Compruebe si es posible crear un polígono con n lados dados

Dada una array arr[] que contiene las longitudes de n lados que pueden o no formar un polígono. La tarea es determinar si es posible formar un polígono con todos los lados dados. Escriba Sí si es posible, de lo contrario, escriba No.
Ejemplos: 
 

Entrada: arr[] = {2, 3, 4} 
Salida: Sí
Entrada: arr[] = {3, 4, 9, 2} 
Salida: No 
 

Enfoque: Para crear un polígono con n lados dados, hay una cierta propiedad que deben cumplir los lados del polígono. 
 

Propiedad: La longitud de cada lado dado debe ser menor que la suma de los otros lados restantes.

 
Encuentra el lado más grande entre los lados dados. Luego, verifica si es más pequeño que la suma de los otros lados o no. Si es más pequeño, escriba Sí , de lo contrario, escriba No.
A continuación se muestra la implementación del enfoque anterior:
 

// 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 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 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# 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 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";
?>

<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>

Yes

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)