# 645. Set Mismatch

`Input: nums = [1,2,2,4]Output: [2,3]`
`Input: nums = [1,1]Output: [1,2]`
• `2 <= nums.length <= 10^4`
• `1 <= nums[i] <= 10^4`
`class Solution:    def findErrorNums(self, nums: List[int]) -> List[int]:        xor_=0        miss,dup=0,0        hash=dict()        for i in range(len(nums)):            if nums[i] not in hash:                hash[nums[i]]=1            else:                hash[nums[i]]+=1        cnt=1        #cnt is number from 1 to len(nums)        for i in hash:            xor_^=i^cnt            cnt+=1            if(hash[i]==2):                dup=i        miss=xor_^cnt        return[dup,miss]`
`class Solution:    def findErrorNums(self, nums: List[int]) -> List[int]:        set_=set()        result=[]        xor_=0        miss,dup=0,0        for i in nums:            set_.add(i)                    for i in range(len(nums)):            xor_^=(i+1)^nums[i]                    for i in range(1,len(nums)+1):            if(i not in set_):                miss=i        dup=miss^xor_            return([dup,miss])`
`class Solution:    def findErrorNums(self, nums: List[int]) -> List[int]:        xor_=0        set_group_xor,unset_group_xor=0,0,        for i in range(len(nums)):            xor_^=(i+1)^nums[i]        right=xor_ & ~(xor_-1)            for i in range(1,len(nums)+1):            if(i&right!=0):                set_group_xor^=i            else:                unset_group_xor^=i            if(nums[i-1]&right!=0):                set_group_xor^=nums[i-1]            else:                unset_group_xor^=nums[i-1]                                        val1=xor_^set_group_xor        val2=xor_^unset_group_xor        for i in nums:            if(i==val1):                return(val1,val2)        return(val2,val1)`

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## Dhanarajappu

Tech-Enthusiast, Coder,Explorer,Geeky,Software Engineer |A piece of code delivers everything that you need. The world is all about codes.