Set Mismatch

Problem

The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

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

Note: The given array size will in the range [2, 10000]. The given array's numbers won't have any order.

Solution

Fixed Size Array Solution: O(n) time and O(1) space

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int a = -1, b = -1; //a is the repeated number and b is the missing number
        int[] counts = new int[nums.length];
        for (int num : nums) counts[num - 1]++;
        for (int i = 0; i < nums.length; i++) {
        //Must loop from i = 0 to the end, otherwise cannot find out the missing number
            if (counts[i] == 2) a = i + 1;
            else if (counts[i] == 0) b = i + 1;
            if (a != -1 && b != -1) return new int[]{a, b};
        }
        throw new IllegalArgumentException();
    }
}

Analysis

A very easy repeated detection problem
Use an array to count the frequency of each number