LeetCode Problem Number: 645

Problem 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]

Solution Code

class Solution {
public:
    vector findErrorNums(vector& nums) {
        unordered_map freq;
        vector ans;
        int n =  nums.size();
        for(int i = 0 ; i < n ; i++){
            freq[nums[i]]++;
        }

        for(auto i: freq){
            if(i.second == 2) {
                ans.push_back(i.first);
                break;
            }
        }
        for(int i = 1 ; i <= n ; i++){
            if(freq[i] == 0){
                ans.push_back(i);
                break;
            }
        }
        return ans;
    }
};
                    

Solution Explanation

The problem involves finding the number that is duplicated and the number that is missing from a set that originally contained all numbers from 1 to n. The solution follows these steps:

• Count Frequency: Use a hash map to count the frequency of each number in the array.
• Find Duplicated Number: Iterate through the hash map to find the number that appears twice.
• Find Missing Number: Iterate from 1 to n to find the number that is missing in the hash map.