645. Set Mismatch

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]

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

Constraints:
2 <= nums.length <= 104
1 <= nums[i] <= 104

Solution:

1. brute force
   traverse over whole array to check if occure twice or doesn’t occure at all

2. sorting -> Arrays.sort(nums)
   numbers lie together, so does duplicate number, missing value can be detect one count apart from duplicate int

3. hashset/hashmap
   only store unique value, duplicate number can be detected, loop through array to detect missing value

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class Solution {
    public int[] findErrorNums(int[] nums) {
        int n = nums.length;
        int duplicate = 0;
        int missing = 0;
        HashSet<Integer> noneDup = new HashSet<Integer>();
        
        for (int i: nums) {
            if (!noneDup.contains(i)) {
                noneDup.add(i);
            } else {
                duplicate = i;
            }
        }
        
        for (int j=1; j<=n; j++) {
            if (!noneDup.contains(j)) {
                missing = j;
            }
        }
        
        return new int[]{duplicate, missing};
    }
}

there are more solutions…

references

• https://leetcode.com/problems/set-mismatch/