Eg: Array : [0,1,0,1,1,0,0]
Final Array: [0,0,1,1,1,0,0] , So swaps required = 1
i need a O(n) or O(nlogn) solution
You can do it in O(n):
In one pass through the data, determine the number of 1s. Call this k (it is just the sum of the elements in the list).
In a second pass through the data, use a sliding window of width k to find the number, m which is the maximum number of 1s in any window of size k. Since this is homework, I'll leave the details to you, but it can be done in O(n).
Then: the minimal number of swaps is k-m.
EDIT This answer assumes that only two neighboring cells can be swapped. If the distance between the two swapped elements is arbitrary, see #JohnColeman's answer.
This can be done easily in linear time.
Suppose that the array is called a and its size is n.
Allocate integer array b of size n. Walk from left to right, save in b[i] the number of ones seen so far in a[0], ..., a[i].
Allocate integer array c of size n. Walk from right to left, save in c[i] the number of ones seen so far in a[i], ..., a[N - 1].
Initialize integer res = 0. Walk through a one last time. For each i with a[i] = 0, add res += min(b[i] c[i])
Output res
Why this works? Each zero must somehow bubble out of the block of ones. So, every zero must either "bubble-up" past all ones to the right of it, or it must "bubble-down" past all ones to the left of it. Swapping zeros with zeros is waste of time, therefore the process of zero-eviction from the homogeneous block of ones must start with those zeros that are as close to the first 1 or the last 1 as possible. This means, that every zero will have to make exactly min(b[i], c[i]) swaps with 1s to exit the homogeneous block of ones.
Example:
a = [0,1,0,1,1,0,1,0,1,0,1,0]
b = [0,1,1,2,3,3,4,4,5,5,6,6]
c = [6,6,5,5,4,3,3,2,2,1,1,0]
now, min(b,c) would be (no need to compute it explicitly):
m = [0,1,1,2,3,3,3,2,2,1,1,0]
^ ^ ^ ^ ^ ^
The interesting values of min(b[i], c[i]) which correspond to 0s are marked with ^. Summing it up yields: 0 + 1 + 3 + 2 + 1 + 0 = 7.
Indeed:
[0,1,0,1,1,0,1,0,1,0,1,0]
[0,0,1,1,1,0,1,0,1,0,1,0] 1
[0,0,1,1,1,0,1,0,1,1,0,0] 2 = 1 + 1
[0,0,1,1,1,0,1,1,0,1,0,0] 3
[0,0,1,1,1,0,1,1,1,0,0,0] 4 = 1 + 1 + 2
[0,0,1,1,0,1,1,1,1,0,0,0] 5
[0,0,1,0,1,1,1,1,1,0,0,0] 6
[0,0,0,1,1,1,1,1,1,0,0,0] 7 = 1 + 1 + 2 + 3
done: block of ones homogeneous.
Runtime for computation of the number res of swaps is obviously O(n). (Note: it does NOT say that the number of swaps is itself O(n)).
Let's consider each 1 as a potential static point. Then the cost for the left side of the static point would be the number of 1's to the left subtracted by the number of 1's already in the section it would naturally extend to, the length of which is the number of 1's on the left. Similarly for the right side.
Now find a way to do it efficiently for each potential static 1 :) Hint: think about how we could update those values as we iterate across the array.
1 0 1 0 1 1 0 0 1 0 1 1
x potential static point
<----- would extend to
-----> would extend to
left cost at x: 3 - 2 = 1
right cost at x: 3 - 1 = 2
Related
I found this problem in a hiring contest(which is over now). Here it is:
You are given two natural numbers N and X. You are required to create an array of N natural numbers such that the bitwise XOR of these numbers is equal to X. The sum of all the natural numbers that are available in the array is as minimum as possible.
If there exist multiple arrays, print the smallest one
Array A< Array B if
A[i] < B[i] for any index i, and A[i]=B[i] for all indices less than i
Sample Input: N=3, X=2
Sample output : 1 1 2
Explanation: We have to print 3 natural numbers having the minimum sum Thus the N-spaced numbers are [1 1 2]
My approach:
If N is odd, I put N-1 ones in the array (so that their xor is zero) and then put X
If N is even, I put N-1 ones again and then put X-1(if X is odd) and X+1(if X is even)
But this algorithm failed for most of the test cases. For example, when N=4 and X=6 my output is
1 1 1 7 but it should be 1 1 2 4
Anyone knows how to make the array sum minimum?
In order to have the minimum sum, you need to make sure that when your target is X, you are not cancelling the bits of X and recreating them again. Because this will increase the sum. For this, you have create the bits of X one by one (ideally) from the end of the array. So, as in your example of N=4 and X=6 we have: (I use ^ to show xor)
X= 7 = 110 (binary) = 2 + 4. Note that 2^4 = 6 as well because these numbers don't share any common bits. So, the output is 1 1 2 4.
So, we start by creating the most significant bits of X from the end of the output array. Then, we also have to handle the corner cases for different values of N. I'm going with a number of different examples to make the idea clear:
``
A) X=14, N=5:
X=1110=8+4+2. So, the array is 1 1 2 4 8.
B) X=14, N=6:
X=8+4+2. The array should be 1 1 1 1 2 12.
C) X=15, N=6:
X=8+4+2+1. The array should be 1 1 1 2 4 8.
D) X=15, N=5:
The array should be 1 1 1 2 12.
E) X=14, N=2:
The array should be 2 12. Because 12 = 4^8
``
So, we go as follows. We compute the number of powers of 2 in X. Let this number be k.
Case 1 - If k <= n (example E): we start by picking the smallest powers from left to right and merge the remaining on the last position in the array.
Case 2 - If k > n (example A, B, C, D): we compute h = n - k. If h is odd we put h = n-k+1. Now, we start by putting h 1's in the beginning of the array. Then, the number of places left is less than k. So, we can follow the idea of Case 1 for the remaining positions. Note that in case 2, instead of having odd number of added 1's we put and even number of 1's and then do some merging at the end. This guarantees that the array is the smallest it can be.
We have to consider that we have to minimize the sum of the array for solution and that is the key point.
First calculate set bits in N suppose if count of setbits are less than or equal to X then divide N in X integers based on set bits like
N = 15, X = 2
setbits in 15 are 4 solution is 1 14
if X = 3 solution is 1 2 12
this minimizes array sum too.
other case if setbits are greater than X
calculate difference = setbits(N) - X
If difference is even then add ones as needed and apply above algorithm all ones will cancel out.
If difference is odd then add ones but now you have take care of that 1 extra one in the answer array.
Check for the corner cases too.
I came across an interesting problem and I can't solve it in a good complexity (better than O(qn)):
There are n persons in a row. Initially every person in this row has some value - lets say that i-th person has value a_i. These values are pairwise distinct.
Every person gets a mark. There are two conditions:
If a_i < a_j then j-th person cant get worse mark than i-th person.
If i < j then j-th person can't get worse mark than i-th person (this condition tells us that sequence of marks is non-decreasing sequence).
There are q operations. In every operation two person are swapped (they swap their values).
After each operation you have tell what is maximal number of diffrent marks that these n persons can get.
Do you have any idea?
Consider any two groups, J and I (j < i and a_j < a_i for all j and i). In any swap scenario, a_i is the new max for J and a_j is the new min for I, and J gets extended to the right at least up to and including i.
Now if there was any group of is to the right of i whos values were all greater than the values in the left segment of I up to i, this group would not have been part of I, but rather its own group or part of another group denoting a higher mark.
So this kind of swap would reduce the mark count by the count of groups between J and I and merge groups J up to I.
Now consider an in-group swap. The only time a mark would be added is if a_i and a_j (j < i), are the minimum and maximum respectively of two adjacent segments, leading to the group splitting into those two segments. Banana123 showed in a comment below that this condition is not sufficient (e.g., 3,6,4,5,1,2 => 3,1,4,5,6,2). We can address this by also checking before the switch that the second smallest i is greater than the second largest j.
Banana123 also showed in a comment below that more than one mark could be added in this instance, for example 6,2,3,4,5,1. We can handle this by keeping in a segment tree a record of min,max and number of groups, which correspond with a count of sequential maxes.
Example 1:
(1,6,1) // (min, max, group_count)
(3,6,1) (1,4,1)
(6,6,1) (3,5,1) (4,4,1) (1,2,1)
6 5 3 4 2 1
Swap 2 and 5. Updates happen in log(n) along the intervals containing 2 and 5.
To add group counts in a larger interval the left group's max must be lower than the right group's min. But if it's not, as in the second example, we must check one level down in the tree.
(1,6,1)
(2,6,1) (1,5,1)
(6,6,1) (2,3,2) (4,4,1) (1,5,1)
6 2 3 4 5 1
Swap 1 and 6:
(1,6,6)
(1,3,3) (4,6,3)
(1,1,1) (2,3,2) (4,4,1) (5,6,2)
1 2 3 4 5 6
Example 2:
(1,6,1)
(3,6,1) (1,4,1)
(6,6,1) (3,5,1) (4,4,1) (1,2,1)
6 5 3 4 2 1
Swap 1 and 6. On the right side, we have two groups where the left group's max is greater than the right group's min, (4,4,1) (2,6,2). To get an accurate mark count, we go down a level and move 2 into 4's group to arrive at a count of two marks. A similar examination is then done in the level before the top.
(1,6,3)
(1,5,2) (2,6,2)
(1,1,1) (3,5,1) (4,4,1) (2,6,2)
1 5 3 4 2 6
Here's an O(n log n) solution:
If n = 0 or n = 1, then there are n distinct marks.
Otherwise, consider the two "halves" of the list, LEFT = [1, n/2] and RIGHT = [n/2 + 1, n]. (If the list has an odd number of elements, the middle element can go in either half, it doesn't matter.)
Find the greatest value in LEFT — call it aLEFT_MAX — and the least value in the second half — call it aRIGHT_MIN.
If aLEFT_MAX < aRIGHT_MIN, then there's no need for any marks to overlap between the two, so you can just recurse into each half and return the sum of the two results.
Otherwise, we know that there's some segment, extending at least from LEFT_MAX to RIGHT_MIN, where all elements have to have the same mark.
To find the leftmost extent of this segment, we can scan leftward from RIGHT_MIN down to 1, keeping track of the minimum value we've seen so far and the position of the leftmost element we've found to be greater than some further-rightward value. (This can actually be optimized a bit more, but I don't think we can improve the algorithmic complexity by doing so, so I won't worry about that.) And, conversely to find the rightmost extent of the segment.
Suppose the segment in question extends from LEFTMOST to RIGHTMOST. Then we just need to recursively compute the number of distinct marks in [1, LEFTMOST) and in (RIGHTMOST, n], and return the sum of the two results plus 1.
I wasn't able to get a complete solution, but here are a few ideas about what can and can't be done.
First: it's impossible to find the number of marks in O(log n) from the array alone - otherwise you could use your algorithm to check if the array is sorted faster than O(n), and that's clearly impossible.
General idea: spend O(n log n) to create any additional data which would let you to compute number of marks in O(log n) time and said data can be updated after a swap in O(log n) time. One possibly useful piece to include is the current number of marks (i.e. finding how number of marks changed may be easier than to compute what it is).
Since update time is O(log n), you can't afford to store anything mark-related (such as "the last person with the same mark") for each person - otherwise taking an array 1 2 3 ... n and repeatedly swapping first and last element would require you to update this additional data for every element in the array.
Geometric interpretation: taking your sequence 4 1 3 2 5 7 6 8 as an example, we can draw points (i, a_i):
|8
+---+-
|7 |
| 6|
+-+---+
|5|
-------+-+
4 |
3 |
2|
1 |
In other words, you need to cover all points by a maximal number of squares. Corollary: exchanging points from different squares a and b reduces total number of squares by |a-b|.
Index squares approach: let n = 2^k (otherwise you can add less than n fictional persons who will never participate in exchanges), let 0 <= a_i < n. We can create O(n log n) objects - "index squares" - which are "responsible" for points (i, a_i) : a*2^b <= i < (a+1)*2^b or a*2^b <= a_i < (a+1)*2^b (on our plane, this would look like a cross with center on the diagonal line a_i=i). Every swap affects only O(log n) index squares.
The problem is, I can't find what information to store for each index square so that it would allow to find number of marks fast enough? all I have is a feeling that such approach may be effective.
Hope this helps.
Let's normalize the problem first, so that a_i is in the range of 0 to n-1 (can be achieved in O(n*logn) by sorting a, but just hast to be done once so we are fine).
function normalize(a) {
let b = [];
for (let i = 0; i < a.length; i++)
b[i] = [i, a[i]];
b.sort(function(x, y) {
return x[1] < y[1] ? -1 : 1;
});
for (let i = 0; i < a.length; i++)
a[b[i][0]] = i;
return a;
}
To get the maximal number of marks we can count how many times
i + 1 == mex(a[0..i]) , i integer element [0, n-1]
a[0..1] denotes the sub-array of all the values from index 0 to i.
mex() is the minimal exclusive, which is the smallest value missing in the sequence 0, 1, 2, 3, ...
This allows us to solve a single instance of the problem (ignoring the swaps for the moment) in O(n), e.g. by using the following algorithm:
// assuming values are normalized to be element [0,n-1]
function maxMarks(a) {
let visited = new Array(a.length + 1);
let smallestMissing = 0, marks = 0;
for (let i = 0; i < a.length; i++) {
visited[a[i]] = true;
if (a[i] == smallestMissing) {
smallestMissing++;
while (visited[smallestMissing])
smallestMissing++;
if (i + 1 == smallestMissing)
marks++;
}
}
return marks;
}
If we swap the values at indices x and y (x < y) then the mex for all values i < x and i > y doesn't change, although it is an optimization, unfortunately that doesn't improve complexity and it is still O(qn).
We can observe that the hits (where mark is increased) are always at the beginning of an increasing sequence and all matches within the same sequence have to be a[i] == i, except for the first one, but couldn't derive an algorithm from it yet:
0 6 2 3 4 5 1 7
*--|-------|*-*
3 0 2 1 4 6 5 7
-|---|*-*--|*-*
This question already has answers here:
Take K elements and maximise the minimum distance
(2 answers)
Closed 7 years ago.
We are given N elements in form of array A , Now we have to choose K indexes from N given indexes such that for any 2 indexes i and j minimum value of |A[i]-A[j]| is as large as possible. We need to tell this maximum value.
Lets take an example : Let N=5 and K=2 and array be [1,5,3,7,11] then here answer is 10 as we can simply choose first and last position and differ = 11-1=10.
Example 2 : Let N=10 and K=3 and array A be [3 9 6 11 15 20 23] then here answer will be 8. As we can select [3,11,23] or [3,15,23].
Now given N , K and Array A we need to find this maximum difference.
We are given that 1 ≤ N ≤ 10^5 and 1 ≤ S ≤ 10^7
Let's sort the array.
Now we can do a binary search over the answer.
For a fixed candidate x, we can just pick the elements greedily(iterating over the sorted array and taking each element if we can). If the number of elements we have picked is not less than K, x is feasible. Otherwise, it is not.
The time complexity is O(N * log N + N * log (MAX_ELEMENT - MIN_ELEMENT))
A pseudo code:
bool isFeasible(int x):
cnt = 1
last = a[0]
for i <- 1 ... n - 1:
if a[i] - last >= x:
last = a[i]
cnt++
return cnt >= k
sort(a)
low = 0
high = a[n - 1] - a[0] + 1
while high - low > 1:
mid = low + (high - low) / 2
if isFeasible(mid):
low = mid
else
high = mid
print(low)
I think this can be dealt with as a dynamic programming problem. Start off by sorting A, and then the problem is to mark K elements in A such that the minimum difference between adjacent marked items is as large as possible. As a starter, you can always mark the first and last elements.
Moving from left to right, at each position for i=1..N work out the largest minimum difference you can get by marking i elements in the sub-array terminating at this position. You can work out the largest minimum difference for k items terminating at this position by considering the largest minimum difference for k-1 items terminating at each position to the left of the position you are working on. The obvious thing to do is to consider each possible position up to the position you are currently working on as ending a stretch of k-1 items with minimum difference, but you may be able to do a binary search here to speed things up.
Once you have worked all the way to the right hand end you know the maximum possible value for the original problem. If you need to know where to put the K elements, you can take notes as you go along so that you can backtrack to find out the elements chosen that lead to this solution, working from right to left.
You are given N and an int K[].
The task at hand is to generate a equal probabilistic random number between 0 to N-1 which doesn't exist in K.
N is strictly a integer >= 0.
And K.length is < N-1. And 0 <= K[i] <= N-1. Also assume K is sorted and each element of K is unique.
You are given a function uniformRand(int M) which generates uniform random number in the range 0 to M-1 And assume this functions's complexity is O(1).
Example:
N = 7
K = {0, 1, 5}
the function should return any random number { 2, 3, 4, 6 } with equal
probability.
I could get a O(N) solution for this : First generate a random number between 0 to N - K.length. And map the thus generated random number to a number not in K. The second step will take the complexity to O(N). Can it be done better in may be O(log N) ?
You can use the fact that all the numbers in K[] are between 0 and N-1 and they are distinct.
For your example case, you generate a random number from 0 to 3. Say you get a random number r. Now you conduct binary search on the array K[].
Initialize i = K.length/2.
Find K[i] - i. This will give you the number of numbers missing from the array in the range 0 to i.
For example K[2] = 5. So 3 elements are missing from K[0] to K[2] (2,3,4)
Hence you can decide whether you have to conduct the remaining search in the first part of array K or the next part. This is because you know r.
This search will give you a complexity of log(K.length)
EDIT: For example,
N = 7
K = {0, 1, 4} // modified the array to clarify the algorithm steps.
the function should return any random number { 2, 3, 5, 6 } with equal probability.
Random number generated between 0 and N-K.length = random{0-3}. Say we get 3. Hence we require the 4th missing number in array K.
Conduct binary search on array K[].
Initial i = K.length/2 = 1.
Now we see K[1] - 1 = 0. Hence no number is missing upto i = 1. Hence we search on the latter part of the array.
Now i = 2. K[2] - 2 = 4 - 2 = 2. Hence there are 2 missing numbers up to index i = 2. But we need the 4th missing element. So we again have to search in the latter part of the array.
Now we reach an empty array. What should we do now? If we reach an empty array between say K[j] & K[j+1] then it simply means that all elements between K[j] and K[j+1] are missing from the array K.
Hence all elements above K[2] are missing from the array, namely 5 and 6. We need the 4th element out of which we have already discarded 2 elements. Hence we will choose the second element which is 6.
Binary search.
The basic algorithm:
(not quite the same as the other answer - the number is only generated at the end)
Start in the middle of K.
By looking at the current value and it's index, we can determine the number of pickable numbers (numbers not in K) to the left.
Similarly, by including N, we can determine the number of pickable numbers to the right.
Now randomly go either left or right, weighted based on the count of pickable numbers on each side.
Repeat in the chosen subarray until the subarray is empty.
Then generate a random number in the range consisting of the numbers before and after the subarray in the array.
The running time would be O(log |K|), and, since |K| < N-1, O(log N).
The exact mathematics for number counts and weights can be derived from the example below.
Extension with K containing a bigger range:
Now let's say (for enrichment purposes) K can also contain values N or larger.
Then, instead of starting with the entire K, we start with a subarray up to position min(N, |K|), and start in the middle of that.
It's easy to see that the N-th position in K (if one exists) will be >= N, so this chosen range includes any possible number we can generate.
From here, we need to do a binary search for N (which would give us a point where all values to the left are < N, even if N could not be found) (the above algorithm doesn't deal with K containing values greater than N).
Then we just run the algorithm as above with the subarray ending at the last value < N.
The running time would be O(log N), or, more specifically, O(log min(N, |K|)).
Example:
N = 10
K = {0, 1, 4, 5, 8}
So we start in the middle - 4.
Given that we're at index 2, we know there are 2 elements to the left, and the value is 4, so there are 4 - 2 = 2 pickable values to the left.
Similarly, there are 10 - (4+1) - 2 = 3 pickable values to the right.
So now we go left with probability 2/(2+3) and right with probability 3/(2+3).
Let's say we went right, and our next middle value is 5.
We are at the first position in this subarray, and the previous value is 4, so we have 5 - (4+1) = 0 pickable values to the left.
And there are 10 - (5+1) - 1 = 3 pickable values to the right.
We can't go left (0 probability). If we go right, our next middle value would be 8.
There would be 2 pickable values to the left, and 1 to the right.
If we go left, we'd have an empty subarray.
So then we'd generate a number between 5 and 8, which would be 6 or 7 with equal probability.
This can be solved by basically solving this:
Find the rth smallest number not in the given array, K, subject to
conditions in the question.
For that consider the implicit array D, defined by
D[i] = K[i] - i for 0 <= i < L, where L is length of K
We also set D[-1] = 0 and D[L] = N
We also define K[-1] = 0.
Note, we don't actually need to construct D. Also note that D is sorted (and all elements non-negative), as the numbers in K[] are unique and increasing.
Now we make the following claim:
CLAIM: To find the rth smallest number not in K[], we need to find right most occurrence of r' in D (which occurs at position defined by j), where r' is the largest number in D, which is < r. Such an r' exists, because D[-1] = 0. Once we find such an r' (and j), the number we are looking for is r-r' + K[j].
Proof: Basically the definition of r' and j tells us that there are exactlyr' numbers missing from 0 to K[j], and more than r numbers missing from 0 to K[j+1]. Thus all the numbers from K[j]+1 to K[j+1]-1 are missing (and these missing are at least r-r' in number), and the number we seek is among them, given by K[j] + r-r'.
Algorithm:
In order to find (r',j) all we need to do is a (modified) binary search for r in D, where we keep moving to the left even if we find r in the array.
This is an O(log K) algorithm.
If you are running this many times, it probably pays to speed up your generation operation: O(log N) time just isn't acceptable.
Make an empty array G. Starting at zero, count upwards while progressing through the values of K. If a value isn't in K add it to G. If it is in K don't add it and progress your K pointer. (This relies on K being sorted.)
Now you have an array G which has only acceptable numbers.
Use your random number generator to choose a value from G.
This requires O(N) preparatory work and each generation happens in O(1) time. After N look-ups the amortized time of all operations is O(1).
A Python mock-up:
import random
class PRNG:
def __init__(self, K,N):
self.G = []
kptr = 0
for i in range(N):
if kptr<len(K) and K[kptr]==i:
kptr+=1
else:
self.G.append(i)
def getRand(self):
rn = random.randint(0,len(self.G)-1)
return self.G[rn]
prng=PRNG( [0,1,5], 7)
for i in range(20):
print prng.getRand()
I found this interview question, and I couldn't come up with an algorithm better than O(N^2 * P):
Given a vector of P natural numbers (1,2,3,...,P) and another vector of length N whose elements are from the first vector, find the longest subsequence in the second vector, such that all elements are uniformly distributed (have the same frequency).
Example : (1,2,3) and (1,2,1,3,2,1,3,1,2,3,1). The longest subsequence is in the interval [2,10], because it contains all the elements from the first sequence with the same frequency (1 appears three times, 2 three times, and 3 three times).
The time complexity should be O(N * P).
"Subsequence" usually means noncontiguous. I'm going to assume that you meant "sublist".
Here's an O(N P) algorithm assuming we can hash (assumption not needed; we can radix sort instead). Scan the array keeping a running total for each number. For your example,
1 2 3
--------
0 0 0
1
1 0 0
2
1 1 0
1
2 1 0
3
2 1 1
2
2 2 1
1
3 2 1
3
3 2 2
1
4 2 2
2
4 3 2
3
4 3 3
1
5 3 3
Now, normalize each row by subtracting the minimum element. The result is
0: 000
1: 100
2: 110
3: 210
4: 100
5: 110
6: 210
7: 100
8: 200
9: 210
10: 100
11: 200.
Prepare two hashes, mapping each row to the first index at which it appears and the last index at which it appears. Iterate through the keys and take the one with maximum last - first.
000: first is at 0, last is at 0
100: first is at 1, last is at 10
110: first is at 2, last is at 5
210: first is at 3, last is at 9
200: first is at 8, last is at 11
The best key is 100, since its sublist has length 9. The sublist is the (1+1)th element to the 10th.
This works because a sublist is balanced if and only if its first and last unnormalized histograms are the same up to adding a constant, which occurs if and only if the first and last normalized histograms are identical.
If the memory usage is not important, it's easy...
You can give the matrix dimensions N*p and save in column (i) the value corresponding to how many elements p is looking between (i) first element in the second vector...
After completing the matrix, you can search for column i that all of the elements in column i are not different. The maximum i is the answer.
With randomization, you can get it down to linear time. The idea is to replace each of the P values with a random integer, such that those integers sum to zero. Now look for two prefix sums that are equal. This allows some small chance of false positives, which we could remedy by checking our output.
In Python 2.7:
# input:
vec1 = [1, 2, 3]
P = len(vec1)
vec2 = [1, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1]
N = len(vec2)
# Choose big enough integer B. For each k in vec1, choose
# a random mod-B remainder r[k], so their mod-B sum is 0.
# Any P-1 of these remainders are independent.
import random
B = N*N*N
r = dict((k, random.randint(0,B-1)) for k in vec1)
s = sum(r.values())%B
r[vec1[0]] = (r[vec1[0]]+B-s)%B
assert sum(r.values())%B == 0
# For 0<=i<=N, let vec3[i] be mod-B sum of r[vec2[j]], for j<i.
vec3 = [0] * (N+1)
for i in range(1,N+1):
vec3[i] = (vec3[i-1] + r[vec2[i-1]]) % B
# Find pair (i,j) so vec3[i]==vec3[j], and j-i is as large as possible.
# This is either a solution (subsequence vec2[i:j] is uniform) or a false
# positive. The expected number of false positives is < N*N/(2*B) < 1/N.
(i, j)=(0, 0)
first = {}
for k in range(N+1):
v = vec3[k]
if v in first:
if k-first[v] > j-i:
(i, j) = (first[v], k)
else:
first[v] = k
# output:
print "Found subsequence from", i, "(inclusive) to", j, "(exclusive):"
print vec2[i:j]
print "This is either uniform, or rarely, it is a false positive."
Here is an observation: you can't get a uniformly distributed sequence that is not a multiplication of P in length. This implies that you only have to check the sub-sequences of N that are P, 2P, 3P... long - (N/P)^2 such sequences.
You can get this down to O(N) time, with no dependence on P by enhancing uty's solution.
For each row, instead of storing the normalized counts of each element, store a hash of the normalized counts while only keeping the normalized counts for the current index. During each iteration, you need to first update the normalized counts, which has an amortized cost of O(1) if each decrement of a count is paid for when it is incremented. Next you recompute the hash. The key here is that the hash needs to be easily updatable following an increment or decrement of one of the elements of the tuple that is being hashed.
At least one way of doing this hashing efficiently, with good theoretical independence guarantees is shown in the answer to this question. Note that the O(lg P) cost for computing the exponential to determine the amount to add to the hash can be eliminated by precomputing the exponentials modulo the prime in with a total running time of O(P) for the precomputation, giving a total running time of O(N + P) = O(N).