3 Sum - Pythagorean Triplet in an array
Last Updated : 21 May, 2025
Given an array of positive integers, the task is to determine if a Pythagorean triplet exists in the given array. A triplet {a, b, c} is considered a Pythagorean triplet if it satisfies the condition a2 + b2 = c2.
Example:
Input: arr[] = [3, 1, 4, 6, 5]
Output: true
Explanation: The array contains a Pythagorean triplet {3, 4, 5}.
Input: arr[] = [10, 4, 6, 12, 5]
Output: false
Explanation: There is no Pythagorean triplet.
[Naive Approach] Explore all the triplets - O(n^3) Time and O(1) Space
The simplest approach is to explore all the triplets(a, b, c) using three nested loops and if any of them satisfies the condition of Pythagorean Triplet, that is a2 + b2 = c2 or a2 + c2 = b2 or b2 + c2 = a2, return true.
C++ // C++ program to find if a Pythagorean triplet exists // by exploring all triplets #include <iostream> #include <vector> using namespace std; bool pythagoreanTriplet(vector<int> &arr) { int n = arr.size(); // Exploring all possible triplets 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 = arr[i] * arr[i]; int y = arr[j] * arr[j]; int z = arr[k] * arr[k]; // If these integers form Pythagorean triplet then // return true if (x == y + z || y == x + z || z == x + y) return true; } } } return false; } int main() { vector<int> arr = {3, 1, 4, 6, 5}; cout << (pythagoreanTriplet(arr) ? "true" : "false"); return 0; } // C program to find if a Pythagorean triplet exists // by exploring all triplets #include <stdio.h> #include <stdbool.h> bool pythagoreanTriplet(int arr[], int n) { // Exploring all possible triplets 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 = arr[i] * arr[i]; int y = arr[j] * arr[j]; int z = arr[k] * arr[k]; // If these integers form Pythagorean triplet then // return true if (x == y + z || y == x + z || z == x + y) return true; } } } return false; } int main() { int arr[] = {3, 1, 4, 6, 5}; int n = sizeof(arr) / sizeof(arr[0]); printf(pythagoreanTriplet(arr, n) ? "true" : "false"); return 0; } // Java program to find if a Pythagorean triplet exists // by exploring all triplets import java.util.*; class GfG { static boolean pythagoreanTriplet(int[] arr) { int n = arr.length; // Exploring all possible triplets 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 = arr[i] * arr[i]; int y = arr[j] * arr[j]; int z = arr[k] * arr[k]; // If these integers form Pythagorean triplet then // return true if (x == y + z || y == x + z || z == x + y) return true; } } } return false; } public static void main(String[] args) { int[] arr = {3, 1, 4, 6, 5}; System.out.println(pythagoreanTriplet(arr)); } } # Python program to find if a Pythagorean triplet exists # by exploring all triplets def pythagoreanTriplet(arr): n = len(arr) # Exploring all possible triplets for i in range(n): for j in range(i + 1, n): for k in range(j + 1, n): # Calculate square of array elements x = arr[i] ** 2 y = arr[j] ** 2 z = arr[k] ** 2 # If these integers form Pythagorean triplet then # return true if x == y + z or y == x + z or z == x + y: return True return False if __name__ == "__main__": arr = [3, 1, 4, 6, 5] if pythagoreanTriplet(arr): print("true") else : print("false") // C# program to find if a Pythagorean triplet exists // by exploring all triplets using System; class GfG { static bool pythagoreanTriplet(int[] arr) { int n = arr.Length; // Exploring all possible triplets 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 = arr[i] * arr[i]; int y = arr[j] * arr[j]; int z = arr[k] * arr[k]; // If these integers form Pythagorean triplet then // return true if (x == y + z || y == x + z || z == x + y) return true; } } } return false; } static void Main() { int[] arr = {3, 1, 4, 6, 5}; Console.WriteLine(pythagoreanTriplet(arr)?"true":"false"); } } // JavaScript program to find if a Pythagorean triplet exists // by exploring all triplets function pythagoreanTriplet(arr) { let n = arr.length; // Exploring all possible triplets 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 = arr[i] * arr[i]; let y = arr[j] * arr[j]; let z = arr[k] * arr[k]; // If these integers form Pythagorean triplet then // return true if (x == y + z || y == x + z || z == x + y) return true; } } } return false; } let arr = [3, 1, 4, 6, 5]; console.log(pythagoreanTriplet(arr)); [Better Approach-1] Two Pointers Technique - O(n^2) Time and O(1) Space
The idea is to square each element of the array and sort it in ascending order. Then, we will traverse the array in reverse and fix the largest element of the triplet as c2. After that, we will use two-pointers technique to find two elements a2 and b2 in remaining array such that a2 + b2 = c2. We initialize two pointers at both ends of the remaining portion of the array (from index 0 to i−1) and adjust them based on the sum of the elements at both pointers as follows:
- If sum = c2, we have found the Pythagorean triplet.
- If sum < c2, we move the left pointer towards right.
- If sum > c2, we move the right pointer towards left.
C++ // C++ program to find if a Pythagorean triplet // exists using the two-pointer technique #include <iostream> #include <algorithm> #include <vector> using namespace std; bool pythagoreanTriplet(vector<int> &arr) { int n = arr.size(); // Taking Square of each element for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; sort(arr.begin(), arr.end()); // fix the largest element of Pythagorean triplet for (int i = n - 1; i > 1; i--) { // Two pointer technique to find remaining two // elements such that a^2 + b^2 = c^2 int l = 0; int r = i - 1; while (l < r) { // A Pythagorean triplet is found if (arr[l] + arr[r] == arr[i]) return true; if (arr[l] + arr[r] < arr[i]) l++; else r--; } } return false; } int main() { vector<int> arr = {3, 1, 4, 6, 5}; cout << (pythagoreanTriplet(arr) ? "true" : "false"); return 0; } // C program to find if a Pythagorean triplet // exists using the two-pointer technique #include <stdio.h> #include <stdlib.h> #include <stdbool.h> int compare(const void *a, const void *b) { return (*(int *)a - *(int *)b); } bool pythagoreanTriplet(int arr[], int n) { // Taking Square of each element for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; // Sort the array qsort(arr, n, sizeof(int), compare); // Fix the largest element of Pythagorean triplet for (int i = n - 1; i > 1; i--) { // Two pointer technique to find remaining two // elements such that a^2 + b^2 = c^2 int l = 0; int r = i - 1; while (l < r) { // A Pythagorean triplet is found if (arr[l] + arr[r] == arr[i]) return true; if (arr[l] + arr[r] < arr[i]) l++; else r--; } } return false; } int main() { int arr[] = {3, 1, 4, 6, 5}; int n = sizeof(arr) / sizeof(arr[0]); printf(pythagoreanTriplet(arr, n) ? "true" : "false"); return 0; } // Java program to find if a Pythagorean triplet // exists using the two-pointer technique import java.util.Arrays; class GfG { static boolean pythagoreanTriplet(int[] arr) { int n = arr.length; // Taking Square of each element for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; Arrays.sort(arr); // Fix the largest element of Pythagorean triplet for (int i = n - 1; i > 1; i--) { // Two pointer technique to find remaining two // elements Such that a^2 + b^2 = c^2 int l = 0; int r = i - 1; while (l < r) { // A Pythagorean triplet is found if (arr[l] + arr[r] == arr[i]) return true; if (arr[l] + arr[r] < arr[i]) l++; else r--; } } return false; } public static void main(String[] args) { int[] arr = {3, 1, 4, 6, 5}; System.out.println(pythagoreanTriplet(arr)); } } # Python program to find if a Pythagorean triplet # exists using the two-pointer technique def pythagoreanTriplet(arr): n = len(arr) # Taking Square of each element for i in range(n): arr[i] = arr[i] * arr[i] arr.sort() # Fix the largest element of Pythagorean triplet for i in range(n - 1, 1, -1): # Two pointer technique to find remaining two # elements such that a^2 + b^2 = c^2 l = 0 r = i - 1 while l < r: # A Pythagorean triplet is found if arr[l] + arr[r] == arr[i]: return True if arr[l] + arr[r] < arr[i]: l += 1 else: r -= 1 return False if __name__ == "__main__": arr = [3, 1, 4, 6, 5] print("true" if pythagoreanTriplet(arr) else "false") // C# program to find if a Pythagorean triplet // exists using the two-pointer technique using System; class GfG { static bool pythagoreanTriplet(int[] arr) { int n = arr.Length; // Taking Square of each element for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; Array.Sort(arr); // Fix the largest element of Pythagorean triplet for (int i = n - 1; i > 1; i--) { // Two pointer technique to find remaining two elements // such that a^2 + b^2 = c^2 int l = 0; int r = i - 1; while (l < r) { // A Pythagorean triplet is found if (arr[l] + arr[r] == arr[i]) return true; if (arr[l] + arr[r] < arr[i]) l++; else r--; } } return false; } static void Main() { int[] arr = {3, 1, 4, 6, 5}; Console.WriteLine(pythagoreanTriplet(arr)?"true":"false"); } } // JavaScript program to find if a Pythagorean triplet // exists using the two-pointer technique function pythagoreanTriplet(arr) { let n = arr.length; // Taking Square of each element for (let i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; arr.sort((a, b) => a - b); // Fix the largest element of Pythagorean triplet for (let i = n - 1; i > 1; i--) { // Two pointer technique to find remaining two // elements such that a^2 + b^2 = c^2 let l = 0; let r = i - 1; while (l < r) { // A Pythagorean triplet is found if (arr[l] + arr[r] === arr[i]) return true; if (arr[l] + arr[r] < arr[i]) l++; else r--; } } return false; } // Driver Code const arr = [3, 1, 4, 6, 5]; console.log(pythagoreanTriplet(arr)); [Better Approach-2] Using Hashing - O(n^2) Time and O(n) Space
First, we will insert all the elements of the given array into a hash set. Then, using two nested loops, we will iterate through all possible combinations of a and b, and check if there exists a third value c that forms a Pythagorean triplet with a and b. If such c exists, we will return true.
C++ // C++ program to find if a Pythagorean triplet exists // using hashing #include <iostream> #include <vector> #include <unordered_set> #include <math.h> using namespace std; bool pythagoreanTriplet(vector<int> &arr) { int n = arr.size(); unordered_set<int> st; for (int i = 0; i < n; i++) st.insert(arr[i]); // Iterate through all possible values of a and b for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int a = arr[i]; int b = arr[j]; // calculate required value for // c to form Pythagorean triplet int c = sqrt(a * a + b * b); // First, verify if c^2 is a perfect square (indicating a // valid c) and check if this c exists in the set. if (c*c == a*a + b*b && st.find(c) != st.end()) return true; } } // No Pythagorean triplet exists in the array return false; } int main() { vector<int> arr = {3, 1, 4, 6, 5}; cout << (pythagoreanTriplet(arr) ? "true" : "false"); return 0; } // Java program to find if a Pythagorean triplet exists // using hashing import java.util.HashSet; class GfG { static boolean pythagoreanTriplet(int[] arr) { int n = arr.length; HashSet<Integer> st = new HashSet<>(); for (int i = 0; i < n; i++) st.add(arr[i]); // Iterate through all possible values of a and b for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int a = arr[i]; int b = arr[j]; // calculate required value for // c to form Pythagorean triplet int c = (int)Math.sqrt(a * a + b * b); // First, verify if c^2 is a perfect square (indicating a // valid c) and check if this c exists in the set. if (c*c == a*a + b*b && st.contains(c)) return true; } } // No Pythagorean triplet exists in the array return false; } public static void main(String[] args) { int[] arr = {3, 1, 4, 6, 5}; System.out.println(pythagoreanTriplet(arr)); } } # Python program to find if a Pythagorean triplet exists # using hashing def pythagoreanTriplet(arr): n = len(arr) st = set() for i in range(n): st.add(arr[i]) # Iterate through all possible values of a and b for i in range(n - 1): for j in range(i + 1, n): a = arr[i] b = arr[j] # calculate required value for # c to form Pythagorean triplet c = (int)((a ** 2 + b ** 2) ** 0.5) # First, verify if c^2 is a perfect square (indicating a # valid c) and check if this c exists in the set. if (c*c == a*a + b*b) and (c in st): return True # No Pythagorean triplet exists in the array return False if __name__ == "__main__": arr = [3, 1, 4, 6, 5] print("true" if pythagoreanTriplet(arr) else "false") // C# program to find if a Pythagorean triplet exists // using hashing using System; using System.Collections.Generic; class GfG { static bool pythagoreanTriplet(int[] arr) { HashSet<int> st = new HashSet<int>(); int n = arr.Length; for (int i = 0; i < n; i++) st.Add(arr[i]); // Iterate through all possible values of a and b for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int a = arr[i]; int b = arr[j]; // calculate required value for // c to form Pythagorean triplet int c = (int)Math.Sqrt(a * a + b * b); // First, verify if c^2 is a perfect square (indicating a // valid c) and check if this c exists in the set. if (c*c == a*a + b*b && st.Contains(c)) return true; } } // No Pythagorean triplet exists in the array return false; } static void Main() { int[] arr = {3, 1, 4, 6, 5}; Console.WriteLine(pythagoreanTriplet(arr)?"true":"false"); } } // JavaScript program to find if a Pythagorean triplet exists // using hashing function pythagoreanTriplet(arr) { const st = new Set(); const n = arr.length; for (let i = 0; i < n; i++) st.add(arr[i]); // Iterate through all possible values of a and b for (let i = 0; i < n - 1; i++) { for (let j = i + 1; j < n; j++) { const a = arr[i] const b = arr[j] // calculate required value for // c to form Pythagorean triplet const c = Math.sqrt(a * a + b * b); // First, verify if c^2 is a perfect square (indicating a // valid c) and check if this c exists in the set. if (c*c === a*a + b*b && st.has(c)) return true; } } // No Pythagorean triplet exists in the array return false; } const arr = [3, 1, 4, 6, 5]; console.log(pythagoreanTriplet(arr) ? "true" : "false"); [Expected Approach] Iterating till max element - O(max(arr)^2) Time and O(max(arr)) Space
The idea is to generate all possible pairs (a, b) within the range of maximum element in the input array using nested loops. For each pair, calculate the value of c required to form a Pythagorean Triplet and check if c exists in the input array. We can check if c exists or not in constant time by marking all the elements of input array in a visited array.
Here, we only need to mark the elements and not store their frequency because for all valid triplets of positive integers (a, b, c), no two elements can be equal, that is a != b, a != c and b != c.
Note: This approach is more efficient than the previous ones, given the constraint (max(arr) < n).
C++ // C++ program to find if a Pythagorean triplet // exists by iterating till max element #include <iostream> #include <vector> #include <cmath> using namespace std; bool pythagoreanTriplet(vector<int> &arr) { int n = arr.size(); int maxEle = 0; for (int ele: arr) maxEle = max(maxEle, ele); // Visited array to mark the elements vector<bool> vis(maxEle + 1, 0); // Marking each element of input array for (int ele: arr) vis[ele] = true; // Iterate for all possible a for (int a = 1; a < maxEle + 1; a++) { // If a is not there if (vis[a] == false) continue; // Iterate for all possible b for (int b = 1; b < maxEle + 1; b++) { // If b is not there if (vis[b] == false) continue; // calculate c value to form a pythagorean triplet int c = sqrt(a * a + b * b); // If c^2 is not a perfect square or c exceeds the // maximum value if ((c * c) != (a * a + b * b) || c > maxEle) continue; // If there exists c in the original array, // we have found the triplet if (vis[c] == true) { return true; } } } return false; } int main() { vector<int> arr = {3, 1, 4, 6, 5}; cout << (pythagoreanTriplet(arr) ? "true" : "false"); return 0; } // Java program to find if a Pythagorean triplet // exists by iterating till max element import java.util.Arrays; class GfG { static boolean pythagoreanTriplet(int[] arr) { int n = arr.length; int maxEle = 0; for (int ele : arr) maxEle = Math.max(maxEle, ele); // Visited array to mark the elements boolean[] vis = new boolean[maxEle + 1]; // Marking each element of input array for (int ele : arr) vis[ele] = true; // Iterate for all possible a for (int a = 1; a < maxEle + 1; a++) { // If a is not there if (!vis[a]) continue; // Iterate for all possible b for (int b = 1; b < maxEle + 1; b++) { // If b is not there if (!vis[b]) continue; // calculate c value to form a pythagorean triplet int c = (int) Math.sqrt(a * a + b * b); // If c^2 is not a perfect square or c exceeds the // maximum value if ((c * c) != (a * a + b * b) || c > maxEle) continue; // If there exists c in the original array, // we have found the triplet if (vis[c]) { return true; } } } return false; } public static void main(String[] args) { int[] arr = {3, 1, 4, 6, 5}; System.out.println(pythagoreanTriplet(arr)); } } # Python program to find if a Pythagorean triplet # exists by iterating till max element import math def pythagoreanTriplet(arr): n = len(arr) maxEle = 0 for ele in arr: maxEle = max(maxEle, ele) # Visited array to mark the elements vis = [False] * (maxEle + 1) # Marking each element of input array for ele in arr: vis[ele] = True # Iterate for all possible a for a in range(1, maxEle + 1): # If a is not there if not vis[a]: continue # Iterate for all possible b for b in range(1, maxEle + 1): # If b is not there if not vis[b]: continue # calculate c value to form a pythagorean triplet c = int(math.sqrt(a * a + b * b)) # If c^2 is not a perfect square or c exceeds the # maximum value if (c * c) != (a * a + b * b) or c > maxEle: continue # If there exists c in the original array, # we have found the triplet if vis[c]: return True return False if __name__ == "__main__": arr = [3, 1, 4, 6, 5] ans = pythagoreanTriplet(arr) if ans: print("true") else : print("false") // C# program to find if a Pythagorean triplet // exists by iterating till max element using System; using System.Collections.Generic; class GfG { static bool pythagoreanTriplet(int[] arr) { int n = arr.Length; int maxEle = 0; foreach (int ele in arr) maxEle = Math.Max(maxEle, ele); // Visited array to mark the elements bool[] vis = new bool[maxEle + 1]; // Marking each element of input array foreach (int ele in arr) vis[ele] = true; // Iterate for all possible a for (int a = 1; a < maxEle + 1; a++) { // If a is not there if (vis[a] == false) continue; // Iterate for all possible b for (int b = 1; b < maxEle + 1; b++) { // If b is not there if (vis[b] == false) continue; // calculate c value to form a pythagorean triplet int c = (int)Math.Sqrt(a * a + b * b); // If c^2 is not a perfect square or c exceeds the // maximum value if ((c * c) != (a * a + b * b) || c > maxEle) continue; // If there exists c in the original array, // we have found the triplet if (vis[c] == true) return true; } } return false; } static void Main() { int[] arr = { 3, 1, 4, 6, 5 }; bool ans = pythagoreanTriplet(arr); if (ans) Console.WriteLine("true"); else Console.WriteLine("false"); } } // JavaScript program to find if a Pythagorean triplet // exists by iterating till max element function pythagoreanTriplet(arr) { let n = arr.length; let maxEle = 0; for (let ele of arr) maxEle = Math.max(maxEle, ele); // Visited array to mark the elements let vis = new Array(maxEle + 1).fill(false); // Marking each element of input array for (let ele of arr) vis[ele] = true; // Iterate for all possible a for (let a = 1; a < maxEle + 1; a++) { // If a is not there if (vis[a] === false) continue; // Iterate for all possible b for (let b = 1; b < maxEle + 1; b++) { // If b is not there if (vis[b] === false) continue; // calculate c value to form a pythagorean triplet let c = Math.floor(Math.sqrt(a * a + b * b)); // If c^2 is not a perfect square or c exceeds the // maximum value if ((c * c) !== (a * a + b * b) || c > maxEle) continue; // If there exists c in the original array, // we have found the triplet if (vis[c] === true) { return true; } } } return false; } const arr = [3, 1, 4, 6, 5]; console.log(pythagoreanTriplet(arr));
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