Possible to form a triangle from array values

2

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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