Set Mismatch

Problem Statement

In this challenge, we begin with a set of integers s which should ideally contain all sequential numbers from 1 to n. A mistake has been made such that one number within this set is erroneously duplicated over another number, leading to one number appearing twice while another is entirely missing.

Given an integer array nums that represents the erroneous set, the task is to identify both the number that has been duplicated (appears twice) and the number that is missing from the set. The result should be returned as an array comprised of these two integers, with the first element being the duplicated number and the second being the missing number.

Examples

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

Approach and Intuition

To solve this problem, we must determine two key elements:

1. The number that appears twice in the array.
2. The number that is missing from the set.

To efficiently find these values, we must consider the properties of the numbers from 1 to n:

1. Calculate Expected Sum: Calculate the sum of the first n natural numbers using the formula n * (n + 1) / 2. This gives us what the sum should be if there were no errors.
2. Sum of Given Array: Compute the sum of the elements in the provided nums array.
3. Track Appearances: As we track the sum, also keep a record of each number's occurrence to identify any number that appears twice.
4. Find the Duplicated and Missing Numbers:
   • The number that appears twice will be directly identified during the scanning of the array based on its frequency.
   • The missing number can be computed by considering the difference between the expected sum and the actual sum of the unique numbers from the array. Adjust this difference by accounting for the repeated number to find the missing one.

For example, in a scenario where nums = [1, 2, 2, 4]:

• Calculate the expected sum for n = 4, which is 10.
• The sum of the array nums is 9.
• From tracking, we see that the number 2 occurs twice.
• Knowing 2 is duplicated, we can calculate the missing number. The missing number should adjust the sum to match the expected sum which leads us to discover that 3 is missing.

The detail of this method allows us to efficiently identify both the duplicated and missing numbers, satisfying the constraints and requirements of the problem.

Solutions

public class Solution {
    public int[] findDuplicatesAndMissing(int[] elements) {
        int combinedXOR = 0, separateXOR1 = 0, separateXOR2 = 0;
        for (int value: elements)
            combinedXOR ^= value;
        for (int i = 1; i <= elements.length; i++)
            combinedXOR ^= i;
        int uniqueBit = combinedXOR & ~(combinedXOR - 1);
        for (int value: elements) {
            if ((value & uniqueBit) != 0)
                separateXOR1 ^= value;
            else
                separateXOR2 ^= value;
        }
        for (int i = 1; i <= elements.length; i++) {
            if ((i & uniqueBit) != 0)
                separateXOR1 ^= i;
            else
                separateXOR2 ^= i;
        }
        for (int i = 0; i < elements.length; i++) {
            if (elements[i] == separateXOR2)
                return new int[]{separateXOR2, separateXOR1};
        }
        return new int[]{separateXOR1, separateXOR2};
    }
}