How to find the missing number in an array in python?

Problem Definition

Find the missing numbers in a given list or array using Python.

For example in the arr = [1,2,4,5] the integer '3' is the missing number.

There are multiple ways to solve this problem using Python. In this article, we will cover the most straightforward ones.

Algorithm

Step 1: Create an empty array for missing items

Step 2: Loop over the elements within the range of the first and last element of the array

Step 3: Compare the loop variable with the given array if the value is not present append it to the missing array

Note: The array must be sorted for this to work. Use arr.sort[] on an unsorted array before feeding it to the program.

Solution 1

arr = [1,2,3,4,5,6,7,9,10]
missing_elements = []
for ele in range[arr[0], arr[-1]+1]:
    if ele not in arr:
        missing_elements.append[ele]
print[missing_elements]

Output:

2. Using List Comprehension

arr = [1,2,3,4,5,7,6,9,10]
missing_elemnts = [item for item in range[arr[0], arr[-1]+1] if item not in arr]
print[missing_elemnts]

Output:

[8]

Using list comprehension we encapsulated the above solution in a single line.

3. Using Set[]

Set[] is a Python unordered mutable datatype that holds only unique values.

arr = [1,2,3,4,5,7,6,9,10]
missing_value = set[range[arr[0], arr[-1]+1]] - set[arr]
print[missing_value]

Output:

{8}

Here we created a set object of having values within the range of initial and final values of the provided array then compared it with the provided array to retrieve the missing value.

Instead of subtraction, we can also use the difference[] method of the set[].

set[range[arr[0], arr[-1]+1]].difference[arr]

PROGRAMS

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Given an array arr[] of size N-1 with integers in the range of [1, N], the task is to find the missing number from the first N integers.

Note: There are no duplicates in the list.

Examples: 

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

Input: arr[] = {1, 2, 3, 5}, N = 5
Output: 4
Explanation: The missing number between 1 to 5 is 4

Approach 1 [Using Hashing]:The idea behind the following approach is

The numbers will be in the range [1, N], an array of size N can be maintained to keep record of the elements present in the given array

• Create a temp array temp[] of size n + 1 with all initial values as 0.
• Traverse the input array arr[], and do following for each arr[i] 
  • if[temp[arr[i]] == 0] temp[arr[i]] = 1
• Traverse temp[] and output the array element having value as 0 [This is the missing element].

Below is the implementation of the above approach:

C++

#include

using namespace std;

void findMissing[int arr[], int N]

{

    int i;

    int temp[N + 1];

    for[int i = 0; i

Javascript

    function getMissingNo[a, n] {

        let total = Math.floor[[n + 1] * [n + 2] / 2];

        for [let i = 0; i < n; i++]

            total -= a[i];

        return total;

    }

    let arr = [ 1, 2, 3, 5 ];

    let N = arr.length;

    let miss = getMissingNo[arr, N];

    document.write[miss];

Time Complexity: O[N]
Auxiliary Space: O[1]

Modification for Overflow: The approach remains the same but there can be an overflow if N is large. 

In order to avoid integer overflow, pick one number from the range [1, N] and subtract a number from the given array [don’t subtract the same number twice]. This way there won’t be any integer overflow.

Algorithm: 

• Create a variable sum = 1 which will store the missing number and a counter variable c = 2.
• Traverse the array from start to end.
  • Update the value of sum as sum = sum – array[i] + c and increment c by 1. This performs the task mentioned in the above idea]
• Print the missing number as a sum.

Below is the implementation of the above approach:

C++

#include

using namespace std;

int getMissingNo[int a[], int n]

{

    int i, total = 1;

    for [i = 2; i

Javascript

      function getMissingNo[a, n]

      {

        var x1 = a[0];

        var x2 = 1;

        for [var i = 1; i < n; i++] x1 = x1 ^ a[i];

        for [var i = 2; i 0]:

            ans = i + 1

    print[ans]

if __name__ == '__main__':

    arr = [1, 3, 7, 5, 6, 2]

    n = len[arr]

    findMissing[arr, n]

Javascript

function findMissing[arr,size]{

  let i;

  for [i = 0; i < size; i++] {

            if [Math.abs[arr[i]] - 1 == size] {

                continue;

            }

            let ind = Math.abs[arr[i]] - 1;

            arr[ind] *= -1;

        }

        let ans = size + 1;

        for [i = 0; i < size; i++] {

            if [arr[i] > 0]

                ans = i + 1;

        }

        console.log[ans];

}

        let arr = [ 1, 3, 7, 5, 6, 2 ];

        let n = arr.length;

       findMissing[arr,n];

Time Complexity: O[N] 
Auxiliary Space: O[1] 

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