Finding a missing number in a sequence of consecutive integers is a common coding interview question used to test a candidate’s logic, mathematical, and Python programming skills. This question evaluates whether the candidate can implement an algorithm to efficiently determine the missing number, without using builtin functions.
In this comprehensive guide, we will examine different methods to solve this technical interview question using Python. We will look at techniques like a mathematical formula, brute force, hash sets, XOR bit manipulation, and mathematical summation  explaining each approach with clear examples and Python code snippets. The advantages and limitations of each method are also analyzed.
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Defining the Problem:
Given an unsorted array ‘arr’ containing consecutive integers from 1 to n, with one number missing, the task is to find the missing number in Python.
For example:
arr = [2, 4, 1, 5]
Here, n = 5, and the numbers from 1 to 5 are 2, 4, 1, 5. So the missing number is 3.
The key aspects of this problem are:
 The array contains consecutive integers from 1 to n with one number missing.
 The numbers may not be in order.
 There will be only one missing number in the sequence.
 We need to find the missing number programmatically, without using builtin functions.
Method 1: Mathematical Formula
A simple mathematical formula can be derived to find the missing number:
Missing Number = Total Sum of Numbers  Sum of Array Elements
Where:
Total Sum of Numbers from 1 to n = n(n+1)/2
To find the missing number using this formula in Python:
def findMissingNumber(arr, n):
# Calculate the expected sum
total = n * (n+1) // 2
# Calculate the sum of array elements
sum_arr = sum(arr)
# Missing number is the difference between the totals
return total  sum_arr
arr = [2, 4, 1, 5]
n = len(arr)+1
print(findMissingNumber(arr, n))
# Output: 3
Analysis:

Time Complexity  O(N): We need to traverse the array of size n to calculate
sum(arr)
. So the time complexity is linear. 
Space Complexity  O(1): No extra space required, as we store sums in variables.
Limitations:

Only works when the numbers in the array are consecutive from 1 to n.

Fails if duplicates are present or if multiple numbers are missing.
Method 2: Brute Force
The brute force technique involves iterating over the numbers from 1 to n, checking if each number is present in the array using linear search. The missing number is the one not found in the array.
def findMissingBruteForce(arr, n):
for i in range(1, n+1):
if i not in arr:
return i
arr = [2, 4, 1, 5]
n = len(arr)+1
print(findMissingBruteForce(arr, n))
# Output: 3
Analysis:

Time Complexity  O(N^2): We iterate from 1 to n, and check if each number is present in the array using linear search. So the time complexity is quadratic.

Space Complexity  O(1): Constant extra space for loop variable.
Limitations:
 Inefficient for large input arrays, due to the nested loop.
Method 3: Using a Hash Set
We can utilize a hash set to keep track of the numbers seen so far. Iterate over the array, adding each element to the hash set. Finally, iterate from 1 to n and print the number not present in the set.
from typing import List
def findMissingHash(arr: List[int], n: int) > int:
# Hash set to store array elements
seen = set()
for num in arr:
seen.add(num)
# Find missing number
for i in range(1, n+1):
if i not in seen:
return i
arr = [2, 4, 1, 5]
n = len(arr) + 1
print(findMissingHash(arr, n))
# Output: 3
Analysis:

Time Complexity  O(N): We iterate over the array of size n once to build the hash set. And iterate from 1 to n again to find the missing number. So overall linear time complexity.

Space Complexity  O(N): The extra space required is linear for the hash set.
Limitations:

Slower than mathematical approaches for smaller arrays.

Extra space needed for the hash set.
Method 4: Using XOR Bit Manipulation
XOR bit manipulation can be leveraged to find the missing number in O(n) time and O(1) extra space.
The key idea is that XOR of a number with itself is 0. And XOR is associative and commutative.
def findMissingXOR(arr, n):
missing = n * (n+1) // 2
for num in arr:
missing ^= num
return missing
Here, we first calculate the XOR of numbers from 1 to n. This can be mathematically derived to n * (n+1) // 2.
Then we XOR this result with each number in the array. The result will be the XOR of the missing number.
Since XOR is commutative and associative, the duplicate numbers will zero out. And we will be left with XOR of just the missing number.
Example:
Arr = [1, 5, 2, 6, 4]
n = 6
missing = 6 * (6 + 1) // 2 = 21
missing ^= 1 ^ 5 ^ 2 ^ 6 ^ 4
= 21 ^ 1 ^ 5 ^ 2 ^ 6 ^ 4
= 3
Analysis:

Time Complexity  O(N): We iterate over the array of size n once.

Space Complexity  O(1): No extra space required.
Limitations:
 Tricky to understand and implement correctly.
Method 5: Mathematical Summation
We can find the missing number by calculating the sum of numbers from 1 to n, and subtracting the sum of array elements.
def findMissingSum(arr, n):
total = (n * (n + 1)) // 2
sum_arr = 0
for num in arr:
sum_arr += num
return total  sum_arr
This avoids overflow for larger values of n, compared to method 1.
Analysis:

Time Complexity  O(N): We traverse the array of size n once.

Space Complexity  O(1): Constant space required.
Limitations:
 Still fails for duplicates or multiple missing numbers.
Based on the above analysis, the recommended approaches are:

Mathematical Formula  Simple logic, fast running time. But fails for duplicates or multiple missing numbers.

XOR Bit Manipulation  Efficient in time and space. But tricky implementation.

Hash Set  Robust for all cases. Needs extra space.
The best method depends on the specific problem constraints and performance requirements.
Handling Duplicates and Multiple Missing Numbers
The above solutions rely on the numbers being consecutive from 1 to n without duplicates. But we may encounter cases where:

The array has duplicate elements.

There are multiple missing numbers.
Here are two techniques to handle these scenarios:
1. Using a Hash Set
We can store all array elements in a hash set. Iterate from 1 to n and print all numbers not present in the set.
This will work seamlessly for both duplicate elements and multiple missing numbers.
def findMissingDuplicates(arr, n):
# Hash set for array elements
seen = set(arr)
# Find all missing numbers
missing = []
for i in range(1, n+1):
if i not in seen:
missing.append(i)
return missing
Analysis:
 Time Complexity  O(N)
 Space Complexity  O(N)
2. Sorting the Array
If the array is sorted, we can traverse the array and find missing elements by comparing consecutive elements:
def findMissingSorted(arr, n):
missing = []
# Sort the array
arr.sort()
# Compare consecutive elements
prev = arr[0]
for i in range(1, len(arr)):
if arr[i]  prev > 1:
missing.append(prev + 1)
prev = arr[i]
# Check for missing number after last element
if prev != n:
missing.append(prev + 1)
return missing
This modifies the input array.
Analysis:
 Time Complexity  O(NlogN) for sorting
 Space Complexity  O(1)
Therefore, using a hash set is preferred for handling duplicates and multiple missing numbers.
Variations of the Problem
Some variations of the missing number problem that you may encounter are:
1. Cyclic Array: The numbers form a cyclic sequence from 1 to n.
For example: arr = [5, 4, 2, 1] and n = 4
Solution: Find the sum of numbers and array elements. The difference is the missing number.
2. Large Sequence: The sequence length n is very large (in billions).
Solution: Use mathematical summation approach to avoid integer overflow.
3. Continous Sequence: The numbers are in a continous sequence but not starting from 1.
For example: arr = [101, 102, 103, 105]
Solution: Maintain the difference between consecutive elements to find the missing number.
4. Duplicates Allowed: The array may contain duplicate elements.
Solution: Use a hash set to store unique elements as discussed above.
5. Multiple Missing Numbers: There can be more than one missing number in the sequence.
Solution: Find missing numbers iteratively using hash set or sorting approaches.
Practical Applications
The techniques to find missing elements in a sequence have several realworld applications:

Missing ID Detection: Finding missing IDs or SKUs in a sequential inventory or product database.

Network Packet Analysis: Analyzing network packet captures to identify missing packet sequence numbers and detect network issues.

Genomic Sequences: Bioinformatics algorithms to find missing genes or nucleotides in a DNA or protein sequence.

Fraud Analysis: Analyze transaction/log datasets and find missing sequence numbers to identify potential anomalies or fraud.

Machine Learning: Can be used as a preprocessing technique for sequencebased ML models like RNNs for imputing missing values.
Summary
To summarize, here are the key points on how to find the missing number in a sequence of consecutive integers in Python:

Derive a mathematical formula based on sum of n integers. Efficient for small arrays but fails for duplicates.

Brute force checking each number has quadratic time complexity.

Use a hash set to store array numbers and find missing by iteration. Handles duplicates and multiple missing numbers.

Leverage XOR bit manipulation to find the missing number in linear time and constant space.

Calculate mathematical summation and subtract array sum to avoid overflow.

Sorting the array can also help identify missing numbers by comparing consecutive elements.

The problem can be extended for cyclic sequences, large n, noncontinuous numbers, and multiple missing values.

This technique has applications in domains like inventory management, network analysis, genomics, fraud detection, and machine learning.
By mastering these algorithms and being able to code the solutions efficiently in Python, you can successfully tackle the “find missing number” problem in coding interviews and data analysis tasks in the real world.