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# 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:

Approach 1: Sorting O(n log n) Time and O(n) Space

class Solution {
public int[] findErrorNums(int[] nums) {
int[] res = new int[2];
res[0]=-1;
res[1]=1;
Arrays.sort(nums);
for(int i=1;i<nums.length;i++)
{
if(nums[i]==nums[i-1])  res[0]=nums[i];
else if(nums[i]>nums[i-1]+1)  res[1]=nums[i-1]+1;
}
if(nums[nums.length-1]!=nums.length) res[1] = nums.length;
return res;
}
}

Approach 2: Using HashMap O(n) Time and Space

class Solution {
public int[] findErrorNums(int[] nums) {
int res[] = new int[2];
Map<Integer,Integer> map = new HashMap<>();
for(int i:nums)
map.put(i,map.getOrDefault(i,0)+1);
for(int i=1;i<=nums.length;i++)
{
if(map.containsKey(i)){
if(map.get(i)==2)
res[0]=i;

}
else
res[1] = i;
}
return res;
}
}

Approach 3: Using Constant space O(n) Time complexity O(1) space

class Solution {
public int[] findErrorNums(int[] nums) {
int res[] = new int[2];
res[0]=-1;
res[1]=1;
for(int i=0;i<nums.length;i++)
{
if(nums[Math.abs(nums[i])-1]<0) res[0]=Math.abs(nums[i]);
else
{
nums[Math.abs(nums[i])-1]*=-1;
}
}
for(int i=0;i<nums.length;i++)
{
if(nums[i]>0)
res[1]=i+1;
}

return res;
}
}