Possible to form a triangle from array values
Last Updated :
19 Sep, 2023
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++
#include <bits/stdc++.h>
using namespace std;
bool isPossibleTriangle(int arr[], int N)
{
if (N < 3)
return false;
sort(arr, arr + N);
for (int i = 0; i < N - 2; i++)
if (arr[i] + arr[i + 1] > arr[i + 2])
return true;
return false;
}
int main()
{
int arr[] = { 5, 4, 3, 1, 2 };
int N = sizeof(arr) / sizeof(int);
isPossibleTriangle(arr, N) ? cout << "Yes" : cout << "No";
return 0;
}
|
C
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
int cmpfunc(const void* a, const void* b)
{
return (*(int*)a - *(int*)b);
}
bool isPossibleTriangle(int arr[], int N)
{
if (N < 3)
return false;
qsort(arr, N, sizeof(int), cmpfunc);
for (int i = 0; i < N - 2; i++)
if (arr[i] + arr[i + 1] > arr[i + 2])
return true;
return false;
}
int main()
{
int arr[] = { 5, 4, 3, 1, 2 };
int N = sizeof(arr) / sizeof(int);
isPossibleTriangle(arr, N) ? printf("Yes") : printf("No");
return 0;
}
|
JAVA
import java.io.*;
import java.util.Arrays;
class GFG {
static boolean isPossibleTriangle(int[] arr, int N)
{
if (N < 3)
return false;
Arrays.sort(arr);
for (int i = 0; i < N - 2; i++)
if (arr[i] + arr[i + 1] > arr[i + 2])
return true;
return false;
}
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");
}
}
|
Python
def isPossibleTriangle (arr , N):
if N < 3:
return False
arr.sort()
for i in range(N - 2):
if arr[i] + arr[i + 1] > arr[i + 2]:
return True
arr = [5, 4, 3, 1, 2]
N = len(arr)
print("Yes" if isPossibleTriangle(arr, N) else "No")
|
C#
using System;
class GFG
{
static bool isPossibleTriangle(int []arr,
int N)
{
if (N < 3)
return false;
Array.Sort(arr);
for (int i = 0; i < N - 2; i++)
if (arr[i] + arr[i + 1] > arr[i + 2])
return true;
return false;
}
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");
}
}
|
PHP
<?php
function isPossibleTriangle( $arr, $N)
{
if ($N < 3)
return false;
sort($arr);
for ( $i = 0; $i < $N - 2; $i++)
if ($arr[$i] + $arr[$i + 1] > $arr[$i + 2])
return true;
}
$arr = array(5, 4, 3, 1, 2);
$N = count($arr);
if(isPossibleTriangle($arr,$N))
echo "Yes" ;
else
echo "No";
?>
|
Javascript
<script>
function isPossibleTriangle(arr, N)
{
if (N < 3)
return false;
arr.sort();
for (let i = 0; i < N - 2; i++)
if (arr[i] + arr[i + 1] > arr[i + 2])
return true;
return false;
}
let arr = [5, 4, 3, 1, 2];
let N = arr.length;
if(isPossibleTriangle(arr, N))
document.write("Yes" );
else
document.write("No");
</script>
|
Time Complexity : O(NlogN), here N is size of array.
Auxiliary Space : O(1),since no extra space is used.
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