# Possible to form a triangle from array values

• Difficulty Level : Easy
• Last Updated : 07 Jul, 2022

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
Recommended Practice

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)

Implementation:

## C++

 // C++ program to find if it is possible to form a triangle// from array values#include using namespace std; // Method prints possible triangle when array values are// taken as sidesbool 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;    return false;} // Driver Codeint main(){    int arr[] = { 5, 4, 3, 1, 2 };    int N = sizeof(arr) / sizeof(int);     isPossibleTriangle(arr, N) ? cout << "Yes" : cout << "No";    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)

## C

 // C program to find if it is possible to form a triangle// from array values#include #include #include  int cmpfunc(const void* a, const void* b){    return (*(int*)a - *(int*)b);} // Method prints possible triangle when array values are// taken as sidesbool 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    qsort(arr, N, sizeof(int), cmpfunc);    // 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;    return false;} // Driver Codeint main(){    int arr[] = { 5, 4, 3, 1, 2 };    int N = sizeof(arr) / sizeof(int);     isPossibleTriangle(arr, N) ? printf("Yes") : printf("No");    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)

## JAVA

 // Java program to find if it is  possible to form a// triangle  from array valuesimport java.io.*;import java.util.Arrays; class GFG {     // Method prints possible triangle when array values are    // taken as sides    static boolean 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        Arrays.sort(arr);        // 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;        return false;    }     // Driver Code    static public void main(String[] args)    {        int[] arr = { 5, 4, 3, 1, 2 };        int N = arr.length;        if (isPossibleTriangle(arr, N))            System.out.println("Yes");        else            System.out.println("No");    }} // This code is contributed by Aditya Kumar (adityakumar129)

## Python

 # 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 sidesdef 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 Codearr = [5, 4, 3, 1, 2]N = len(arr)print("Yes" if isPossibleTriangle(arr, N) else "No") # This code is contributed# by "Sharad_Bhardwaj".

## C#

 // C# program to find if// it is possible to form// a triangle from array valuesusing System; class GFG{             // Method prints possible    // triangle when array values    // are taken as sides    static 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        Array.Sort(arr);             // 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;                         return false;    }         // Driver Code    static public void Main ()    {        int []arr = {5, 4, 3, 1, 2};        int N = arr.Length;                 if(isPossibleTriangle(arr, N))            Console.WriteLine("Yes" );        else            Console.WriteLine("No");    }} // This code is contributed by vt_m.

## PHP

 \$arr[\$i + 2])            return true;} // Driver Code\$arr = array(5, 4, 3, 1, 2);\$N = count(\$arr); if(isPossibleTriangle(\$arr,\$N))echo "Yes" ;else               echo "No"; // This code is contributed by vt_m?>

## Javascript



Output

Yes

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