How about using hashtable?

10 Answers

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How about using hashtable?

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Use the function n(n+1)/2 = sum(0,n). Sum up all of the numbers in the array. Subtract the number from the function from the number in given by the sum. That will be your duplicate entry.

public static int dupeNum ( int [] array ){

int arraySum = 0;

int arraylength = array.length;

int knownSum = (arrayLength * ( arrayLength + 1 ) ) / 2;

for (int i : array ){

arraySum += array[i];

}

return (arraySum - knownSum) ;

}

Should be O(n).

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^^ person who replied above: Your solution fails if the numbers aren't sequential - for all you know, 'a list of n numbers' could be 'n' random numbers

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Merge sort it and then it iterate through the list. This takes nlogn time.

public in getDuplicate(List list) {

List sortedList = Mergesort(list);

for(int i = 0; i < sortedList.length-1; i++)

if(sortedList[i] == sortedList[i+1])

return SortedList[i];

Throw exception;

}

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take XOR of all the numbers.You will get the sum with out the duplicated number.

(sum of all n - above sum) will give you the number

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put the numbers into hashmap while traversing the list. Before placing the key into hashmap check whether it is null or not. if it isnot you've found it. worst case O(n). extra hashmap in the memory.

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i would sort them in n log and then traverse them. while traversing, chech two adjacent numbers are different. if not, that is the number.

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Construct a binary search tree with the given number.when u find a duplicate number return from the method as in binary search tree one has to put element either left or right.

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Construct a binary search tree with the given number.when u find a duplicate number return from the method as in binary search tree one has to put element either left or right.

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I gave an nlogn solution, where I said we will heap sort / quick sort the array, and then do a linear traversal to find out the duplicate entry. The interviewer was okay with the solution, and then she asked me code it, and then to write test cases for it.