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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 this set got duplicated which caused another number to be missing. You need to return an array of two numbers, one of which is the duplicated number and the other is the missing number.

Clarifying Questions

1. Input format:
   • Is the input always valid (i.e., there will always be exactly one duplicate and one missing number)?
   • What is the range of n? Are we dealing with very large numbers?
2. Constraints:
   • Are there any constraints on the memory or time complexity?
3. Output format:
   • Should the output array follow a particular order (e.g., duplicated number first)?

Without loss of generality, we’ll assume the following for the problem:

• Input: An array nums containing n integers.
• Output: An array with two integers [duplicated, missing].

Strategy

1. Initialize Data Structures:
   • Use a list or dictionary to keep track of the frequency of each number.
   • Alternatively, use mathematical properties to determine the missing and duplicate numbers.
2. Identify the Duplicate:
   • By iterating through the array and using a frequency count or set, identify the number that appears twice.
3. Identify the Missing Number:
   • Calculate the expected sum of the first n natural numbers and the actual sum of the given array.
   • The difference between the expected sum and the actual sum, adjusted with the duplicate number, will give the missing number.

Code

def findErrorNums(nums):
    n = len(nums)
    duplicate = -1
    missing = -1
    
    # Use a frequency array to track the occurrences of each number
    count = [0] * (n + 1)
    
    for num in nums:
        count[num] += 1
    
    for i in range(1, n + 1):
        if count[i] == 2:
            duplicate = i
        elif count[i] == 0:
            missing = i
    
    return [duplicate, missing]

# Example usage:
nums = [1, 2, 2, 4]
print(findErrorNums(nums))  # Output: [2, 3]

Time Complexity

• The time complexity of this solution is O(n), where n is the length of the input array nums.
  • First, we iterate through the array once to fill in the frequency array.
  • Then, we iterate through the frequency array to find the duplicate and missing numbers.
• The space complexity is O(n) due to the additional frequency array used, but the range of this array is well-defined and limited from 1 to n.

This method ensures that we efficiently identify the duplicate and missing numbers using a straightforward approach.

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