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.
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 <iostream>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 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 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. |
PHP
<?php// A PHP 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]function isTriplet($ar, $n){ for ($i = 0; $i < $n; $i++) { for ($j = $i + 1; $j < $n; $j++) { for ($k = $j + 1; $k < $n; $k++) { // 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 true; } } } // If we reach here, // no triplet found return false;} // Driver Code $ar = array(3, 1, 4, 6, 5); $ar_size = count($ar); if(isTriplet($ar, $ar_size)) echo "Yes"; else echo "No";// This code is contributed by anuj_67.?> |
Javascript
<script>// A Javascript 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]function isTriplet( ar, n){ for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { for (let k = j + 1; k < n; k++) { // Calculate square of array elements let 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 let ar = [ 3, 1, 4, 6, 5 ]; let ar_size = ar.length; if (isTriplet(ar, ar_size) == true) document.write("Yes"); else document.write("No"); </script> |
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.
- Fix ‘a’ as the last element of the sorted array.
- 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.
- 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 <algorithm>#include <iostream>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
<?php// A PHP program that returns // true if there is a Pythagorean// Triplet in a given array.// 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 arrayfunction isTriplet( $arr, $n){ // Square array elements for ($i = 0; $i < $n; $i++) $arr[$i] = $arr[$i] * $arr[$i]; // Sort array elements sort($arr); // Now fix one element one by // one and find the other two // elements for($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] $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
<script>// A javascript program that returns true if there is a Pythagorean// Triplet in a given array. // 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 function isTriplet(arr , n) { // Square array elements for (i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; // Sort array elements arr.sort((a,b)=>a-b); // Now fix one element one by one and find the other two // elements for (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 var l = 0; // index of the first element in arr[0..i-1] var 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 var arr = [ 3, 1, 4, 6, 5 ]; var arr_size = arr.length; if (isTriplet(arr, arr_size) == true) document.write("Yes"); else document.write("No");// This code is contributed by umadevi9616</script> |
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 <bits/stdc++.h>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 mathdef 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
<script> // Function to check if the // Pythagorean triplet exists or not function checkTriplet(arr , n) { var maximum = 0; // Find the maximum element for (i = 0; i < n; i++) { maximum = Math.max(maximum, arr[i]); } // Hashing array var hash = Array(maximum + 1).fill(0); // Increase the count of array elements // in hash table for (i = 0; i < n; i++) hash[arr[i]]++; // Iterate for all possible a for (i = 1; i < maximum + 1; i++) { // If a is not there if (hash[i] == 0) continue; // Iterate for all possible b for (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 var val = parseInt( 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 Code var arr = [ 3, 2, 4, 6, 5 ]; var n = arr.length; if (checkTriplet(arr, n)) document.write("Yes"); else document.write("No");// This code is contributed by gauravrajput1 </script> |
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 <bits/stdc++.h>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<int, int> 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<Integer,Integer> 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 mathdef 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<int, int> umap = new Dictionary<int,int>(); // 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
<script> // Returns true if there is Pythagorean triplet in // ar[0..n-1] function checkTriplet(arr , n) { // initializing unordered map with key and value as // integers var umap = new Map(); // Increase the count of array elements in unordered map for (i = 0; i < n; i++) if (umap.has(arr[i])) umap.set(arr[i], umap.get(arr[i]) + 1); else umap.set(arr[i], 1); for (i = 0; i < n - 1; i++) { for (j = i + 1; j < n; j++) { // calculating the squares of two elements as // integer and float var p = parseInt( Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j])); var q = Math.sqrt(arr[i] * arr[i] + arr[j] * arr[j]); // Condition is true if the value is same in // integer and var 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 var arr = [ 3, 2, 4, 6, 5 ]; var n = arr.length; if (checkTriplet(arr, n)) document.write("Yes"); else document.write("No");// This code contributed by gauravrajput1</script> |
Output
Yes
Time Complexity:O(n2)
Auxiliary Space:O(n)
This article is contributed by Harshit Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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] – arr[innerLoopIndex]. If yes, then return “True”.
C++
#include <bits/stdc++.h>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<int> 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<Integer> 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<int> s = new HashSet<int>(); 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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