Question

How can we find a repeated number in array in O(n) time and O(1) complexity?
eg
array 2,1,4,3,3,10
output is 3

EDIT:
I tried in following way.
i found that if no is oddly repeated then we can achieve the result by doing xor . so i thought to make the element which is odd no repeating to even no and every evenly repeating no to odd.but for that i need to find out unique element array from input array in O(n) but couldn’t find the way.

Answer 1

Assuming that there is an upped bound for the values of the numbers in the array (which is the case with all built-in integer types in all programming languages I ‘ve ever used — for example, let’s say they are 32-bit integers) there is a solution that uses constant space:

1. Create an array of N elements, where N is the upper bound for the integer values in the input array and initialize all elements to 0 or false or some equivalent. I ‘ll call this the lookup array.
2. Loop over the input array, and use each number to index into the lookup array. If the value you find is 1 or true (etc), the current number in the input array is a duplicate.
3. Otherwise, set the corresponding value in the lookup array to 1 or true to remember that we have seen this particular input number.

Technically, this is O(n) time and O(1) space, and it does not destroy the input array. Practically, you would need things to be going your way to have such a program actually run (e.g. it’s out of the question if talking about 64-bit integers in the input).