Using XOR operator for finding duplicate elements in a array fails in many cases

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野的像风
野的像风 2021-02-01 10:36

I came across a post How to find a duplicate element in an array of shuffled consecutive integers? but later realized that this fails for many input.

For ex:
a

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  •  执笔经年
    2021-02-01 11:21

    Remember these two properties of XOR operator :

    (1) If you take xor of a number with 0 ( zero ) , it would return the same number again.

    Means , n ^ 0 = n

    (2) If you take xor of a number with itself , it would return 0 ( zero ).

    Means , n ^ n = 0

    Now , Coming to the problem :

       Let    Input_arr = { 23 , 21 , 24 , 27 , 22 , 27 , 26 , 25 }    
    
       Output should be 27 ( because 27 is the duplicate element in the Input_arr ).
    

    Solution :

    Step 1 : Find “min” and “max” value in the given array. It will take O(n).

    Step 2 : Find XOR of all integers from range “min” to “max” ( inclusive ).

    Step 3 : Find XOR of all elements of the given array.

    Step 4 : XOR of Step 2 and Step 3 will give the required duplicate number.

    Description :

    Step1 : min = 21 , max = 27
    
    Step 2 : Step2_result = 21 ^ 22 ^ 23 ^ 24 ^ 25 ^ 26 ^ 27 = 20
    
    Step 3 : Step3_result = 23 ^ 21 ^ 24 ^ 27 ^ 22 ^ 27 ^ 26 ^ 25 = 15
    
    Step 4 : Final_Result = Step2_result ^ Step3_result = 20 ^ 15 = 27
    
    But , How Final_Result calculated the duplicate number ?
    
    Final_Result = ( 21 ^ 22 ^ 23 ^ 24 ^ 25 ^ 26 ^ 27 ) ^ ( 23 ^ 21 ^ 24 ^ 27 ^ 22 ^ 27 ^ 26 ^ 25 )
    
    Now , Remember above two properties : n ^ n = 0 AND n ^ 0 = n
    
    So , here ,
    
    Final_Result = ( 21 ^ 21 ) ^ ( 22 ^ 22 ) ^ ( 23 ^ 23 ) ^ ( 24 ^ 24 ) ^ ( 25 ^ 25 ) ^ ( 26 ^ 26 ) ^ ( 27 ^ 27 ^ 27 )
    
                 = 0 ^ 0 ^ 0 ^ 0 ^ 0 ^ 0 ^ ( 27 ^ 0 ) ( property applied )
    
                 = 0 ^ 27 ( because we know 0 ^ 0 = 0 )
    
                 = 27 ( Required Result )
    

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