Missing Number LeetCode

Key takeaways:
The missing number in an array of distinct numbers from the range can be found using the XOR operation, which cancels out all existing numbers, leaving only the missing one.
The algorithm runs in O(n) time and uses O(1) extra space, making it an optimal solution for this problem.
This problem has real-world relevance in areas like database index reconstruction and sequence alignment in bioinformatics, where data integrity is critical.

In the field of algorithmic challenges, the missing number problem on LeetCode is a great way to learn more about how to work with lists and find things within them. This kind of problem is especially important in situations where data integrity is important, such as database index reconstruction or sequence alignment in bioinformatics.

Solution

To find the missing number in an array containing $n$ distinct numbers taken from the range $[0, n]$ , we use an efficient algorithm that uses the properties of the XOR operation. The algorithm follows these key steps:

We start by initializing a variable, missing, to the length of the input array nums. This value represents $n$ , the upper bound of the range from which numbers are taken, and ensures that all numbers from $0$ to $n$ are included in the XOR operation.
We iterate through nums, and in each iteration, we update the missing variable by applying the XOR operation with the index of the current element and its corresponding value. The result of this operation is saved in missing. The XOR operation is applied to each index and each element, ensuring that every number in the complete range $[0, n]$ is XORed exactly once.
After iterating through nums, we return missing, which holds the value of the missing number. This is because the XOR operation cancels out all numbers that appear twice (once as an index and once as an array element), leaving only the missing number.

This algorithm works efficiently due to the properties of the XOR operation: any number XORed with itself results in $0$ , and any number XORed with $0$ remains unchanged. By XORing all indexes and all elements, all present numbers cancel out, and the remaining value is the missing number.

Let’s look at the following illustration for a better understanding:

def find_missing_number(nums):
    # Initialize the missing number with the length of the array
    missing = len(nums)
    # Iterate over the array
    for i in range(len(nums)):
        # Update the missing number by XORing the current index, 
        # the current element and the missing variable
        missing ^= i ^ nums[i]
    return missing
def main():
  inputnumbers = [[4, 0, 3, 1], 
    [8, 3, 5, 2, 4, 6, 0, 1], 
    [1, 2, 3, 4, 6, 7, 8, 9, 10, 5], 
    [0], 
    [1, 2, 3, 0, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23]
  ]
  i = 1
  for x in inputnumbers:
    print(i, ".\tnums: ", x, sep = "")
    print("\n\tMissing number: ", find_missing_number(x), sep = "")
    print("-"*100, "\n", sep = "")
    i+=1
if __name__ == "__main__":
    main()

Missing number LeetCode

Now, let’s look at the time and space complexity of the solution.

Time complexity

The time complexity of this solution is $O(n)$ , since it iterates over the array only $n$ times, where $n$ is the number of elements in nums.

Space complexity

The space complexity of this solution is $O(1)$ , since it uses constant amount of additional space.

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