# Pythagorean Triplet in an array

• Difficulty Level : Easy
• Last Updated : 09 Mar, 2023

Given an array of integers, write a function that returns true if there is a triplet (a, b, c) that satisfies a2 + b2 = c2.

Example

Input: arr[] = {3, 1, 4, 6, 5}
Output: True
There is a Pythagorean triplet (3, 4, 5).

Input: arr[] = {10, 4, 6, 12, 5}
Output: False
There is no Pythagorean triplet.

Recommended Practice

Method 1 (Naive)
A simple solution is to run three loops, three loops pick three array elements, and check if the current three elements form a Pythagorean Triplet.

Below is the implementation of the above idea :

## C++

 // A C++ program that returns true if there is a Pythagorean// Triplet in a given array.#include  using namespace std; // Returns true if there is Pythagorean triplet in ar[0..n-1]bool isTriplet(int ar[], int n){    for (int i = 0; i < n; i++) {        for (int j = i + 1; j < n; j++) {            for (int k = j + 1; k < n; k++) {                // Calculate square of array elements                int x = ar[i] * ar[i], y = ar[j] * ar[j], z = ar[k] * ar[k];                 if (x == y + z || y == x + z || z == x + y)                    return true;            }        }    }     // If we reach here, no triplet found    return false;} /* Driver program to test above function */int main(){    int ar[] = { 3, 1, 4, 6, 5 };    int ar_size = sizeof(ar) / sizeof(ar[0]);    isTriplet(ar, ar_size) ? cout << "Yes" : cout << "No";    return 0;}

## Java

 // A Java program that returns true if there is a Pythagorean// Triplet in a given array.import java.io.*; class PythagoreanTriplet {     // Returns true if there is Pythagorean triplet in ar[0..n-1]    static boolean isTriplet(int ar[], int n)    {        for (int i = 0; i < n; i++) {            for (int j = i + 1; j < n; j++) {                for (int k = j + 1; k < n; k++) {                    // Calculate square of array elements                    int x = ar[i] * ar[i], y = ar[j] * ar[j], z = ar[k] * ar[k];                     if (x == y + z || y == x + z || z == x + y)                        return true;                }            }        }         // If we reach here, no triplet found        return false;    }     // Driver program to test above function    public static void main(String[] args)    {        int ar[] = { 3, 1, 4, 6, 5 };        int ar_size = ar.length;        if (isTriplet(ar, ar_size) == true)            System.out.println("Yes");        else            System.out.println("No");    }}/* This code is contributed by Devesh Agrawal */

## Python3

 # Python program to check if there is Pythagorean# triplet in given array # Returns true if there is Pythagorean# triplet in ar[0..n-1] def isTriplet(ar, n):    j = 0         for i in range(n - 2):        for j in range(i + 1, n):            for k in range(j + 1, n - 1):                # Calculate square of array elements                x = ar[i]*ar[i]                y = ar[j]*ar[j]                z = ar[k]*ar[k]                if (x == y + z or y == x + z or z == x + y):                    return 1         # If we reach here, no triplet found    return 0  # Driver program to test above functionar = [3, 1, 4, 6, 5]ar_size = len(ar) if(isTriplet(ar, ar_size)):    print("Yes")else:    print("No") # This code is contributed by Anvesh Govind Saxena

## C#

 // A C# program that returns true// if there is a Pythagorean// Triplet in a given array.using System; class GFG {     // Returns true if there is Pythagorean    // triplet in ar[0..n-1]    static bool isTriplet(int[] ar, int n)    {        for (int i = 0; i < n; i++) {            for (int j = i + 1; j < n; j++) {                for (int k = j + 1; k < n; k++) {                     // Calculate square of array elements                    int x = ar[i] * ar[i], y = ar[j] * ar[j], z = ar[k] * ar[k];                     if (x == y + z || y == x + z || z == x + y)                        return true;                }            }        }         // If we reach here,        // no triplet found        return false;    }     // Driver code    public static void Main()    {        int[] ar = { 3, 1, 4, 6, 5 };        int ar_size = ar.Length;        if (isTriplet(ar, ar_size) == true)            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by shiv_bhakt.



## Javascript



Output:

Yes

The Time Complexity of the above solution is O(n3).

Auxiliary Space: O(1)

Method 2 (Use Sorting)
We can solve this in O(n2) time by sorting the array first.
1) Do the square of every element in the input array. This step takes O(n) time.
2) Sort the squared array in increasing order. This step takes O(nLogn) time.
3) To find a triplet (a, b, c) such that a2 = b2 + c2, do following.

1. Fix ‘a’ as the last element of the sorted array.
2. Now search for pair (b, c) in subarray between the first element and ‘a’. A pair (b, c) with a given sum can be found in O(n) time using the meet in middle algorithm discussed in method 1 of this post.
3. If no pair is found for current ‘a’, then move ‘a’ one position back and repeat step 3.2.

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

## C++

 // A C++ program that returns true if there is a Pythagorean// Triplet in a given array.#include #include  using namespace std; // Returns true if there is a triplet with following property// A[i]*A[i] = A[j]*A[j] + A[k]*[k]// Note that this function modifies given arraybool isTriplet(int arr[], int n){    // Square array elements    for (int i = 0; i < n; i++)        arr[i] = arr[i] * arr[i];     // Sort array elements    sort(arr, arr + n);     // Now fix one element one by one and find the other two    // elements    for (int i = n - 1; i >= 2; i--) {        // To find the other two elements, start two index        // variables from two corners of the array and move        // them toward each other        int l = 0; // index of the first element in arr[0..i-1]        int r = i - 1; // index of the last element in arr[0..i-1]        while (l < r) {            // A triplet found            if (arr[l] + arr[r] == arr[i])                return true;             // Else either move 'l' or 'r'            (arr[l] + arr[r] < arr[i]) ? l++ : r--;        }    }     // If we reach here, then no triplet found    return false;} /* Driver program to test above function */int main(){    int arr[] = { 3, 1, 4, 6, 5 };    int arr_size = sizeof(arr) / sizeof(arr[0]);    isTriplet(arr, arr_size) ? cout << "Yes" : cout << "No";    return 0;}

## Java

 // A Java program that returns true if there is a Pythagorean// Triplet in a given array.import java.io.*;import java.util.*; class PythagoreanTriplet {    // Returns true if there is a triplet with following property    // A[i]*A[i] = A[j]*A[j] + A[k]*[k]    // Note that this function modifies given array    static boolean isTriplet(int arr[], int n)    {        // Square array elements        for (int i = 0; i < n; i++)            arr[i] = arr[i] * arr[i];         // Sort array elements        Arrays.sort(arr);         // Now fix one element one by one and find the other two        // elements        for (int i = n - 1; i >= 2; i--) {            // To find the other two elements, start two index            // variables from two corners of the array and move            // them toward each other            int l = 0; // index of the first element in arr[0..i-1]            int r = i - 1; // index of the last element in arr[0..i-1]            while (l < r) {                // A triplet found                if (arr[l] + arr[r] == arr[i])                    return true;                 // Else either move 'l' or 'r'                if (arr[l] + arr[r] < arr[i])                    l++;                else                    r--;            }        }         // If we reach here, then no triplet found        return false;    }     // Driver program to test above function    public static void main(String[] args)    {        int arr[] = { 3, 1, 4, 6, 5 };        int arr_size = arr.length;        if (isTriplet(arr, arr_size) == true)            System.out.println("Yes");        else            System.out.println("No");    }}/*This code is contributed by Devesh Agrawal*/

## Python3

 # Python program that returns true if there is# a Pythagorean Triplet in a given array. # Returns true if there is Pythagorean# triplet in ar[0..n-1]def isTriplet(ar, n):    # Square all the elements    for i in range(n):        ar[i] = ar[i] * ar[i]     # sort array elements    ar.sort()     # fix one element    # and find other two    # i goes from n - 1 to 2    for i in range(n-1, 1, -1):        # start two index variables from        # two corners of the array and        # move them toward each other        j = 0        k = i - 1        while (j < k):            # A triplet found            if (ar[j] + ar[k] == ar[i]):                return True            else:                if (ar[j] + ar[k] < ar[i]):                    j = j + 1                else:                    k = k - 1    # If we reach here, then no triplet found    return False # Driver program to test above function */ar = [3, 1, 4, 6, 5]ar_size = len(ar)if(isTriplet(ar, ar_size)):    print("Yes")else:    print("No") # This code is contributed by Aditi Sharma

## C#

 // C# program that returns true// if there is a Pythagorean// Triplet in a given array.using System; class GFG {     // Returns true if there is a triplet    // with following property A[i]*A[i]    // = A[j]*A[j]+ A[k]*[k] Note that    // this function modifies given array    static bool isTriplet(int[] arr, int n)    {         // Square array elements        for (int i = 0; i < n; i++)            arr[i] = arr[i] * arr[i];         // Sort array elements        Array.Sort(arr);         // Now fix one element one by one        // and find the other two elements        for (int i = n - 1; i >= 2; i--) {            // To find the other two elements,            // start two index variables from            // two corners of the array and            // move them toward each other            // index of the first element            // in arr[0..i-1]            int l = 0;             // index of the last element            // in arr[0..i - 1]            int r = i - 1;            while (l < r) {                 // A triplet found                if (arr[l] + arr[r] == arr[i])                    return true;                 // Else either move 'l' or 'r'                if (arr[l] + arr[r] < arr[i])                    l++;                else                    r--;            }        }         // If we reach here, then        // no triplet found        return false;    }     // Driver Code    public static void Main()    {        int[] arr = { 3, 1, 4, 6, 5 };        int arr_size = arr.Length;        if (isTriplet(arr, arr_size) == true)            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by shiv_bhakt.

## PHP

 = 2; \$i--)    {                 // To find the other two        // elements, start two index        // variables from two corners        // of the array and move        // them toward each other                 // index of the first element        // in arr[0..i-1]        \$l = 0;                 // index of the last element        // in arr[0..i-1]        \$r = \$i - 1;        while (\$l < \$r)        {                         // A triplet found            if (\$arr[\$l] + \$arr[\$r] == \$arr[\$i])                return true;             // Else either move 'l' or 'r'            (\$arr[\$l] + \$arr[\$r] < \$arr[\$i])? \$l++: \$r--;        }    }     // If we reach here,    // then no triplet found    return false;}     // Driver Code    \$arr = array(3, 1, 4, 6, 5);    \$arr_size = count(\$arr);    if(isTriplet(\$arr, \$arr_size))        echo "Yes";    else        echo "No"; // This code is contributed by anuj_67.?>

## Javascript



Output:

Yes

The time complexity of this method is O(n2).

Auxiliary Space: O(1)

Method 3: (Using Hashing)
The problem can also be solved using hashing. We can use a hash map to mark all the values of the given array. Using two loops, we can iterate for all the possible combinations of a and b, and then check if there exists the third value c. If there exists any such value, then there is a Pythagorean triplet.

Below is the implementation of the above approach:

## C++

 #include using namespace std; // Function to check if the// Pythagorean triplet exists or notbool checkTriplet(int arr[], int n){    int maximum = 0;     // Find the maximum element    for (int i = 0; i < n; i++) {        maximum = max(maximum, arr[i]);    }     // Hashing array    int hash[maximum + 1] = { 0 };     // Increase the count of array elements    // in hash table    for (int i = 0; i < n; i++)        hash[arr[i]]++;     // Iterate for all possible a    for (int i = 1; i < maximum + 1; i++) {         // If a is not there        if (hash[i] == 0)            continue;         // Iterate for all possible b        for (int j = 1; j < maximum + 1; j++) {             // If a and b are same and there is only one a            // or if there is no b in original array            if ((i == j && hash[i] == 1) || hash[j] == 0)                continue;             // Find c            int val = sqrt(i * i + j * j);             // If c^2 is not a perfect square            if ((val * val) != (i * i + j * j))                continue;             // If c exceeds the maximum value            if (val > maximum)                continue;             // If there exists c in the original array,            // we have the triplet            if (hash[val]) {                return true;            }        }    }    return false;}// Driver Codeint main(){    int arr[] = { 3, 2, 4, 6, 5 };    int n = sizeof(arr) / sizeof(arr[0]);    if (checkTriplet(arr, n))        cout << "Yes";    else        cout << "No";}

## Java

 import java.util.*; class GFG{ // Function to check if the// Pythagorean triplet exists or notstatic boolean checkTriplet(int arr[], int n){    int maximum = 0;     // Find the maximum element    for (int i = 0; i < n; i++)    {        maximum = Math.max(maximum, arr[i]);    }     // Hashing array    int []hash = new int[maximum + 1];     // Increase the count of array elements    // in hash table    for (int i = 0; i < n; i++)        hash[arr[i]]++;     // Iterate for all possible a    for (int i = 1; i < maximum + 1; i++)    {         // If a is not there        if (hash[i] == 0)            continue;         // Iterate for all possible b        for (int j = 1; j < maximum + 1; j++)        {             // If a and b are same and there is only one a            // or if there is no b in original array            if ((i == j && hash[i] == 1) || hash[j] == 0)                continue;             // Find c            int val = (int) Math.sqrt(i * i + j * j);             // If c^2 is not a perfect square            if ((val * val) != (i * i + j * j))                continue;             // If c exceeds the maximum value            if (val > maximum)                continue;             // If there exists c in the original array,            // we have the triplet            if (hash[val] == 1)            {                return true;            }        }    }    return false;} // Driver Codepublic static void main(String[] args){    int arr[] = { 3, 2, 4, 6, 5 };    int n = arr.length;    if (checkTriplet(arr, n))        System.out.print("Yes");    else        System.out.print("No");}} // This code is contributed by Rajput-Ji

## Python3

 # Function to check if the# Pythagorean triplet exists or notimport math def checkTriplet(arr, n):    maximum = 0     # Find the maximum element    maximum = max(arr)         # Hashing array    hash = [0]*(maximum+1)     # Increase the count of array elements    # in hash table    for i in range(n):        hash[arr[i]] += 1         # Iterate for all possible a    for i in range(1, maximum+1):        # If a is not there        if (hash[i] == 0):            continue         # Iterate for all possible b        for j in range(1, maximum+1):            # If a and b are same and there is only one a            # or if there is no b in original array            if ((i == j and hash[i] == 1) or hash[j] == 0):                continue             # Find c            val = int(math.sqrt(i * i + j * j))             # If c^2 is not a perfect square            if ((val * val) != (i * i + j * j)):                continue             # If c exceeds the maximum value            if (val > maximum):                continue             # If there exists c in the original array,            # we have the triplet            if (hash[val]):                return True    return False  # Driver Codearr = [3, 2, 4, 6, 5]n = len(arr)if (checkTriplet(arr, n)):    print("Yes")else:    print("No")  # This code is contributed by ankush_953

## C#

 using System; class GFG{ // Function to check if the// Pythagorean triplet exists or notstatic bool checkTriplet(int []arr, int n){    int maximum = 0;     // Find the maximum element    for (int i = 0; i < n; i++)    {        maximum = Math.Max(maximum, arr[i]);    }     // Hashing array    int []hash = new int[maximum + 1];     // Increase the count of array elements    // in hash table    for (int i = 0; i < n; i++)        hash[arr[i]]++;     // Iterate for all possible a    for (int i = 1; i < maximum + 1; i++)    {         // If a is not there        if (hash[i] == 0)            continue;         // Iterate for all possible b        for (int j = 1; j < maximum + 1; j++)        {             // If a and b are same and there is only one a            // or if there is no b in original array            if ((i == j && hash[i] == 1) || hash[j] == 0)                continue;             // Find c            int val = (int) Math.Sqrt(i * i + j * j);             // If c^2 is not a perfect square            if ((val * val) != (i * i + j * j))                continue;             // If c exceeds the maximum value            if (val > maximum)                continue;             // If there exists c in the original array,            // we have the triplet            if (hash[val] == 1)            {                return true;            }        }    }    return false;} // Driver Codepublic static void Main(String[] args){    int []arr = { 3, 2, 4, 6, 5 };    int n = arr.Length;    if (checkTriplet(arr, n))        Console.Write("Yes");    else        Console.Write("No");}} // This code is contributed by Rajput-Ji

## Javascript



Output

Yes

Thanks to Striver for suggesting the above approach.
Time Complexity: O( max * max ), where max is the maximum element in the array.

Auxiliary Space: O(max)

Method -4:Using STL

Approach:

The problem can be solved using ordered maps and unordered maps. There is no need to store the elements in an ordered manner so implementation by an unordered map is faster. We can use the unordered map to mark all the values of the given array. Using two loops, we can iterate for all the possible combinations of a and b, and then check if there exists the third value c. If there exists any such value, then there is a Pythagorean triplet.

Below is the implementation of the above approach:

## C++

 #include using namespace std; //  Returns true if there is Pythagorean triplet in//  ar[0..n-1]bool checkTriplet(int arr[], int n){    // initializing unordered map with key and value as    // integers    unordered_map umap;         // Increase the count of array elements in unordered map    for (int i = 0; i < n; i++)        umap[arr[i]] = umap[arr[i]] + 1;     for (int i = 0; i < n - 1; i++)    {        for (int j = i + 1; j < n; j++)        {               // calculating the squares of two elements as            // integer and float            int p = sqrt(arr[i] * arr[i] + arr[j] * arr[j]);            float q                = sqrt(arr[i] * arr[i] + arr[j] * arr[j]);                         // Condition is true if the value is same in            // integer and float and also the value is            // present in unordered map            if (p == q && umap[p] != 0)                return true;        }    }        // If we reach here, no triplet found    return false;} // Driver Codeint main(){    int arr[] = { 3, 2, 4, 6, 5 };    int n = sizeof(arr) / sizeof(arr[0]);    if (checkTriplet(arr, n))        cout << "Yes";    else        cout << "No";}// This code is contributed by Vikkycirus

## Java

 import java.util.*;class GFG{   //  Returns true if there is Pythagorean triplet in  //  ar[0..n-1]  static boolean checkTriplet(int arr[], int n)  {     // initializing unordered map with key and value as    // integers    HashMap umap = new HashMap<>();     // Increase the count of array elements in unordered map    for (int i = 0; i < n; i++)      if(umap.containsKey(arr[i]))        umap.put(arr[i] , umap.get(arr[i]) + 1);    else      umap.put(arr[i], 1);     for (int i = 0; i < n - 1; i++)    {      for (int j = i + 1; j < n; j++)      {            // calculating the squares of two elements as        // integer and float        int p =(int) Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]);        float q          =(float) Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]);         // Condition is true if the value is same in        // integer and float and also the value is        // present in unordered map        if (p == q && umap.get(p) != 0)          return true;      }    }     // If we reach here, no triplet found    return false;  }   // Driver Code  public static void main(String[] args)  {    int arr[] = { 3, 2, 4, 6, 5 };    int n = arr.length;    if (checkTriplet(arr, n))      System.out.print("Yes");    else      System.out.print("No");  }} // This code is contributed by umadevi9616

## Python3

 # Function to check if the# Pythagorean triplet exists or notimport math def checkTriplet(arr, n):       # creating dictionary/unordered map    h = {arr[i]: 1 for i in range(n)}    for i in range(n-1):        for j in range(i+1, n):                       # Calculating the squares of 2 elements            q = math.sqrt(arr[i]*arr[i] + arr[j]*arr[j])                         # Checking if squareroot is integer and present in map            if q == int(q) and int(q) in h:                return True    return False # Driver Codearr = [3, 2, 4, 6, 5]n = len(arr)if (checkTriplet(arr, n)):    print("Yes")else:    print("No") # This code is contributed by Anvesh Govind Saxena

## C#

 using System;using System.Collections.Generic; public class GFG {   // Returns true if there is Pythagorean triplet in  // ar[0..n-1]  static bool checkTriplet(int []arr, int n) {     // initializing unordered map with key and value as    // integers    Dictionary umap = new Dictionary();     // Increase the count of array elements in unordered map    for (int i = 0; i < n; i++)      if (umap.ContainsKey(arr[i]))        umap.Add(arr[i], umap[arr[i]] + 1);    else      umap.Add(arr[i], 1);     for (int i = 0; i < n - 1; i++) {      for (int j = i + 1; j < n; j++) {         // calculating the squares of two elements as        // integer and float        int p = (int) Math.Sqrt(arr[i] * arr[i] + arr[j] * arr[j]);        float q = (float) Math.Sqrt(arr[i] * arr[i] + arr[j] * arr[j]);         // Condition is true if the value is same in        // integer and float and also the value is        // present in unordered map        if (p == q && umap[p] != 0)          return true;      }    }     // If we reach here, no triplet found    return false;  }   // Driver Code  public static void Main(String[] args) {    int []arr = { 3, 2, 4, 6, 5 };    int n = arr.Length;    if (checkTriplet(arr, n))      Console.Write("Yes");    else      Console.Write("No");  }} // This code is contributed by umadevi9616

## Javascript



Output

Yes

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

Method 5 – A better hashing based approach

This approach uses Set. Firstly, we’ll square the elements of the array and then sort the array in increasing order. Run two loops where the outer loop starts from the last index of the array to the second index (0 based indexing is assumed) and the inner loop starts from outerLoopIndex – 1 to the start. Create a set to store the elements in between outerLoopIndex and innerLoopIndex. Check if there is a number in the set which is equal to arr[outerLoopIndex] â€“ arr[innerLoopIndex]. If yes, then return “True”.

## C++

 #include using namespace std;bool checkTriplet(int arr[],int n){  for(int i = 0; i < n; i++)    arr[i] = arr[i]*arr[i];   sort(arr, arr + n);   for(int i = n - 1; i > 1; i--)  {    unordered_set s;    for(int j = i - 1; j >- 1; j--)    {      if(s.count(arr[i] - arr[j]))        return true;       s.insert(arr[j]);    }  }   return false;}int main(){  int arr[] = {3, 2, 4, 6, 5};  int n = sizeof(arr)/sizeof(arr[0]);   if (checkTriplet(arr, n))    cout << "Yes";  else    cout << "No";   return 0;} // This code is contributed by aditya942003patil

## Python3

 # Python program to check if there exists a pythagorean tripletdef checkTriplet(arr, n):    for i in range(n):        arr[i] = arr[i] * arr[i]     arr.sort()     for i in range(n - 1, 1, -1):        s = set()        for j in range(i - 1, -1, -1):            if (arr[i] - arr[j]) in s:                return True            s.add(arr[j])    return False # Driver Programarr = [3, 2, 4, 6, 5]n = len(arr)if (checkTriplet(arr, n)):    print("Yes")else:    print("No") # This is contributed by Manvi Pandey

## Java

 import java.util.Arrays;import java.util.HashSet; public class Triplet {    static boolean checkTriplet(int[] arr, int n) {        // loop through each element in the array and square it        for (int i = 0; i < n; i++) {            arr[i] = arr[i] * arr[i];        }         // sort the array        Arrays.sort(arr);         // loop through each element in the array starting from the last index        for (int i = n - 1; i > 1; i--) {            HashSet s = new HashSet<>();            // loop through each element from the current index to the first index            for (int j = i - 1; j >= 0; j--) {                // check if the difference between the current element and the previous element is present in the set                if (s.contains(arr[i] - arr[j])) {                    return true;                }                // add the previous element to the set                s.add(arr[j]);            }        }        return false;    }     public static void main(String[] args) {        int[] arr = {3, 2, 4, 6, 5};        int n = arr.length;         if (checkTriplet(arr, n)) {            System.out.println("Yes");        } else {            System.out.println("No");        }    }}

## C#

 using System;using System.Collections.Generic;using System.Linq; class Program{    static bool CheckTriplet(int[] arr, int n)    {        for (int i = 0; i < n; i++)            arr[i] = arr[i] * arr[i];         Array.Sort(arr);         for (int i = n - 1; i > 1; i--)        {            HashSet s = new HashSet();            for (int j = i - 1; j >= 0; j--)            {                if (s.Contains(arr[i] - arr[j]))                    return true;                 s.Add(arr[j]);            }        }         return false;    }     static void Main()    {        int[] arr = { 3, 2, 4, 6, 5 };        int n = arr.Length;         if (CheckTriplet(arr, n))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }}

## Javascript

 // JavaScript program to check if there exists a pythagorean tripletfunction checkTriplet(arr, n) {// Square all array elementsfor (let i = 0; i < n; i++) {arr[i] = arr[i] * arr[i];}// Sort the array in non-decreasing orderarr.sort((a, b) => a - b); // Check for Pythagorean tripletfor (let i = n - 1; i >= 2; i--) {    let s = new Set();    for (let j = i - 1; j >= 0; j--) {        if (s.has(arr[i] - arr[j])) {            return true;        }        s.add(arr[j]);    }}return false;} // Driver programlet arr = [3, 2, 4, 6, 5];let n = arr.length;if (checkTriplet(arr, n)) {console.log("Yes");} else {console.log("No");}

Output

Yes

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

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