Encuentra el único número que falta en una array ordenada

Se le da una array ordenada de N enteros de 1 a N con un número que falta encontrar el número que falta Complejidad de tiempo esperada O (logn)  
Ejemplos: 
 

Input :ar[] = {1, 3, 4, 5}
Output : 2

Input : ar[] = {1, 2, 3, 4, 5, 7, 8}
Output : 6

Una solución simple es atravesar linealmente la array dada. Encuentre el punto donde el elemento actual no es uno más que el anterior.
Una solución eficiente es utilizar la búsqueda binaria . Usamos el índice para buscar el elemento que falta y la búsqueda binaria modificada. Si el elemento en mid != index+1 y este es el primer elemento faltante, mid + 1 es el elemento faltante. De lo contrario, si este no es el primer elemento faltante sino ar[mid] != mid+1 busca en la mitad izquierda. De lo contrario, busque en la mitad derecha y, si izquierda>derecha, no falta ningún elemento. 
 

// CPP program to find the only missing element.
#include <iostream>
using namespace std;
 
int findmissing(int ar[], int N)
{
    int l = 0, r = N - 1;
    while (l <= r) {
 
        int mid = (l + r) / 2;
 
        // If this is the first element
        // which is not index + 1, then
        // missing element is mid+1
        if (ar[mid] != mid + 1 &&
                        ar[mid - 1] == mid)
            return mid + 1;
 
        // if this is not the first missing
        // element search in left side
        if (ar[mid] != mid + 1)
            r = mid - 1;
 
        // if it follows index+1 property then
        // search in right side
        else
            l = mid + 1;
    }
 
    // if no element is missing
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5, 7, 8};
    int N = sizeof(arr)/sizeof(arr[0]);
    cout << findmissing(arr, N);
    return 0;
}

// Java program to find
// the only missing element.
class GFG
{
static int findmissing(int [] ar, int N)
{
     
    int l = 0, r = N - 1;
    while (l <= r)
    {
        int mid = (l + r) / 2;
     
        // If this is the first element
        // which is not index + 1, then
        // missing element is mid+1
        if (ar[mid] != mid + 1 &&
            ar[mid - 1] == mid)
            return (mid + 1);
     
        // if this is not the first
        // missing element search
        // in left side
        if (ar[mid] != mid + 1)
            r = mid - 1;
     
        // if it follows index+1
        // property then search
        // in right side
        else
            l = mid + 1;
    }
 
// if no element is missing
return -1;
}
 
// Driver code
public static void main(String [] args)
{
    int arr[] = {1, 2, 3, 4, 5, 7, 8};
    int N = arr.length;
    System.out.println(findmissing(arr, N));
}
}
 
// This code is contributed
// by Shivi_Aggarwal

# PYTHON 3 program to find
# the only missing element.
def findmissing(ar, N):
    l = 0
    r = N - 1
    while (l <= r):
        mid = (l + r) / 2
        mid= int (mid)
 
    # If this is the first element
    # which is not index + 1, then
    # missing element is mid+1
        if(ar[mid] != mid + 1 and
           ar[mid - 1] == mid):
            return (mid + 1)
 
    # if this is not the first
    # missing element search
    # in left side
        elif(ar[mid] != mid + 1):
            r = mid - 1
 
    # if it follows index+1
    # property then search
    # in right side
        else:
            l = mid + 1
     
    # if no element is missing
    return (-1)
     
def main():
    ar= [1, 2, 3, 4, 5, 7, 8]
    N = len(ar)
    res= findmissing(ar, N)
    print (res)
if __name__ == "__main__":
    main()
 
# This code is contributed
# by Shivi_Aggarwal

// C# program to find
// the only missing element.
using System;
 
class GFG
{
static int findmissing(int []ar,
                       int N)
{
     
    int l = 0, r = N - 1;
    while (l <= r)
    {
        int mid = (l + r) / 2;
     
        // If this is the first element
        // which is not index + 1, then
        // missing element is mid+1
        if (ar[mid] != mid + 1 &&
            ar[mid - 1] == mid)
            return (mid + 1);
     
        // if this is not the first
        // missing element search
        // in left side
        if (ar[mid] != mid + 1)
            r = mid - 1;
     
        // if it follows index+1
        // property then search
        // in right side
        else
            l = mid + 1;
    }
 
// if no element is missing
return -1;
}
 
// Driver code
public static void Main()
{
    int []arr = {1, 2, 3, 4, 5, 7, 8};
    int N = arr.Length;
    Console.WriteLine(findmissing(arr, N));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

<?php
// PHP program to find
// the only missing element.
function findmissing(&$ar, $N)
{
    $r = $N - 1;
    $l = 0;
    while ($l <= $r)
    {
        $mid = ($l + $r) / 2;
 
        // If this is the first element
        // which is not index + 1, then
        // missing element is mid+1
        if ($ar[$mid] != $mid + 1 &&
            $ar[$mid - 1] == $mid)
            return ($mid + 1);
 
        // if this is not the first
        // missing element search
        // in left side
        if ($ar[$mid] != $mid + 1)
            $r = $mid - 1;
 
        // if it follows index+1
        // property then search
        // in right side
        else
            $l = $mid + 1;
    }
 
    // if no element is missing
    return (-1);
}
 
// Driver Code
$ar = array(1, 2, 3, 4, 5, 7, 8);
$N = sizeof($ar);
echo(findmissing($ar, $N));
 
// This code is contributed
// by Shivi_Aggarwal
?>

<script>
 
// JavaScript program to find the only missing element.
 
      function findmissing(ar, N) {
        var l = 0,
          r = N - 1;
        while (l <= r) {
          var mid = parseInt((l + r) / 2);
 
          // If this is the first element
          // which is not index + 1, then
          // missing element is mid+1
          if (ar[mid] != mid + 1 && ar[mid - 1] == mid)
          return mid + 1;
 
          // if this is not the first missing
          // element search in left side
          if (ar[mid] != mid + 1) r = mid - 1;
          // if it follows index+1 property then
          // search in right side
          else l = mid + 1;
        }
 
        // if no element is missing
        return -1;
      }
 
      // Driver code
      var arr = [1, 2, 3, 4, 5, 7, 8];
      var N = arr.length;
      document.write(findmissing(arr, N));
       
</script>

6

Complejidad de Tiempo: O(Log n) 
Espacio Auxiliar: O(1)