Given an array of integers, we need to find out whether it is possible to construct at least one non-degenerate triangle using array values as its sides. In other words, we need to find out 3 such array indices which can become sides of a non-degenerate triangle.

Examples:

Input : [4, 1, 2] Output : No No triangle is possible from given array values Input : [5, 4, 3, 1, 2] Output : Yes Sides of possible triangle are 2 3 4

For a non-degenerate triangle, its sides should follow these constraints,

A + B > C and B + C > A and C + A > B where A, B and C are length of sides of the triangle.

The task is to find any triplet from array that satisfies above condition.

A **Simple Solution** is to generate all triplets and for every triplet check if it forms a triangle or not by checking above three conditions.

An **Efficient Solution **is use sorting. First, we sort the array then we loop once and we will check three consecutive elements of this array if any triplet satisfies arr[i] + arr[i+1] > arr[i+2], then we will output that triplet as our final result.

**Why checking only 3 consecutive elements will work instead of trying all possible triplets of sorted array?**

Let we are at index i and 3 line segments are arr[i], arr[i + 1] and arr[i + 2] with relation arr[i] < arr[i+1] < arr[i+2], If they can't form a non-degenerate triangle, Line segments of lengths arr[i-1], arr[i+1] and arr[i+2] or arr[i], arr[i+1] and arr[i+3] can't form a non-degenerate triangle also because sum of arr[i-1] and arr[i+1] will be even less than sum of arr[i] and arr[i+1] in first case and sum of arr[i] and arr[i+1] must be less than arr[i+3] in second case, So we don't need to try all the combinations, we will try just 3 consecutive indices of array in sorted form.

The total complexity of below solution is O(n log n)

## C++

// C++ program to find if it is possible to form // a triangle from array values #include <bits/stdc++.h> using namespace std; // Method prints possible triangle when array // values are taken as sides bool isPossibleTriangle(int arr[], int N) { // If number of elements are less than 3, // then no triangle is possible if (N < 3) return false; // first sort the array sort(arr, arr + N); // loop for all 3 consecutive triplets for (int i=0; i<N-2; i++) // If triplet satisfies triangle // condition, break if (arr[i] + arr[i+1] > arr[i+2]) return true; } // Driver code to test above method int main() { int arr[] = {5, 4, 3, 1, 2}; int N = sizeof(arr) / sizeof(int); isPossibleTriangle(arr, N)? cout << "Yes" : cout << "No"; return 0; }

## Python3

# Python3 code to find if it is possible # to form a triangle from array values # Method prints possible triangle when # array values are taken as sides def isPossibleTriangle (arr , N): # If number of elements are less # than 3, then no triangle is possible if N < 3: return False # first sort the array arr.sort() # loop for all 3 consecutive triplets for i in range(N - 2): # If triplet satisfies triangle # condition, break if arr[i] + arr[i+1] > arr[i+2]: return True # Driver code to test above method arr = [5, 4, 3, 1, 2] N = len(arr) print("Yes" if isPossibleTriangle(arr, N) else "No") # This code is contributed by "Sharad_Bhardwaj".

Output:

Yes

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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