The numbers in the array can just be placed in the appropriate place in the array, just use the number as it's own position index. If when you go to place a number i in position i you notice that a[i]

*already*contains i then you know you found the duplicate. If not, place it and use the number that was there as your new index.

#!/usr/bin/python

from Numeric import *

def find(a):

i = 0

while True:

print a

if (a[i] == i):

i += 1

continue

if (a[a[i]] == a[i]):

print "Repeat is %d" % a[i]

return a[i]

c = a[a[i]]

a[a[i]] = a[i]

a[i] = c

i = c

#Test cases

a = array([0,3,2,1,4,4,6])

a = array([0,1,2,3,4,4,6])

a = array([4,4,2,3,0,1,6])

a = array([4,4,0,1,2,3,6])

find(a)

## 7 comments:

How would you do it if your array was immutable.

Moreover, there is a simpler solution to above problem than one you have given using mathematical serieses.

I don't know of a way to do it without writing to an array. If you have a better way, I'd love to see it!

Sorry, i was looking at a modified problem, if you are given that the numbers are only from 0 to n-2 (n-1 excluded).

Then you can use series formula along with the sum of the given array.

For the given problem, your solution is the optimal.

There is one more similar qu.

If you are told that there is more than 1 duplicates and numbers are from 0 to n-2. Then how would you find one of the duplicates ?

instead of i = c in the last line, why can't we do i++ ? If I replace i=c with i++, the program finds the duplicate with less number of iteration in all the test cases I could try.

I would like to know if there is a case where i = c will find the duplicate but the program would fail for i++?

Thanks

What if numbers are signed?

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