# Find three element from given three arrays such that their sum is X | Set 2

Given three sorted integer arrays **A[]**, **B[]** and **C[]**, the task is to find three integers, one from each array such that they sum up to a given target value **X**. Print **Yes** or **No** depending on whether such triplet exists or not.

**Examples:**

Input:A[] = {2}, B[] = {1, 6, 7}, C[] = {4, 5}, X = 12

Output:Yes

A[0] + B[1] + C[0] = 2 + 6 + 4 = 12

Input:A[] = {2}, B[] = {1, 6, 7}, C[] = {4, 5}, X = 14

Output:Yes

A[0] + B[2] + C[1] = 2 + 7 + 5 = 14

**Approach:** We have already discusses a hash based approach in this article which takes O(N) extra space.

In this article, we will solve this problem using space efficient method that takes O(1) extra space. The idea is using two pointer technique.

We will iterate through the smallest of all the arrays and for each index **i**, we will use two-pointer on the larger two arrays to find a pair with sum equal to **X – A[i]** (assuming **A[]** is the smallest in length among the three arryas).

Now, what is the idea behind using two pointer on larger two arrays? We will try to understand the same from an example.

Let’s assume

len(A) = 100000

len(B) = 10000

len(C) = 10

Case 1:Applying two pointer on larger two arrays

Number of iterations will be of order = len(C) * (len(A) + len(B)) = 10 * (110000) = 1100000

Case 2:Applying two pointer on smaller two arrays

Number of iterations will be of order = len(A) * (len(B) + len(C)) = 100000 * (10010) = 1001000000

Case 3:Applying two pointer on smallest and largest array

Number of iterations will be of order = len(B) * (len(A) + len(C)) = 10000 * (100000 + 10) = 1000100000As we can see,

Case 1is the most optimal for this example and it can be easily proved that its most optimal in general as well.

**Algorithm:**

- Sort the arrays in increasing order of there lengths.
- Let’s say the smallest array after sorting is A[]. Then, iterate through all the elements of A[] and for each index ‘i’, apply two-pointer on the other two arrays. We will put a pointer on the beginning of array B[] and a pointer to end of array C[]. Let’s call the pointer ‘j’ and ‘k’ respectively.
- If B[j] + C[k] = X – A[i], we found a match.
- If B[j] + C[k] less than X – A[i], we increase value of ‘j’ by 1.
- If B[j] + C[k] greater than X – A[i], we decrease value of ‘k’ by 1.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function that returns true if there ` `// exists a triplet with sum x ` `bool` `existsTriplet(` `int` `a[], ` `int` `b[], ` ` ` `int` `c[], ` `int` `x, ` `int` `l1, ` ` ` `int` `l2, ` `int` `l3) ` `{ ` ` ` `// Sorting arrays such that a[] ` ` ` `// represents smallest array ` ` ` `if` `(l2 <= l1 and l2 <= l3) ` ` ` `swap(l2, l1), swap(a, b); ` ` ` `else` `if` `(l3 <= l1 and l3 <= l2) ` ` ` `swap(l3, l1), swap(a, c); ` ` ` ` ` `// Iterating the smallest array ` ` ` `for` `(` `int` `i = 0; i < l1; i++) { ` ` ` ` ` `// Two pointers on second and third array ` ` ` `int` `j = 0, k = l3 - 1; ` ` ` `while` `(j < l2 and k >= 0) { ` ` ` ` ` `// If a valid triplet is found ` ` ` `if` `(a[i] + b[j] + c[k] == x) ` ` ` `return` `true` `; ` ` ` `if` `(a[i] + b[j] + c[k] < x) ` ` ` `j++; ` ` ` `else` ` ` `k--; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `false` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `a[] = { 2, 7, 8, 10, 15 }; ` ` ` `int` `b[] = { 1, 6, 7, 8 }; ` ` ` `int` `c[] = { 4, 5, 5 }; ` ` ` `int` `l1 = ` `sizeof` `(a) / ` `sizeof` `(` `int` `); ` ` ` `int` `l2 = ` `sizeof` `(b) / ` `sizeof` `(` `int` `); ` ` ` `int` `l3 = ` `sizeof` `(c) / ` `sizeof` `(` `int` `); ` ` ` ` ` `int` `x = 14; ` ` ` ` ` `if` `(existsTriplet(a, b, c, x, l1, l2, l3)) ` ` ` `cout << ` `"Yes"` `; ` ` ` `else` ` ` `cout << ` `"No"` `; ` ` ` ` ` `return` `0; ` `} ` |

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## Python3

`# Function that returns True if there ` `# exists a triplet with sum x ` `def` `existsTriplet(a, b,c, x, l1,l2, l3): ` ` ` ` ` `# Sorting arrays such that a ` ` ` `# represents smallest array ` ` ` `if` `(l2 <` `=` `l1 ` `and` `l2 <` `=` `l3): ` ` ` `l1, l2 ` `=` `l2,l1 ` ` ` `a, b ` `=` `b,a ` ` ` `elif` `(l3 <` `=` `l1 ` `and` `l3 <` `=` `l2): ` ` ` `l1, l3 ` `=` `l3,l1 ` ` ` `a, c ` `=` `c,a ` ` ` ` ` `# Iterating the smallest array ` ` ` `for` `i ` `in` `range` `(l1): ` ` ` ` ` `# Two pointers on second and third array ` ` ` `j ` `=` `0` ` ` `k ` `=` `l3 ` `-` `1` ` ` `while` `(j < l2 ` `and` `k >` `=` `0` `): ` ` ` ` ` `# If a valid triplet is found ` ` ` `if` `(a[i] ` `+` `b[j] ` `+` `c[k] ` `=` `=` `x): ` ` ` `return` `True` ` ` `if` `(a[i] ` `+` `b[j] ` `+` `c[k] < x): ` ` ` `j ` `+` `=` `1` ` ` `else` `: ` ` ` `k ` `-` `=` `1` ` ` ` ` `return` `False` ` ` `# Driver code ` `a ` `=` `[ ` `2` `, ` `7` `, ` `8` `, ` `10` `, ` `15` `] ` `b ` `=` `[ ` `1` `, ` `6` `, ` `7` `, ` `8` `] ` `c ` `=` `[ ` `4` `, ` `5` `, ` `5` `] ` `l1 ` `=` `len` `(a) ` `l2 ` `=` `len` `(b) ` `l3 ` `=` `len` `(c) ` ` ` `x ` `=` `14` ` ` `if` `(existsTriplet(a, b, c, x, l1, l2, l3)): ` ` ` `print` `(` `"Yes"` `) ` `else` `: ` ` ` `print` `(` `"No"` `) ` ` ` `# This code is contributed by mohit kumar 29 ` ` ` |

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**Output:**

Yes

**Time complexity:** O(min(len(A), len(B), len(C)) * max(len(A), len(B), len(C)))

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