplets
// that sum up to a given value.
#include <bits/stdc++.h>
using namespace std;
 
// Structure to define a triplet.
struct triplet 
{
    int first, second, third;
};
 
// Function to find unique triplets that
// sum up to a given value.
int findTriplets(int nums[], int n, int sum)
{
    int i, j, k;
 
    // Vector to store all unique triplets.
    vector<triplet> triplets;
 
    // Set to store already found triplets
    // to avoid duplication.
    unordered_set<string> uniqTriplets;
 
    // Variable used to hold triplet
    // converted to string form.
    string temp;
 
    // Variable used to store current
    // triplet which is stored in vector
    // if it is unique.
    triplet newTriplet;
 
    // Sort the input array.
    sort(nums, nums + n);
 
    // Iterate over the array from the
    // start and consider it as the
    // first element.
    for (i = 0; i < n - 2; i++) 
    {
        // index of the first element in
        // the remaining elements.
        j = i + 1;
 
        // index of the last element.
        k = n - 1;
 
        while (j < k) {
 
            // If sum of triplet is equal to
            // given value, then check if
            // this triplet is unique or not.
            // To check uniqueness, convert
            // triplet to string form and
            // then check if this string is
            // present in set or not. If
            // triplet is unique, then store
            // it in vector.
            if (nums[i] + nums[j] + nums[k] == sum) 
            {
                temp = to_string(nums[i]) + " : "
                       + to_string(nums[j]) + " : "
                       + to_string(nums[k]);
                if (uniqTriplets.find(temp)
                    == uniqTriplets.end()) {
                    uniqTriplets.insert(temp);
                    newTriplet.first = nums[i];
                    newTriplet.second = nums[j];
                    newTriplet.third = nums[k];
                    triplets.push_back(newTriplet);
                }
 
                // Increment the first index
                // and decrement the last
                // index of remaining elements.
                j++;
                k--;
            }
 
            // If sum is greater than given
            // value then to reduce sum
            // decrement the last index.
            else if (nums[i] + nums[j] + nums[k] > sum)
                k--;
 
            // If sum is less than given value
            // then to increase sum increment
            // the first index of remaining
            // elements.
            else
                j++;
        }
    }
 
    // If no unique triplet is found, then
    // return 0.
    if (triplets.size() == 0)
        return 0;
 
    // Print all unique triplets stored in
    // vector.
    for (i = 0; i < triplets.size(); i++) {
        cout << "[" << triplets[i].first << ", "
             << triplets[i].second << ", "
             << triplets[i].third << "], ";
    }
}
 
// Driver Code
int main()
{
    int nums[] = { 12, 3, 6, 1, 6, 9 };
    int n = sizeof(nums) / sizeof(nums[0]);
    int sum = 24;
   
    // Function call
    if (!findTriplets(nums, n, sum))
        cout << "No triplets can be formed.";
 
    return 0;
}
 
// This code is contributed by NIKHIL JINDAL.

Java

// Java program to find all triplets with given sum
import java.util.*;
 
public class triplets {
 
    // returns all triplets whose sum is
    //  equal to sum value
    public static List<List<Integer> >
    findTriplets(int[] nums, int sum)
    {
 
        /* Sort the elements */
        Arrays.sort(nums);
 
        List<List<Integer> > pair 
          = new ArrayList<>();
        TreeSet<String> set 
          = new TreeSet<String>();
        List<Integer> triplets 
          = new ArrayList<>();
 
        /* Iterate over the array from the start 
        and consider it as the first element*/
        for (int i = 0;
             i < nums.length - 2; 
             i++) {
 
            // index of the first element in the
            // remaining elements
            int j = i + 1;
 
            // index of the last element
            int k = nums.length - 1;
 
            while (j < k) {
 
                if (nums[i] + nums[j] 
                    + nums[k] == sum) {
 
                    String str 
                      = nums[i] + ":" + nums[j]
                                 + ":" + nums[k];
 
                    if (!set.contains(str))
                    {
 
                        // To check for the unique triplet
                        triplets.add(nums[i]);
                        triplets.add(nums[j]);
                        triplets.add(nums[k]);
                        pair.add(triplets);
                        triplets = new ArrayList<>();
                        set.add(str);
                    }
 
                     // increment the second value index
                    j++;
                    
                     // decrement the third value index
                    k--;
                }
                else if (nums[i] 
                         + nums[j] 
                         + nums[k] < sum)
                    j++;
 
                else
                    k--;
            }
        }
        return pair;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] nums = { 12, 3, 6, 1, 6, 9 };
        int sum = 24;
 
        List<List<Integer> > triplets
            = findTriplets(nums, sum);
 
        // Function call
        if (!triplets.isEmpty()) {
            System.out.println(triplets);
        }
        else {
            System.out.println("No triplets can be formed");
        }
    }
}

Python3

# Python program to find unique triplets
# that sum up to a given value.
 
# Function to find unique triplets that
# sum up to a given value.
def findTriplets(nums, n, Sum):
    i = 0
    j = 0
    k = 0
 
    # list to store all unique triplets.
    triplet = []
 
    # list to store already found triplets
    # to avoid duplication.
    uniqTriplets = []
 
    # Variable used to hold triplet
    # converted to string form.
    temp = ""
 
    # Variable used to store current
    # triplet which is stored in vector
    # if it is unique.
    newTriplet = [0, 0, 0]
 
    # Sort the input array.
    nums.sort()
 
    # Iterate over the array from the
    # start and consider it as the
    # first element.
    for i in range(n - 2):
         
        # index of the first element in
        # the remaining elements.
        j = i + 1
 
        # index of the last element.
        k = n - 1
 
        while(j < k):
           
            # If sum of triplet is equal to
            # given value, then check if
            # this triplet is unique or not.
            # To check uniqueness, convert
            # triplet to string form and
            # then check if this string is
            # present in set or not. If
            # triplet is unique, then store
            # it in list.
            if(nums[i] + nums[j] + nums[k] == Sum):
                temp = str(nums[i]) + ":" + str(nums[j]) + ":" + str(nums[k])
                if temp not in uniqTriplets:
                    uniqTriplets.append(temp)
                    newTriplet[0] = nums[i]
                    newTriplet[1] = nums[j]
                    newTriplet[2] = nums[k]
                    triplet.append(newTriplet)
                    newTriplet = [0, 0, 0]
 
                # Increment the first index
                # and decrement the last
                # index of remaining elements.
                j += 1
                k -= 1
                 
            # If sum is greater than given
            # value then to reduce sum
            # decrement the last index.
            elif(nums[i] + nums[j] + nums[k] > Sum):
                k -= 1
                 
            # If sum is less than given value
            # then to increase sum increment
            # the first index of remaining
            # elements.
            else:
                j += 1
 
    # If no unique triplet is found, then
       # return 0.
    if(len(triplet) == 0):
        return 0
     
    # Print all unique triplets stored in
    # list.
    for i in range(len(triplet)):
        print(triplet[i], end = ", ")
    return 1
 
# Driver Code
nums = [12, 3, 6, 1, 6, 9]
n = len(nums)
Sum = 24
 
# Function call
if(not findTriplets(nums, n, Sum)):
    print("No triplets can be formed.")
 
# This code is contributed by rag2127

C#

// C# program to find all triplets with given sum
using System;
using System.Collections.Generic;
 
class GFG {
 
    // returns all triplets whose sum is 
    // equal to sum value
    public static List<List<int>> 
      findTriplets(int[] nums,int sum)
    {
 
        /* Sort the elements */
        Array.Sort(nums);
 
        List<List<int> > pair 
          = new List<List<int> >();
        SortedSet<String> set
          = new SortedSet<String>();
        List<int> triplets 
          = new List<int>();
 
        /* Iterate over the array from the start and
        consider it as the first element*/
        for (int i = 0; 
             i < nums.Length - 2; 
             i++) {
 
            // index of the first element in the
            // remaining elements
            int j = i + 1;
 
            // index of the last element
            int k = nums.Length - 1;
 
            while (j < k) {
                if (nums[i] 
                    + nums[j] 
                    + nums[k] == sum) {
                    String str
                      = nums[i] + ":" + nums[j]
                                 + ":" + nums[k];
 
                    if (!set.Contains(str)) 
                    {
                       
                        // To check for the unique triplet
                        triplets.Add(nums[i]);
                        triplets.Add(nums[j]);
                        triplets.Add(nums[k]);
                        pair.Add(triplets);
                        triplets = new List<int>();
                        set.Add(str);
                    }
 
                    j++; // increment the second value index
                    k--; // decrement the third value index
                }
                else if (nums[i] 
                         + nums[j] 
                         + nums[k] < sum)
                    j++;
 
                else
                    k--;
            }
        }
        return pair;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] nums = { 12, 3, 6, 1, 6, 9 };
        int sum = 24;
 
        List<List<int> > triplets 
          = findTriplets(nums, sum);
       
        // Function call
        if (triplets.Count != 0) {
            Console.Write("[");
            for (int i = 0; i < triplets.Count; i++) 
            {
                List<int> l = triplets[i];
                Console.Write("[");
                for (int j = 0; j < l.Count; j++)
                {
                    Console.Write(l[j]);
                    if (l.Count != j + 1)
                        Console.Write(", ");
                }
                Console.Write("]");
                if (triplets.Count != i + 1)
                    Console.Write(",");
            }
            Console.Write("]");
        }
        else
        {
            Console.WriteLine("No triplets can be formed");
        }
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript program to find all triplets with given sum
     
    // returns all triplets whose sum is
    //  equal to sum value
    function findTriplets(nums,sum)
    {
        /* Sort the elements */
        nums.sort(function(a,b){return a-b;});
         
        let pair = [];
         
        let set = new Set();
         
        let  triplets= [];
         
        //let newTriplet = [0, 0, 0];
         
        for (let i = 0;
             i < nums.length - 2;
             i++)
        {
            // index of the first element in the
            // remaining elements
            let j = i + 1;
  
            // index of the last element
            let k = nums.length - 1;
  
            while (j < k) {
  
                if (nums[i] + nums[j]
                    + nums[k] == sum) {
  
                     
                     let str= nums[i] + ":" + nums[j]
                                 + ":" + nums[k];
  
                    if (!set.has(str))
                    {
  
                        // To check for the unique triplet
                        triplets.push(nums[i]);
                        triplets.push(nums[j]);
                        triplets.push(nums[k]);
                        pair.push(triplets);
                        triplets =[];
                        set.add(str);
                    }
  
                     // increment the second value index
                    j++;
                     
                     // decrement the third value index
                    k--;
                }
                else if (nums[i]
                         + nums[j]
                         + nums[k] < sum)
                    j++;
  
                else
                    k--;
            }
        }
        return pair;
         
         
    }
     
    // Driver code
    let nums=[12, 3, 6, 1, 6, 9];
    let sum = 24;
    let triplets
            = findTriplets(nums, sum);
    // Function call
    if (triplets.length!=0) {
        document.write("[");
            for (let i = 0; i < triplets.length; i++)
            {
                let l = triplets[i];
                document.write("[");
                for (let j = 0; j < l.length; j++)
                {
                    document.write(l[j]);
                    if (l.length != j + 1)
                        document.write(",  ");
                }
                document.write("]");
                if (triplets.length != i + 1)
                    document.write(", ");
            }
            document.write("]");
    }
    else {
        document.write("No triplets can be formed");
    }
     
    //This code is contributed by avanitrachhadiya2155
     
</script>

Output

[3, 9, 12], [6, 6, 12],

Time Complexity: O(n²)
Space Complexity: O(n)

Method 2

In this method, we will see another way to solve this problem which will use only a constant amount of space. Here in this method, we would check if the current element is the same as the previous one then we will not consider it and just simply skip that element. By using this we will be able to find only unique triplets only.

Algorithm

Sort the input array.
For the first element iterate i from 0 to n-2.
For each iteration we will have two indexes starts and end where the start would be equal to i+1 and end would be equal to n-1.
Now we will look for a target whose value is sum-a[i] in the range from start to end using two pointer techniques.
In this we will look for the previous elements if they are the same then we would not work for them to find all unique duplicates.
If we found target==a[start]+a[end] then we would print it and increment start and decrement end.
If target>a[start]+a[end] then we would increment start.
Else we would decrement end

Below is the implementation of the above idea:

C++

// C++ program to find all unique triplets
// without using any extra space.
#include <bits/stdc++.h>
using namespace std;
 
// Function to all find unique triplets 
// without using extra space
void findTriplets(int a[], int n, int sum)
{
    int i;
 
    // Sort the input array
    sort(a, a + n);
 
    // For handling the cases when no such 
    // triplets exits.
    bool flag = false;
 
    // Iterate over the array from start to n-2.
    for (i = 0; i < n - 2; i++)
    {
        if (i == 0 || a[i] > a[i - 1])
        {
            // Index of the first element in 
            // remaining range.
            int start = i + 1;
 
            // Index of the last element
            int end = n - 1;
 
            // Setting our new target
            int target = sum - a[i];
 
            while (start < end) 
            {
                // Checking if current element 
                // is same as previous
                if (start > i + 1
                    && a[start] == a[start - 1]) 
                {
                    start++;
                    continue;
                }
 
                // Checking if current element is 
                // same as previous
                if (end < n - 1 
                    && a[end] == a[end + 1]) 
                {
                    end--;
                    continue;
                }
 
                // If we found the triplets then print it
                // and set the flag
                if (target == a[start] + a[end]) 
                {
                    cout << "[" << a[i] 
                         << "," << a[start]
                         << "," << a[end] << "] ";
                    flag = true;
                    start++;
                    end--;
                }
                // If target is greater then
                //  increment the start index
                else if (target > (a[start] + a[end]))
                {
                    start++;
                }
                // If target is smaller than 
                // decrement the end index
                else {
                    end--;
                }
            }
        }
    }
 
    // If no such triplets found
    if (flag == false) {
        cout << "No Such Triplets Exist"
             << "\n";
    }
}
 
// Driver code
int main()
{
    int a[] = { 12, 3, 6, 1, 6, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    int sum = 24;
 
    // Function call
    findTriplets(a, n, sum);
    return 0;
}
 
// This code is contributed by Rohit

Java

// Java program to find all unique triplets
// without using any extra space.
import java.util.*;
public class Main
{
    // Function to all find unique triplets 
    // without using extra space
    public static void findTriplets(int a[], int n, int sum)
    {
        int i;
       
        // Sort the input array
        Arrays.sort(a);
       
        // For handling the cases when no such 
        // triplets exits.
        boolean flag = false;
       
        // Iterate over the array from start to n-2.
        for (i = 0; i < n - 2; i++)
        {
            if (i == 0 || a[i] > a[i - 1])
            {
                // Index of the first element in 
                // remaining range.
                int start = i + 1;
       
                // Index of the last element
                int end = n - 1;
       
                // Setting our new target
                int target = sum - a[i];
       
                while (start < end) 
                {
                    // Checking if current element 
                    // is same as previous
                    if (start > i + 1
                        && a[start] == a[start - 1]) 
                    {
                        start++;
                        continue;
                    }
       
                    // Checking if current element is 
                    // same as previous
                    if (end < n - 1
                        && a[end] == a[end + 1]) 
                    {
                        end--;
                        continue;
                    }
       
                    // If we found the triplets then print it
                    // and set the flag
                    if (target == a[start] + a[end]) 
                    {
                        System.out.print("[" + a[i] 
                             + "," + a[start]
                             + "," + a[end] + "] ");
                        flag = true;
                        start++;
                        end--;
                    }
                    
                    // If target is greater then
                    //  increment the start index
                    else if (target > (a[start] + a[end]))
                    {
                        start++;
                    }
                    
                    // If target is smaller than 
                    // decrement the end index
                    else {
                        end--;
                    }
                }
            }
        }
       
        // If no such triplets found
        if (flag == false) {
            System.out.print("No Such Triplets Exist");
        }
    }
     
    public static void main(String[] args) {
        int a[] = { 12, 3, 6, 1, 6, 9 };
        int n = a.length;
        int sum = 24;
       
        // Function call
        findTriplets(a, n, sum);
    }
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find all 
# unique triplets without using 
# any extra space.
 
# Function to all find unique 
# triplets without using extra 
# space
def findTriplets(a, n, sum):
 
    # Sort the input array
    a.sort()
 
    # For handling the cases 
    # when no such triplets exits.
    flag = False
 
    # Iterate over the array from 
    # start to n-2.
    for i in range(n - 2):
        if (i == 0 or
            a[i] > a[i - 1]):
 
            # Index of the first 
            # element in remaining 
            # range.
            start = i + 1
 
            # Index of the last 
            # element
            end = n - 1
 
            # Setting our new target
            target = sum - a[i]
 
            while (start < end):
 
                # Checking if current element
                # is same as previous
                if (start > i + 1 and
                    a[start] == a[start - 1]):
 
                    start += 1
                    continue
 
                # Checking if current
                # element is same as 
                # previous
                if (end < n - 1 and
                    a[end] == a[end + 1]):
                    end -= 1
                    continue
 
                # If we found the triplets 
                # then print it and set the 
                # flag
                if (target == a[start] + a[end]):
                    print("[", a[i], ",", 
                          a[start], ",", 
                          a[end], "]",
                          end = " ")
                    flag = True
                    start += 1
                    end -= 1
 
                # If target is greater then
                #  increment the start index
                elif (target >
                     (a[start] + a[end])):
                    start += 1
 
                # If target is smaller than
                # decrement the end index
                else:
                    end -= 1
 
    # If no such triplets found
    if (flag == False):
        print("No Such Triplets Exist")
 
# Driver code
if __name__ == "__main__":
 
    a = [12, 3, 6, 1, 6, 9]
    n = len(a)
    sum = 24
 
    # Function call
    findTriplets(a, n, sum)
 
# This code is contributed by Chitranayal

C#

// C# program to find all unique triplets
// without using any extra space.
using System;
using System.Collections.Generic; 
class GFG {
     
    // Function to all find unique triplets 
    // without using extra space
    static void findTriplets(int[] a, int n, int sum)
    {
        int i;
      
        // Sort the input array
        Array.Sort(a);
      
        // For handling the cases when no such 
        // triplets exits.
        bool flag = false;
      
        // Iterate over the array from start to n-2.
        for (i = 0; i < n - 2; i++)
        {
            if (i == 0 || a[i] > a[i - 1])
            {
                // Index of the first element in 
                // remaining range.
                int start = i + 1;
      
                // Index of the last element
                int end = n - 1;
      
                // Setting our new target
                int target = sum - a[i];
      
                while (start < end) 
                {
                    // Checking if current element 
                    // is same as previous
                    if (start > i + 1
                        && a[start] == a[start - 1]) 
                    {
                        start++;
                        continue;
                    }
      
                    // Checking if current element is 
                    // same as previous
                    if (end < n - 1 
                        && a[end] == a[end + 1]) 
                    {
                        end--;
                        continue;
                    }
      
                    // If we found the triplets then print it
                    // and set the flag
                    if (target == a[start] + a[end]) 
                    {
                        Console.Write("[" + a[i] 
                             + "," + a[start]
                             + "," + a[end] + "] ");
                        flag = true;
                        start++;
                        end--;
                    }
                   
                    // If target is greater then
                    //  increment the start index
                    else if (target > (a[start] + a[end]))
                    {
                        start++;
                    }
                   
                    // If target is smaller than 
                    // decrement the end index
                    else {
                        end--;
                    }
                }
            }
        }
      
        // If no such triplets found
        if (flag == false) {
            Console.WriteLine("No Such Triplets Exist");
        }
    }
 
  static void Main() {
    int[] a = { 12, 3, 6, 1, 6, 9 };
    int n = a.Length;
    int sum = 24;
  
    // Function call
    findTriplets(a, n, sum);
  }
}
 
// This code is contributed by divyesh072019

Javascript

<script>
    // Javascript program to find all unique triplets
    // without using any extra space.
     
    // Function to all find unique triplets
    // without using extra space
    function findTriplets(a, n, sum)
    {
        let i;
 
        // Sort the input array
        a.sort(function(a, b){return a - b});
 
        // For handling the cases when no such
        // triplets exits.
        let flag = false;
 
        // Iterate over the array from start to n-2.
        for (i = 0; i < n - 2; i++)
        {
            if (i == 0 || a[i] > a[i - 1])
            {
                // Index of the first element in
                // remaining range.
                let start = i + 1;
 
                // Index of the last element
                let end = n - 1;
 
                // Setting our new target
                let target = sum - a[i];
 
                while (start < end)
                {
                    // Checking if current element
                    // is same as previous
                    if (start > i + 1
                        && a[start] == a[start - 1])
                    {
                        start++;
                        continue;
                    }
 
                    // Checking if current element is
                    // same as previous
                    if (end < n - 1
                        && a[end] == a[end + 1])
                    {
                        end--;
                        continue;
                    }
 
                    // If we found the triplets then print it
                    // and set the flag
                    if (target == a[start] + a[end])
                    {
                        document.write("[" + a[i]
                             + "," + a[start]
                             + "," + a[end] + "] ");
                        flag = true;
                        start++;
                        end--;
                    }
                    // If target is greater then
                    //  increment the start index
                    else if (target > (a[start] + a[end]))
                    {
                        start++;
                    }
                    // If target is smaller than
                    // decrement the end index
                    else {
                        end--;
                    }
                }
            }
        }
 
        // If no such triplets found
        if (flag == false) {
            document.write("No Such Triplets Exist");
        }
    }
     
    let a = [ 12, 3, 6, 1, 6, 9 ];
    let n = a.length;
    let sum = 24;
  
    // Function call
    a.sort();
    findTriplets(a, n, sum);
     
    // This code is contributed by suresh07.
</script>

Output

[3,9,12] [6,6,12]

Complexity Analysis:

Time Complexity: O(n²).
Since two nested loops is required, so the time complexity is O(n²).
Auxiliary Space: O(1).
Since we need no extra space for solving this.

Anonymous

Improve

Largest subarray with equal number of 0s and 1s

Range Queries for Frequencies of array elements

Easy problems on Hashing

Intermediate problems on Hashing

Hard problems on Hashing

All unique triplets that sum up to a given value

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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