# 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 : 2Input : 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# ```// 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
`6`

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

### Find the only missing number in a sorted array Using Direct Formula approach

In this approach we will create Function to find the missing number using the sum of natural numbers formula. First we will Calculate the total sum of the first N natural numbers using formula n * (n + 1) / 2. Now we calculate sum of all elements in given array. Subtract the total sum with sum of all elements in given array and return the missing number.

Below is the implementation of above approach:

C++ ```#include <iostream> #include <vector> using namespace std; int findMissingNumber(const vector<int>& nums) { // Calculate the total sum int n = nums.size() + 1; int totalSum = n * (n + 1) / 2; // Calculate sum of all elements in the given array int arraySum = 0; for (int num : nums) { arraySum += num; } // Subtract and return the total sum with the sum of // all elements in the array int missingNumber = totalSum - arraySum; return missingNumber; } int main() { vector<int> numbers = { 1, 2, 3, 4, 5, 7, 8 }; int missing = findMissingNumber(numbers); cout << "The only missing number is: " << missing << endl; return 0; } ``` Java ```import java.util.ArrayList; import java.util.List; public class Main { public static int findMissingNumber(List<Integer> nums) { // Calculate the total sum int n = nums.size() + 1; int totalSum = n * (n + 1) / 2; // Calculate sum of all elements in the given array int arraySum = 0; for (int num : nums) { arraySum += num; } // Subtract and return the total sum with the sum of // all elements in the array int missingNumber = totalSum - arraySum; return missingNumber; } public static void main(String[] args) { List<Integer> numbers = new ArrayList<>(); numbers.add(1); numbers.add(2); numbers.add(3); numbers.add(4); numbers.add(5); numbers.add(7); numbers.add(8); int missing = findMissingNumber(numbers); System.out.println("The only missing number is: " + missing); } } ``` Python ```def find_missing_number(nums): # Calculate the total sum n = len(nums) + 1 total_sum = n * (n + 1) // 2 # Calculate sum of all elements in the given array array_sum = sum(nums) # Subtract and return the total sum with the sum of # all elements in the array missing_number = total_sum - array_sum return missing_number if __name__ == "__main__": numbers = [1, 2, 3, 4, 5, 7, 8] missing = find_missing_number(numbers) print("The only missing number is:", missing) # This code is contributed by Ayush Mishra ``` JavaScript ```function findMissingNumber(nums) { // Calculate the total sum const n = nums.length + 1; const totalSum = (n * (n + 1)) / 2; // Calculate sum of all elements in the given array let arraySum = 0; for (let num of nums) { arraySum += num; } // Subtract and return the total sum with the sum of // all elements in the array const missingNumber = totalSum - arraySum; return missingNumber; } function main() { const numbers = [1, 2, 3, 4, 5, 7, 8]; const missing = findMissingNumber(numbers); console.log("The only missing number is: " + missing); } main(); ```

Output
```The only missing number is: 6
```

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