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 : 6A 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 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)
<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 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
?>
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;
}
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);
}
}
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
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();
OutputThe only missing number is: 6
Time Complexity: O(1)
Auxiliary Space: O(1)
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