645. Set Mismatch

LeetCode 645. Set Mismatch

Description

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

Tags

Hash Table, Math

Solution

We try to visit all nums[nums[i]-1]. emulating that we move nums[i] to ith position. Every time we invert the element we visited. If this number is encountered the second time, we find the duplicate, which is the first element of the output.

Then we traverse the array after inversion and find the only positive element. This element's index is the missing number, a.k.a the second element of the output.

Complexity

• Time complexity: $O(n)$

• Space complexity: $O(1)$

Code

func findErrorNums(nums []int) []int {
	dup, missing := -1, 1
	for _, n := range nums {
		if nums[abs(n)-1] < 0 {
			dup = abs(n)
		} else {
			nums[abs(n)-1] *= -1
		}
	}
	for i := 1; i < len(nums); i++ {
		if nums[i] > 0 {
			missing = i + 1
		}
	}
	return []int{dup, missing}
}

func abs(a int) int {
	if a < 0 {
		return -1 * a
	}
	return a
}

Reference

1. 错误的集合