Problem:

You have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number.

You are given an integer array nums representing the data status of this set after the error.

Find the number that occurs twice and the number that is missing and return them in the form of an array.

Example 1:

Input: nums = [1,2,2,4] Output: [2,3]

Example 2:

Input: nums = [1,1] Output: [1,2]

Constraints:

• 2 <= nums.length <= 104
• 1 <= nums[i] <= 104

Problem Analysis:

Solution 1

1. High-Level Strategy:
   • The code uses a Counter to count the occurrences of each element in the given list nums.
   • It then identifies the duplicate by finding the element with the maximum count using the max function and the key parameter.
   • The missing element is determined by iterating over the range from 1 to the length of nums + 1 and finding the first element not present in the Counter.
2. Complexity Analysis:
   • The time complexity of the solution is dominated by the loop that iterates over the range from 1 to the length of nums + 1. In the worst case, this loop runs in O(n) time, where n is the length of the input list.
   • The Counter construction takes O(n) time, and finding the maximum count takes O(1) time.
   • The overall time complexity is O(n).
   • The space complexity is O(n) due to the Counter, which stores the count of each unique element in the input list.

Solution 2

1. High-Level Strategy:

   • This solution takes advantage of the properties of the sum of natural numbers from 1 to n.
   • It calculates the expected sum using the formula n * (n + 1) // 2, which is the sum of all unique elements if there were no duplicates or missing elements.
   • It then calculates the actual sum of the given list nums.
   • The difference between the actual sum and the sum of unique elements gives the duplicate, and the difference between the expected sum and the actual sum gives the missing element.
   • The solution directly computes the duplicate and missing elements without explicitly counting occurrences using loops or data structures.

2. Complexity Analysis:

   • Calculating the expected sum takes constant time, O(1).
   • Computing the actual sum of nums takes O(n) time, where n is the length of the input list.
   • Constructing a set from nums takes O(n) time as well.
   • The overall time complexity is dominated by the summation and set operations, resulting in O(n) time complexity.
   • The space complexity is O(n) due to the space required for the set, which stores the unique elements of the input list.

Solutions:

Solution 1:

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        count = Counter(nums)
        duplicate = max(count, key=count.get)
        missing = 0
        for i in range(1, len(nums)+1):
            if i not in count.keys():
                missing = i
        return [duplicate, missing]

Solution 2:

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        n = len(nums)
        expected_sum = n * (n + 1) // 2  # Sum of natural numbers from 1 to n

        actual_sum = sum(nums)
        duplicate = actual_sum - sum(set(nums))  # Duplicate is the difference between actual sum and sum of unique elements

        missing = expected_sum - (actual_sum - duplicate)  # Missing is the difference between expected sum and actual sum

        return [duplicate, missing]

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