Encuentre el intervalo más grande que contenga exactamente uno de los N enteros dados.

Dada una array arr[] de N enteros distintos, la tarea es encontrar el elemento máximo en un intervalo [L, R] tal que el intervalo contenga exactamente uno de los N enteros dados y 1 ≤ L ≤ R ≤ 10 ⁵
 

Entrada: arr[] = {5, 10, 200} 
Salida: 99990 
Todos los intervalos posibles son [1, 9], [6, 199] y [11, 100000]. 
[11, 100000] tiene los números enteros máximos, es decir, 99990.
Entrada: arr[] = {15000, 25000, 40000, 70000, 80000} 
Salida: 44999 
 

Enfoque: La idea es fijar el elemento que queremos que contenga nuestro intervalo. Ahora estamos interesados ​​en cuánto podemos extender nuestro intervalo hacia la izquierda y hacia la derecha sin superponernos con otros elementos. 
Entonces, primero ordenamos nuestra array. Luego, para un elemento fijo, determinamos sus extremos utilizando los elementos anterior y siguiente. También debemos cuidar los casos de esquina que son cuando arreglamos nuestro primer y último intervalo. De esta manera para cada elemento i, encontramos la longitud máxima del intervalo.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum
// size of the required interval
int maxSize(vector<int>& v, int n)
{
    // Insert the borders for array
    v.push_back(0);
    v.push_back(100001);
    n += 2;
 
    // Sort the elements in ascending order
    sort(v.begin(), v.end());
 
    // To store the maximum size
    int mx = 0;
    for (int i = 1; i < n - 1; i++) {
 
        // To store the range [L, R] such that
        // only v[i] lies within the range
        int L = v[i - 1] + 1;
        int R = v[i + 1] - 1;
 
        // Total integers in the range
        int cnt = R - L + 1;
        mx = max(mx, cnt);
    }
 
    return mx;
}
 
// Driver code
int main()
{
    vector<int> v = { 200, 10, 5 };
    int n = v.size();
 
    cout << maxSize(v, n);
 
    return 0;
}

// Java implementation of the approach
import static java.lang.Integer.max;
import java.util.*;
 
class GFG
{
 
    // Function to return the maximum
    // size of the required interval
    static int maxSize(Vector<Integer> v, int n)
    {
        // Insert the borders for array
        v.add(0);
        v.add(100001);
        n += 2;
 
        // Sort the elements in ascending order
        Collections.sort(v);
 
        // To store the maximum size
        int mx = 0;
        for (int i = 1; i < n - 1; i++)
        {
 
            // To store the range [L, R] such that
            // only v[i] lies within the range
            int L = v.get(i - 1) + 1;
            int R = v.get(i + 1) - 1;
 
            // Total integers in the range
            int cnt = R - L + 1;
            mx = max(mx, cnt);
        }
 
        return mx;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Integer arr[] = {200, 10, 5};
        Vector v = new Vector(Arrays.asList(arr));
        int n = v.size();
 
        System.out.println(maxSize(v, n));
    }
}
 
// This code is contributed by Princi Singh

# Python3 implementation of the approach
 
# Function to return the maximum
# size of the required interval
def maxSize(v, n):
 
    # Insert the borders for array
    v.append(0)
    v.append(100001)
    n += 2
 
    # Sort the elements in ascending order
    v = sorted(v)
 
    # To store the maximum size
    mx = 0
    for i in range(1, n - 1):
 
        # To store the range [L, R] such that
        # only v[i] lies within the range
        L = v[i - 1] + 1
        R = v[i + 1] - 1
 
        # Total integers in the range
        cnt = R - L + 1
        mx = max(mx, cnt)
     
 
    return mx
 
 
# Driver code
v = [ 200, 10, 5]
n = len(v)
 
print(maxSize(v, n))
 
# This code is contributed by mohit kumar 29

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to return the maximum
    // size of the required interval
    static int maxSize(List<int> v, int n)
    {
        // Insert the borders for array
        v.Add(0);
        v.Add(100001);
        n += 2;
 
        // Sort the elements in ascending order
        v.Sort();
 
        // To store the maximum size
        int mx = 0;
        for (int i = 1; i < n - 1; i++)
        {
 
            // To store the range [L, R] such that
            // only v[i] lies within the range
            int L = v[i - 1] + 1;
            int R = v[i + 1] - 1;
 
            // Total integers in the range
            int cnt = R - L + 1;
            mx = Math.Max(mx, cnt);
        }
 
        return mx;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {200, 10, 5};
        List<int> v = new List<int>(arr);
        int n = v.Count;
 
        Console.WriteLine(maxSize(v, n));
    }
}
 
// This code is contributed by 29AjayKumar

<script>
 
// Javascript implementation of the approach
 
// Function to return the maximum
// size of the required interval
function maxSize(v, n)
{
    // Insert the borders for array
    v.push(0);
    v.push(100001);
    n += 2;
 
    // Sort the elements in ascending order
    v.sort((a, b) => a - b);
 
    // To store the maximum size
    let mx = 0;
    for (let i = 1; i < n - 1; i++) {
 
        // To store the range [L, R] such that
        // only v[i] lies within the range
        let L = v[i - 1] + 1;
        let R = v[i + 1] - 1;
 
        // Total integers in the range
        let cnt = R - L + 1;
        mx = Math.max(mx, cnt);
    }
 
    return mx;
}
 
// Driver code
    let v = [ 200, 10, 5 ];
    let n = v.length;
 
    document.write(maxSize(v, n));
 
</script>

99990

Complejidad de tiempo: O(nlog(n))
Espacio auxiliar: O(1)