Find the only missing number in a sorted array

Last Updated : 12 May, 2024

You are given a sorted array of N integers from 1 to N with one number missing find the missing number

Examples:

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

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

A simple solution is to linearly traverse the given array. Find the point where current element is not one more than previous.

An efficient solution is to use binary search. We use the index to search for the missing element and modified binary search. If element at mid != index+1 and this is first missing element then mid + 1 is the missing element. Else if this is not first missing element but ar[mid] != mid+1 search in left half. Else search in right half and if left>right then no element is missing.

C++

// 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

// 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

# 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)

JavaScript

<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>

PHP

<?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 ?>

Output

Time Complexity: O(Log n)
Auxiliary Space: O(1)