[Leetcode] Interview question 17.19: Two numbers that disappear

daily progress

• Find the absolute value:Math.abs()

topic

Given an array containing all integers from 1 to N, but two numbers are missing. Can you find them in O(N) time and only O(1) space?
Both numbers can be returned in any order.

• Example 1:
  • input: [1]
  • output: [2,3]
• Example 2:
  • Input: [2,3]
  • Output: [1,4]
• hint:
  • nums.length <= 30000

Source: LeetCode
Link: https://leetcode.cn/problems/missing-two-lcci
The copyright belongs to Leetcode.com. This article is for personal study only, non-commercial use.

answer

The solution starting from the constraints:
I use the bit subscript of the array iand the sign of the corresponding value nums[i]to record numswhether the array contains integers i+1.

1. We can use an N-bit array to record whether N data appear. Since numsthe array is missing two values, the length is N-2, so we need to create an array of length 2 last2to record whether the last two elements are present.
   • We use nums[0: n-3]and last2[0: 1]to record whether N data appear, and if so, modify the value to the opposite number. Since the original values ​​are all greater than 0, when the final tmpbit is less than 0, the surface array numscontains integerstmp+1
2. Record existence information, traverse numsthe array:
   • First, we extract numsthe raw data that the array contains. The absolute value of the array value is calculated because some of the values ​​will be changed to opposite numbers later: tmp = Math.abs(nums[i]).
   • Then, modify nums[tmp-1]or last2[tmp-nums.length-1]to the original opposite number, so that the original information contained in the array is not lost, and whether the array includes integers is recorded tmp.
3. Retrieve stored information, iterate over numsarrays and last2arrays:
   • If ithe value at the position nums[i]is less than 0, it means that numsthe array contains integersi+1
   • If ithe value at the position nums[i]is greater than 0, it means that numsthere is a lack of integers in the array i+1, so one of the answers has been found

Advantages: Compared with other problem solutions on the Internet, this method can be more easily extended to the case of missing m data. That is, replace an array of length 2 last2with an array of length m lastm.

code:

class Solution {
    
    
    public int[] missingTwo(int[] nums) {
    
    
        int[] last2 = {
    
    1, 1};
        int[] ans = {
    
    0, 0};

        for (int i = 0; i < nums.length; i++){
    
    
            int tmp = Math.abs(nums[i]);

            if (tmp <= nums.length){
    
    
                nums[tmp-1] = -nums[tmp-1];
            }else{
    
    
                last2[tmp-nums.length-1] = -1;
            }
        }

        for (int i = 0; i < nums.length; i++){
    
    
            if (nums[i] > 0){
    
    
                if (ans[0] == 0)
                    ans[0] = i+1;
                else
                    ans[1] = i+1;
            }
        }
        for (int i = 0; i < last2.length; i++){
    
    
            if (last2[i] > 0){
    
    
                if (ans[0] == 0)
                    ans[0] = i+1+nums.length;
                else
                    ans[1] = i+1+nums.length;
            }
        }
        return ans;
    }
}

The execution time and memory consumption are as follows: