The Problem

You are given an array of integers nums that originally contains all the numbers from 1 to n. Due to an error, one number is duplicated, and another number is missing. The task is to find the duplicated number and the missing number.

Brute Force Approach

A brute force solution involves iterating through the expected range of numbers from 1 to n and checking for each number if it exists in the input array nums. We can use nested loops or auxiliary data structures to accomplish this.

Algorithm

1. Create a set to store the numbers in the input array nums.
2. Iterate from 1 to n.
3. For each number i, check if it exists in the set.
   • If it does not exist, then i is the missing number.
   • While iterating, also determine the duplicate number by checking if the current number has already been added to the set. If it has, it's the duplicate.
4. Return an array containing the duplicate and missing numbers.

Code

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int[] findErrorNums(int[] nums) {
        Set<Integer> seen = new HashSet<>();
        int duplicate = -1;
        int missing = -1;
        int n = nums.length;

        for (int num : nums) {
            if (seen.contains(num)) {
                duplicate = num;
            }
            seen.add(num);
        }

        for (int i = 1; i <= n; i++) {
            if (!seen.contains(i)) {
                missing = i;
                break;
            }
        }

        return new int[] {duplicate, missing};
    }
}

Time and Space Complexity

• Time Complexity: O(n), where n is the length of the input array, due to iterating through the array once to populate the set and again to find the missing number.
• Space Complexity: O(n), where n is the length of the input array, due to storing the numbers in a set.

Optimal Approach

An optimal solution can be achieved by using the input array itself as a hash table. This approach utilizes the indices of the array to store information about the presence or absence of numbers.

Algorithm

1. Iterate through the input array nums.
2. For each number num, take its absolute value and use it as an index. Negate the value at that index.
   • If the value at that index is already negative, then num is the duplicate number.
3. Iterate through the input array nums again.
4. Find the index where the value is positive. The index + 1 is the missing number.
5. Return an array containing the duplicate and missing numbers.

Code

class Solution {
    public int[] findErrorNums(int[] nums) {
        int duplicate = -1;
        int missing = -1;
        int n = nums.length;

        // Find the duplicate number
        for (int i = 0; i < n; i++) {
            int index = Math.abs(nums[i]) - 1;
            if (nums[index] < 0) {
                duplicate = Math.abs(nums[i]);
            } else {
                nums[index] *= -1;
            }
        }

        // Find the missing number
        for (int i = 0; i < n; i++) {
            if (nums[i] > 0) {
                missing = i + 1;
                break;
            }
        }

        // Restore the array to its original state (optional)
        for (int i = 0; i < n; i++) {
            nums[i] = Math.abs(nums[i]);
        }

        return new int[] {duplicate, missing};
    }
}

Time and Space Complexity

• Time Complexity: O(n), where n is the length of the input array, due to iterating through the array twice.
• Space Complexity: O(1), as the algorithm uses constant extra space.

Edge Cases

• Empty Array: The problem statement specifies that the length of the array is between 2 and 10⁴, so we do not need to handle the case where the array is empty.
• Array with One Element: The problem statement specifies that the length of the array is between 2 and 10⁴, so we do not need to handle the case where the array has only one element.
• Large Input: The problem statement specifies that the elements of the array are between 1 and 10⁴, so we do not need to worry about integer overflow issues.