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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 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]

Question

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]

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Solution

To solve this problem, you can use the mathematical properties of sets and the Python language. Here is a step-by-step solution:

  1. First, calculate the sum of the numbers from 1 to n. This can be done using the formula n*(n+1)/2. Let's call this sum expected_sum.

  2. Then, calculate the sum of the numbers in the nums array. Let's call this sum actual_sum.

  3. Also, calculate the sum of the squares of the numbers from 1 to n. This can be done using the formula n*(n+1)*(2n+1)/6. Let's call this sum expected_sum_squares.

  4. Then, calculate the sum of the squares of the numbers in the nums array. Let's call this sum actual_sum_squares.

  5. Now, you have two equations:

    • expected_sum - actual_sum = missing_number - duplicate_number (equation 1)
    • expected_sum_squares - actual_sum_squares = missing_number^2 - duplicate_number^2 (equation 2)

  6. Equation 2 can be rewritten as (missing_number + duplicate_number) * (missing_number - duplicate_number). So, you can substitute equation 1 into equation 2 to get missing_number + duplicate_number.

  7. Now, you have two equations with two variables, which can be solved to get the values of missing_number and duplicate_number.

Here is a Python code snippet that implements this solution:

def findErrorNums(nums):
    n = len(nums)
    expected_sum = n*(n+1)//2
    actual_sum = sum(nums)
    expected_sum_squares = n*(n+1)*(2*n+1)//6
    actual_sum_squares = sum(x*x for x in nums)
    diff = expected_sum - actual_sum
    sum_ = (expected_sum_squares - actual_sum_squares) // diff
    return [(sum_ + diff) // 2, (sum_ - diff) // 2]

In this function, findErrorNums, it takes in the nums array, calculates the expected_sum, actual_sum, expected_sum_squares, actual_sum_squares, diff, and sum_, and then returns the duplicate_number and missing_number in an array.

This problem has been solved

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