Possible to form a triangle from array values

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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). And auxiliary space is O(1).

Implementation:

// 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;
    return false;
}

// Driver Code
int 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 program to find if it is possible to form a triangle
// from array values
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>

int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}

// 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
    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 Code
int 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 program to find if it is  possible to form a
// triangle  from array values
import 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)

# 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
arr = [5, 4, 3, 1, 2]
N = len(arr) 
print("Yes" if isPossibleTriangle(arr, N) else "No")

# This code is contributed 
# by "Sharad_Bhardwaj".

// C# program to find if 
// it is possible to form
// a triangle from array values
using 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
// PHP program to find if 
// it is possible to form
// a triangle from array values


// Method prints possible 
// triangle when array values
// are taken as sides
function 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
    sort($arr);

    // loop for all 3 
    // consecutive triplets
    for ( $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
$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
?>

<script>
// Javascript program to find if it is 
// possible to form a triangle 
// from array values

    // Method prints possible 
    // triangle when array values
    // are taken as sides
    function 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 (let 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    

        let arr = [5, 4, 3, 1, 2];
        let N = arr.length;
          
        if(isPossibleTriangle(arr, N))
            document.write("Yes" );
        else
            document.write("No");
    
    // This code is contributed by susmitakundugoaldanga.
</script>

Yes

Time Complexity : O(NlogN), here N is size of array.

Auxiliary Space : O(1),since no extra space is used.