Missing Number Leetcode Problem 268 [Python Solution]

Welcome to another exciting coding tutorial!

In this blog post, we’ll tackle the Missing Number problem, which is categorized under Bit Manipulation on LeetCode.

This problem is considered easy and provides a great opportunity to explore different solutions, making it ideal for beginners.

We will also optimize this blog post for SEO to help you find it easily when you need assistance with this problem.

So, let’s dive in!

Question

Here’s the problem statement:
Given an array nums containing n distinct numbers in the range [0, n], your task is to find and return the only number in this range that is missing from the array.

Example 1:
Input: nums = [3, 0, 1]
Output: 2
Explanation: In this case, n = 3 since there are 3 numbers, so all numbers are in the range [0, 3].

The missing number is 2 because it does not appear in nums.

Example 2:
Input: nums = [0, 1]
Output: 2
Explanation: Here, n = 2, and the missing number is 2 because it’s absent from nums.

Example 3:
Input: nums = [9, 6, 4, 2, 3, 5, 7, 0, 1]
Output: 8
Explanation: In this case, n = 9, and the missing number is 8 as it’s not included in nums.

Constraints:
– n == nums.length
– 1 <= n <= 104
– 0 <= nums[i] <= n
– All the numbers in nums are unique.

Now that we understand the problem, let’s explore different approaches to solving it.

Problem Overview

Brute-force Approach

One simple way to find the missing number is through brute force.

You can iterate through all numbers in the range [0, n] and check if each number exists in the nums array.

If you find a number that is not in the array, that’s your missing number.

This approach has a time complexity of O(n^2), as you iterate through the range and search for each number in the array.

An Efficient Approach with XOR

A more efficient approach takes advantage of bitwise XOR (^).

We know that XOR is commutative and associative, and XORing a number with itself results in 0. So, if we XOR all the numbers in the range [0, n] and XOR them with all the numbers in the nums array, the missing number will be the result.

Here’s a Python solution that demonstrates this approach:

def missingNumber(nums):
    result = len(nums)

    for i in range(len(nums)):
        result ^= i ^ nums[i]
    
    return result

This code calculates the missing number with a time complexity of O(n) and uses only O(1) extra memory, which is quite efficient.

It’s the approach we’ll focus on in this blog post.

Efficient Python Code Solution

Let’s dive into the efficient Python code solution mentioned earlier:

def missingNumber(nums):
    result = len(nums)

    for i in range(len(nums)):
        result ^= i ^ nums[i]

    return result

This function takes a list of numbers, nums, as input and returns the missing number.

Here’s how it works:

1. Initialize the result variable with the length of the nums array.
2. This is because the missing number can be at most n, so initializing result with n is a reasonable starting point.
3. Iterate through the nums array using a for loop and an index i.
4. In each iteration, perform bitwise XOR (^) on the result with both i and nums[i].
5. The XOR operation cancels out elements that exist in both sets and preserves the missing element.
6. Finally, return the result, which will hold the missing number after the loop completes.

By applying this approach, you can find the missing number efficiently, even in large arrays.

Time and Space Complexity

Now, let’s discuss the time and space complexity of the efficient solution:

• Time Complexity: The time complexity of this solution is O(n).

We iterate through the nums array once, performing a constant number of operations in each iteration.

• Space Complexity: The space complexity is O(1) because we use a single variable, result, to store the missing number.

We don’t use any additional data structures that grow with the input size.

Reasoning Behind Our Approach

The efficient approach leverages the properties of XOR, a bitwise operation that allows us to find the missing number without requiring additional memory.

Here’s the reasoning:

1. When we XOR a number with itself, it results in 0 (e.g., x ^ x = 0).
2. XOR is commutative and associative, meaning the order of operations does not matter (e.g., a ^ b ^ c is equivalent to c ^ a ^ b).
3. If we XOR all numbers from 0 to n and XOR them with all numbers in nums, the missing number will remain because it doesn’t have a pair to cancel it out.
4. The final result variable will hold the missing number.

This approach simplifies the problem by reducing it to a series of XOR operations, making it both efficient and elegant.

Edge Cases a Valid Solution Must Consider

Before we conclude, it’s essential to discuss some edge cases that a valid solution must consider.

These cases ensure that the solution works correctly for all scenarios:

1. An Empty Input: When nums is an empty list, the missing number should be 0. This edge case should return 0.
2. Missing Number at the Beginning: If the missing number is at the beginning of the range (0), the solution should correctly identify it.
3. Missing Number at the End: Similarly, if the missing number is at the end of the range (n), the solution should handle this case accurately.
4. No Missing Number: When there is no missing number, meaning that the nums list contains all numbers from 0 to n, the solution should return n + 1.

By considering these edge cases, you can ensure that your solution is robust and handles all possible scenarios.

Related Interview Questions By Company:

• Min Cost Climbing Stairs LeetCode
• Majority Element LeetCode
• Number Of 1 Bits LeetCode

Related Interview Questions By Difficulty:

• Counting Bits LeetCode
• Sum Of Two Integers LeetCode
• Missing Number LeetCode

Conclusion

In this blog post, we’ve explored the Missing Number problem on LeetCode and provided an efficient Python solution that leverages bitwise XOR to find the missing number.

This approach has a time complexity of O(n) and uses O(1) extra memory, making it an optimal solution for this problem.

We’ve also discussed the reasoning behind our approach, highlighted important edge cases to consider, and explained the time and space complexity of the solution.

Whether you’re a beginner or an experienced programmer, understanding this problem and its efficient solution can be valuable for your coding journey.

Feel free to comment, ask questions, make suggestions, or share this content with others.

Your engagement and feedback are highly encouraged and appreciated.

If you’d like to explore the problem further or practice your skills, you can find it on LeetCode.

Thank you for reading, and happy coding!