Open In App

Check if it is possible to create a polygon with given n sides

Last Updated : 17 Feb, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

Given an array arr[] that contains the lengths of n sides that may or may not form a polygon. The task is to determine whether it is possible to form a polygon with all the given sides. Print Yes if possible else print No.
Examples: 
 

Input: arr[] = {2, 3, 4} 
Output: Yes
Input: arr[] = {3, 4, 9, 2} 
Output: No 
 

 

Approach: In order to create a polygon with given n sides, there is a certain property that must be satisfied by the sides of the polygon. 
 

Property: The length of the every given side must be less than the sum of the other remaining sides.

 
Find the largest side among the given sides. Then, check whether it is smaller than the sum of the other sides or not. If it is smaller than print Yes else print No.
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if it is possible
// to form a polygon with the given sides
bool isPossible(int a[], int n)
{
 
    // Sum stores the sum of all the sides
    // and maxS stores the length of
    // the largest side
    int sum = 0, maxS = 0;
    for (int i = 0; i < n; i++) {
        sum += a[i];
        maxS = max(a[i], maxS);
    }
 
    // If the length of the largest side
    // is less than the sum of the
    // other remaining sides
    if ((sum - maxS) > maxS)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int a[] = { 2, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    if (isPossible(a, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}


Java




// Java implementation of the approach
class GFG {
 
    // Function that returns true if it is possible
    // to form a polygon with the given sides
    static boolean isPossible(int a[], int n)
    {
        // Sum stores the sum of all the sides
        // and maxS stores the length of
        // the largest side
        int sum = 0, maxS = 0;
        for (int i = 0; i < n; i++) {
            sum += a[i];
            maxS = Math.max(a[i], maxS);
        }
 
        // If the length of the largest side
        // is less than the sum of the
        // other remaining sides
        if ((sum - maxS) > maxS)
            return true;
 
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 2, 3, 4 };
        int n = a.length;
 
        if (isPossible(a, n))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}


Python




# Python 3 implementation of the approach
 
# Function to check whether
# it is possible to create a
# polygon with given sides length
def isPossible(a, n):
    # Sum stores the sum of all the sides
    # and maxS stores the length of
    # the largest side
    sum = 0
    maxS = 0
    for i in range(n):
        sum += a[i]
        maxS = max(a[i], maxS)
 
    # If the length of the largest side
    # is less than the sum of the
    # other remaining sides
    if ((sum - maxS) > maxS):
        return True
     
    return False
 
# Driver code
a =[2, 3, 4]
n = len(a)
 
if(isPossible(a, n)):
    print("Yes")
else:
    print("No")


C#




// C# implementation of the approach
using System;
class GFG {
 
    // Function that returns true if it is possible
    // to form a polygon with the given sides
    static bool isPossible(int[] a, int n)
    {
        // Sum stores the sum of all the sides
        // and maxS stores the length of
        // the largest side
        int sum = 0, maxS = 0;
        for (int i = 0; i < n; i++) {
            sum += a[i];
            maxS = Math.Max(a[i], maxS);
        }
 
        // If the length of the largest side
        // is less than the sum of the
        // other remaining sides
        if ((sum - maxS) > maxS)
            return true;
 
        return false;
    }
 
    // Driver code
    static void Main()
    {
        int[] a = { 2, 3, 4 };
        int n = a.Length;
 
        if (isPossible(a, n))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}


PHP




<?php
// PHP implementation of the approach
 
// Function that returns true if it is possible
// to form a polygon with the given sides
function isPossible($a, $n)
{
    // Sum stores the sum of all the sides
    // and maxS stores the length of
    // the largest side
    $sum = 0;
    $maxS = 0;
    for ($i = 0; $i < $n; $i++) {
        $sum += $a[$i];
        $maxS = max($a[$i], $maxS);
    }
 
    // If the length of the largest side
    // is less than the sum of the
    // other remaining sides
    if (($sum - $maxS) > $maxS)
        return true;
     
    return false;
}
 
// Driver code
$a = array(2, 3, 4);
$n = count($a);
 
if(isPossible($a, $n))
    echo "Yes";
else
    echo "No";
?>


Javascript




<script>
 
// Javascript implementation of the approach
 
 
// Function that returns true if it is possible
// to form a polygon with the given sides
function isPossible( a, n)
{
 
    // Sum stores the sum of all the sides
    // and maxS stores the length of
    // the largest side
    let sum = 0, maxS = 0;
    for (let i = 0; i < n; i++) {
        sum += a[i];
        maxS = Math.max(a[i], maxS);
    }
 
    // If the length of the largest side
    // is less than the sum of the
    // other remaining sides
    if ((sum - maxS) > maxS)
        return true;
 
    return false;
}
 
    // Driver Code
     
    let a = [ 2, 3, 4 ];
    let n = a.length;
 
    if (isPossible(a, n))
        document.write("Yes");
    else
        document.write("No");
     
     
</script>


Output: 

Yes

 

Time Complexity: O(n)

Auxiliary Space: O(1)



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads