# Find three element from different three arrays such that a + b + c = sum

• Difficulty Level : Medium
• Last Updated : 01 Feb, 2023

Given three integer arrays and a “sum”, the task is to check if there are three elements a, b, c such that a + b + c = sum and a, b and c belong to three different arrays.

Examples :

```Input : a1[] = { 1 , 2 , 3 , 4 , 5 };
a2[] = { 2 , 3 , 6 , 1 , 2 };
a3[] = { 3 , 2 , 4 , 5 , 6 };
sum = 9
Output : Yes
1  + 2  + 6 = 9  here 1 from a1[] and 2 from
a2[] and 6 from a3[]

Input : a1[] = { 1 , 2 , 3 , 4 , 5 };
a2[] = { 2 , 3 , 6 , 1 , 2 };
a3[] = { 3 , 2 , 4 , 5 , 6 };
sum = 20
Output : No ```

A naive approach is to run three loops and check sum of three element form different arrays equal to given number if find then print exist and otherwise print not exist.

Implementation:

## C++

 `// C++ program to find three element``// from different three arrays such``// that a + b + c is equal to``// given sum``#include``using` `namespace` `std;` `// Function to check if there is``// an element from each array such``// that sum of the three elements``// is equal to given sum.``bool` `findTriplet(``int` `a1[], ``int` `a2[],``                 ``int` `a3[], ``int` `n1,``                 ``int` `n2, ``int` `n3, ``int` `sum)``{``    ``for` `(``int` `i = 0; i < n1; i++)``    ``for` `(``int` `j = 0; j < n2; j++)``        ``for` `(``int` `k = 0; k < n3; k++)``            ``if` `(a1[i] + a2[j] + a3[k] == sum)``            ``return` `true``;` `    ``return` `false``;``}` `// Driver Code``int` `main()``{``    ``int` `a1[] = { 1 , 2 , 3 , 4 , 5 };``    ``int` `a2[] = { 2 , 3 , 6 , 1 , 2 };``    ``int` `a3[] = { 3 , 2 , 4 , 5 , 6 };``    ``int` `sum = 9;``    ``int` `n1 = ``sizeof``(a1) / ``sizeof``(a1[0]);``    ``int` `n2 = ``sizeof``(a2) / ``sizeof``(a2[0]);``    ``int` `n3 = ``sizeof``(a3) / ``sizeof``(a3[0]);``    ``findTriplet(a1, a2, a3, n1, n2, n3, sum)?``                ``cout << ``"Yes"` `: cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program to find three element``// from different three arrays such``// that a + b + c is equal to``// given sum``class` `GFG``{``        ` `    ``// Function to check if there is``    ``// an element from each array such``    ``// that sum of the three elements``    ``// is equal to given sum.``    ``static` `boolean` `findTriplet(``int` `a1[], ``int` `a2[],``                               ``int` `a3[], ``int` `n1,``                               ``int` `n2, ``int` `n3, ``int` `sum)``    ``{``        ``for` `(``int` `i = ``0``; i < n1; i++)``            ``for` `(``int` `j = ``0``; j < n2; j++)``                ``for` `(``int` `k = ``0``; k < n3; k++)``                    ``if` `(a1[i] + a2[j] + a3[k] == sum)``                    ``return` `true``;``    ` `        ``return` `false``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `a1[] = { ``1` `, ``2` `, ``3` `, ``4` `, ``5` `};``        ``int` `a2[] = { ``2` `, ``3` `, ``6` `, ``1` `, ``2` `};``        ``int` `a3[] = { ``3` `, ``2` `, ``4` `, ``5` `, ``6` `};``        ``int` `sum = ``9``;``        ` `        ``int` `n1 = a1.length;``        ``int` `n2 = a2.length;``        ``int` `n3 = a3.length;``        ` `        ``if``(findTriplet(a1, a2, a3, n1, n2, n3, sum))``            ``System.out.print(``"Yes"``);``        ``else``            ``System.out.print(``"No"``);``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to find``# three element from different``# three arrays such that``# a + b + c is equal to``# given sum` `# Function to check if there``# is an element from each``# array such that sum of the``# three elements is equal to``# given sum.``def` `findTriplet(a1, a2, a3,``                ``n1, n2, n3, ``sum``):` `    ``for` `i ``in` `range``(``0` `, n1):``        ``for` `j ``in` `range``(``0` `, n2):``            ``for` `k ``in` `range``(``0` `, n3):``                ``if` `(a1[i] ``+` `a2[j] ``+``                    ``a3[k] ``=``=` `sum``):``                    ``return` `True` `    ``return` `False` `# Driver Code``a1 ``=` `[ ``1` `, ``2` `, ``3` `, ``4` `, ``5` `]``a2 ``=` `[ ``2` `, ``3` `, ``6` `, ``1` `, ``2` `]``a3 ``=` `[ ``3` `, ``2` `, ``4` `, ``5` `, ``6` `]``sum` `=` `9``n1 ``=` `len``(a1)``n2 ``=` `len``(a2)``n3 ``=` `len``(a3)``print``(``"Yes"``) ``if` `findTriplet(a1, a2, a3,``                            ``n1, n2, n3,``                            ``sum``) ``else` `print``(``"No"``)` `# This code is contributed``# by Smitha`

## C#

 `// C# program to find three element``// from different three arrays such``// that a + b + c is equal to``// given sum``using` `System;` `public` `class` `GFG``{` `// Function to check if there is an``// element from each array such that``// sum of the three elements is``// equal to given sum.``static` `bool` `findTriplet(``int` `[]a1, ``int` `[]a2,``                        ``int` `[]a3, ``int` `n1,``                        ``int` `n2, ``int` `n3,``                        ``int` `sum)``{``    ` `    ``for` `(``int` `i = 0; i < n1; i++)``    ` `        ``for` `(``int` `j = 0; j < n2; j++)``        ` `            ``for` `(``int` `k = 0; k < n3; k++)``            ``if` `(a1[i] + a2[j] + a3[k] == sum)``            ``return` `true``;` `    ``return` `false``;``}` `    ``// Driver Code``    ``static` `public` `void` `Main ()``    ``{``        ``int` `[]a1 = {1 , 2 , 3 , 4 , 5};``        ``int` `[]a2 = {2 , 3 , 6 , 1 , 2};``        ``int` `[]a3 = {3 , 2 , 4 , 5 , 6};``        ``int` `sum = 9;``        ``int` `n1 = a1.Length;``        ``int` `n2 = a2.Length;``        ``int` `n3 = a3.Length;``        ``if``(findTriplet(a1, a2, a3, n1,``                       ``n2, n3, sum))``        ``Console.WriteLine(``"Yes"``);``        ``else``        ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`Yes`

Time Complexity : O(n3
Auxiliary Space: O(1)

An efficient solution is to store all elements of first array in hash table (unordered_set in C++) and calculate sum of two elements last two array elements one by one and subtract from given number k and check in hash table if it exists in the hash table then print exist and otherwise not exist.

```1. Store all elements of first array in hash table
2. Generate all pairs of elements from two arrays using
nested loop. For every pair (a1[i], a2[j]), check if
sum - (a1[i] + a2[j]) exists in hash table. If yes
return true.      ```

Below is the implementation of above idea.

## C++

 `// C++ program to find three element``// from different three arrays such``// that a + b + c is equal to``// given sum``#include``using` `namespace` `std;` `// Function to check if there is``// an element from each array such``// that sum of the three elements is``// equal to given sum.``bool` `findTriplet(``int` `a1[], ``int` `a2[],``                 ``int` `a3[], ``int` `n1,``                 ``int` `n2, ``int` `n3,``                 ``int` `sum)``{``    ``// Store elements of``    ``// first array in hash``    ``unordered_set <``int``> s;``    ``for` `(``int` `i = 0; i < n1; i++)``        ``s.insert(a1[i]);` `    ``// sum last two arrays``    ``// element one by one``    ``for` `(``int` `i = 0; i < n2; i++)``    ``{``        ``for` `(``int` `j = 0; j < n3; j++)``        ``{``            ``// Consider current pair and``            ``// find if there is an element``            ``// in a1[] such that these three``            ``// form a required triplet``            ``if` `(s.find(sum - a2[i] - a3[j]) !=``                                       ``s.end())``                ``return` `true``;``        ``}``    ``}` `    ``return` `false``;``}` `// Driver Code``int` `main()``{``    ``int` `a1[] = { 1 , 2 , 3 , 4 , 5 };``    ``int` `a2[] = { 2 , 3 , 6 , 1 , 2 };``    ``int` `a3[] = { 3 , 2 , 4 , 5 , 6 };``    ``int` `sum = 9;``    ``int` `n1 = ``sizeof``(a1) / ``sizeof``(a1[0]);``    ``int` `n2 = ``sizeof``(a2) / ``sizeof``(a2[0]);``    ``int` `n3 = ``sizeof``(a3) / ``sizeof``(a3[0]);``    ``findTriplet(a1, a2, a3, n1, n2, n3, sum)?``    ``cout << ``"Yes"` `: cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java program to find three element``// from different three arrays such``// that a + b + c is equal to``// given sum``import` `java.util.*;` `class` `GFG``{` `    ``// Function to check if there is``    ``// an element from each array such``    ``// that sum of the three elements is``    ``// equal to given sum.``    ``static` `boolean` `findTriplet(``int` `a1[], ``int` `a2[], ``int` `a3[],``                                ``int` `n1, ``int` `n2, ``int` `n3,``                                ``int` `sum)``    ``{``        ``// Store elements of``        ``// first array in hash``        ``HashSet s = ``new` `HashSet();``        ``for` `(``int` `i = ``0``; i < n1; i++)``        ``{``            ``s.add(a1[i]);``        ``}` `        ``// sum last two arrays``        ``// element one by one``        ``ArrayList al = ``new` `ArrayList<>(s);``        ``for` `(``int` `i = ``0``; i < n2; i++)``        ``{``            ``for` `(``int` `j = ``0``; j < n3; j++)``            ``{``                ` `                ``// Consider current pair and``                ``// find if there is an element``                ``// in a1[] such that these three``                ``// form a required triplet``                ``if` `(al.contains(sum - a2[i] - a3[j]) &``                            ``al.indexOf(sum - a2[i] - a3[j])``                            ``!= al.get(al.size() - ``1``))``                ``{``                    ``return` `true``;``                ``}``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a1[] = {``1``, ``2``, ``3``, ``4``, ``5``};``        ``int` `a2[] = {``2``, ``3``, ``6``, ``1``, ``2``};``        ``int` `a3[] = {``3``, ``2``, ``4``, ``5``, ``6``};``        ``int` `sum = ``9``;``        ``int` `n1 = a1.length;``        ``int` `n2 = a2.length;``        ``int` `n3 = a3.length;``        ``if` `(findTriplet(a1, a2, a3, n1, n2, n3, sum))``        ``{``            ``System.out.println(``"Yes"``);``        ``}``        ``else``        ``{``            ``System.out.println(``"No"``);``        ``}``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to find three element``# from different three arrays such``# that a + b + c is equal to``# given sum` `# Function to check if there is``# an element from each array such``# that sum of the three elements is``# equal to given sum.``def` `findTriplet(a1, a2, a3,``                ``n1, n2, n3, ``sum``):` `    ``# Store elements of first``    ``# array in hash``    ``s ``=` `set``()` `    ``# sum last two arrays element``    ``# one by one``    ``for` `i ``in` `range``(n1):``        ``s.add(a1[i])` `    ``for` `i ``in` `range``(n2):``        ``for` `j ``in` `range``(n3):` `            ``# Consider current pair and``            ``# find if there is an element``            ``# in a1[] such that these three``            ``# form a required triplet``            ``if` `sum` `-` `a2[i] ``-` `a3[j] ``in` `s:``                ``return` `True``    ``return` `False` `# Driver code``a1 ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]``a2 ``=` `[``2``, ``3``, ``6``, ``1``, ``2``]``a3 ``=` `[``3``, ``24``, ``5``, ``6``]``n1 ``=` `len``(a1)``n2 ``=` `len``(a2)``n3 ``=` `len``(a3)``sum` `=` `9``if` `findTriplet(a1, a2, a3,``               ``n1, n2, n3, ``sum``) ``=``=` `True``:``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by Shrikant13`

## C#

 `// C# program to find three element``// from different three arrays such``// that a + b + c is equal to``// given sum``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``// Function to check if there is``    ``// an element from each array such``    ``// that sum of the three elements is``    ``// equal to given sum.``    ``static` `bool` `findTriplet(``int` `[]a1, ``int` `[]a2, ``int` `[]a3,``                                ``int` `n1, ``int` `n2, ``int` `n3,``                                ``int` `sum)``    ``{``        ``// Store elements of``        ``// first array in hash``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();``        ``for` `(``int` `i = 0; i < n1; i++)``        ``{``            ``s.Add(a1[i]);``        ``}` `        ``// sum last two arrays``        ``// element one by one``        ``List<``int``> al = ``new` `List<``int``>(s);``        ``for` `(``int` `i = 0; i < n2; i++)``        ``{``            ``for` `(``int` `j = 0; j < n3; j++)``            ``{``                ` `                ``// Consider current pair and``                ``// find if there is an element``                ``// in a1[] such that these three``                ``// form a required triplet``                ``if` `(al.Contains(sum - a2[i] - a3[j]) &``                            ``al.IndexOf(sum - a2[i] - a3[j])``                            ``!= al[al.Count - 1])``                ``{``                    ``return` `true``;``                ``}``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]a1 = {1, 2, 3, 4, 5};``        ``int` `[]a2 = {2, 3, 6, 1, 2};``        ``int` `[]a3 = {3, 2, 4, 5, 6};``        ``int` `sum = 9;``        ``int` `n1 = a1.Length;``        ``int` `n2 = a2.Length;``        ``int` `n3 = a3.Length;``        ``if` `(findTriplet(a1, a2, a3, n1, n2, n3, sum))``        ``{``            ``Console.WriteLine(``"Yes"``);``        ``}``        ``else``        ``{``            ``Console.WriteLine(``"No"``);``        ``}``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`Yes`

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

Another efficient approach ( Space optimization ) : we will run two loops, then we will search for required sum in third loop using binary search .

Below is the implementation of the above approach :

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `//Function to check if x is present in the array or not``bool` `binarysearch(``int` `arr[], ``int` `N, ``int` `x)``{``    ``int` `l = 0, r = N - 1;` `    ``while` `(l <= r) {``        ``int` `mid = (l + r) / 2;` `        ``// Checking if the middle element is equal to x``        ``if` `(arr[mid] == x) {``            ``return` `true``;``        ``}``        ``else` `if` `(arr[mid] < x) {``            ``l = mid + 1;``        ``}``        ``else` `{``            ``r = mid - 1;``        ``}``    ``}``    ``// return true , if element x is present in the array``    ``// else false``    ``return` `false``;``}``// Function to check is such triplet(a+b+c=sum) exist that``bool` `findTriplet(``int` `a1[], ``int` `a2[], ``int` `a3[], ``int` `n1, ``int` `n2, ``int` `n3,``int` `sum)``{   sort(a3,a3+n3);``//sort third array(a3) in ascending order``                  ``// for binary search``   ` `    ``// Iterate each element of array a1``    ``for` `(``int` `i = 0; i < n1; i++)``    ``{``        ``// Iterate each element of array a2``    ``for` `(``int` `j = 0; j < n2; j++)``    ``{   ``int` `requiredsum= sum-a1[i]-a2[j];``        ``if` `(binarysearch(a3, n3,requiredsum))``        ``{ ``            ``return` `true``;;``//return true if Triplet exist``        ``}``    ``}``      ` `    ``}``   ``return` `false``;``//return false if Triplet doesn't exist``}` `// Driver Code``int` `main()``{``    ``int` `a1[] = { 1 , 2 , 3 , 4 , 5 };``    ``int` `a2[] = { 2 , 3 , 6 , 1 , 2 };``    ``int` `a3[] = { 3 , 2 , 4 , 5 , 6 };``    ``int` `sum=9;``    ``int` `n1 = ``sizeof``(a1) / ``sizeof``(``int``);``    ``int` `n2 = ``sizeof``(a2) / ``sizeof``(``int``);``    ``int` `n3 = ``sizeof``(a3) / ``sizeof``(``int``);``  ` `    ``//Function call``    ``if``(findTriplet(a1, a2, a3, n1, n2, n3,sum))``    ``{``        ``cout<<``"YES"``<

Output

`YES`

Time Complexity: O(n1*n2*log n3)
Auxiliary Space: O(1)

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