Find the Missing Number

Software Development
1

You are given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in the list. One of the integers is missing in the list. Write an efficient code to find the missing integer.
Example:

Input: arr[] = {1, 2, 4, 6, 3, 7, 8} Output: 5 Explanation: The missing number from 1 to 8 is 5 Input: arr[] = {1, 2, 3, 5} Output: 4 Explanation: The missing number from 1 to 5 is 4.

Method 1: This method uses the technique of the summation formula.

  • Approach: The length of the array is n-1. So the sum of all n elements, i.e sum of numbers from 1 to n can be calculated using the formula n(n+1)/2*. Now find the sum of all the elements in the array and subtract it from the sum of first n natural numbers, it will be the value of the missing element.
  • Algorithm:
    1. Calculate the sum of first n natural numbers as sumtotal= n(n+1)/2*
    2. Create a variable sum to store the sum of array elements.
    3. Traverse the array from start to end.
    4. Update the value of sum as sum = sum + array[i]
    5. Print the missing number as sumtotal – sum
  • Implementation:

#include <bits/stdc++.h>

using namespace std;

// Function to get the missing number

int getMissingNo( int a[], int n)

{

int total = (n + 1) * (n + 2) / 2;

for ( int i = 0; i < n; i++)

total -= a[i];

return total;

}

// Driver Code

int main()

{

int arr[] = { 1, 2, 4, 5, 6 };

int n = sizeof (arr) / sizeof (arr[0]);

int miss = getMissingNo(arr, n);

cout << miss;

}

Output

3

  • Complexity Analysis:
    • Time Complexity: O(n).
      Only one traversal of the array is needed.
    • Space Complexity: O(1).
      No extra space is needed

Modification for Overflow

  • Approach: The approach remains the same but there can be overflow if n is large. In order to avoid integer overflow, pick one number from known numbers and subtract one number from given numbers. This way there won’t have Integer Overflow ever.
  • Algorithm:
    1. Create a variable sum = 1 which will store the missing number and a counter variable c = 2.
    2. Traverse the array from start to end.
    3. Update the value of sum as sum = sum – array[i] + c and update c as c++.
    4. Print the missing number as a sum.

#include <bits/stdc++.h>

using namespace std;

// a represents the array

// n : Number of elements in array a

int getMissingNo( int a[], int n)

{

int i, total=1;

for ( i = 2; i<= (n+1); i++)

{

total+=i;

total -= a[i-2];

}

return total;

}

//Driver Program

int main() {

int arr[] = {1, 2, 3, 5};

cout<<getMissingNo(arr, sizeof (arr)/ sizeof (arr[0]));

return 0;

}

Output:

4

  • Complexity Analysis:
    • Time Complexity: O(n).
      Only one traversal of the array is needed.
    • **Space Complexity:**O(1).
      No extra space is needed

Method 2: This method uses the technique of XOR to solve the problem.

  • Approach:
    XOR has certain properties
    • Assume a1 ^ a2 ^ a3 ^ …^ an = a and a1 ^ a2 ^ a3 ^ …^ an-1 = b
    • Then a ^ b = an
  • Algorithm:
    1. Create two variables a = 0 and b = 0
    2. Run a loop from 1 to n with i as counter.
    3. For every index update a as a = a ^ i
    4. Now traverse the array from start to end.
    5. For every index update b as b = b ^ array[i]
    6. Print the missing number as a ^ b.
  • Implementation:

#include <bits/stdc++.h>

using namespace std;

// Function to get the missing number

int getMissingNo( int a[], int n)

{

// For xor of all the elements in array

int x1 = a[0];

// For xor of all the elements from 1 to n+1

int x2 = 1;

for ( int i = 1; i < n; i++)

x1 = x1 ^ a[i];

for ( int i = 2; i <= n + 1; i++)

x2 = x2 ^ i;

return (x1 ^ x2);

}

// Driver Code

int main()

{

int arr[] = { 1, 2, 4, 5, 6 };

int n = sizeof (arr) / sizeof (arr[0]);

int miss = getMissingNo(arr, n);

cout << miss;

}

Output:

3

  • Complexity Analysis:
    • Time Complexity: O(n).
      Only one traversal of the array is needed.
    • Space Complexity: O(1).
      No extra space is needed.

Method 3: This method will work only in python.
Approach:
Take the sum of all elements in the array and subtract that from the sum of n+1 elements. For Eg:
If my elements are li=[1,2,4,5] then take the sum of all elements in li and subtract that from the sum of len(li)+1 elements. The following code shows the demonstration.

// C++ program to find

// the mising Number

#include<bits/stdc++.h>

using namespace std;

// getMissingNo takes list

// as argument

int getMissingNo( int a[] ,

int n)

{

int n_elements_sum = n * (n + 1) / 2 ;

int sum = 0;

for ( int i = 0; i < n - 1; i++)

sum += a[i];

return n_elements_sum-sum;

}

// Driver code

int main()

{

int a[] = {1, 2, 4, 5, 6};

int n = sizeof (a) /

sizeof (a[0]) + 1;

int miss = getMissingNo(a, n);

cout << (miss);

return 0;

}

Output:

3

Time Complexity: O(n)

Auxiliary Space: O(1)