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Practice
Contest

DIFFICULTY

EASY

PREREQUISITES

Data Structure, Double Ended Queue

PROBLEM

Given a permutation of numbers from 1 to N, how many segments of length W exist such that

• Suppose the segment S of length W starts from L and ends at R, where R-L+1 = W
• We sort the items in S in increasing order
• Now, S should satisfy S[i] = S[i-1]+1, for i between 2 and K, inclusive (assuming S was 1-based)

QUICK EXPLANATION

It is easy to see that a segment is valid iff
The largest value in the segment - the smallest value in the segment + 1 = W.

Now, we can find the largest and the smallest value in amortised O(1) time for all the N-W+1 segments and then in one parse, find the number of valid segments. The overall complexity of this algorithm is O(N).

EXPLANATION

We have to find the smallest value in all N-W+1 segments of size W. To do so, we want to build a Queue with getMax capability.

We wish to use a queue to mimic the sliding window behaviour of considering segments, one after the other. And if the queue can efficiently tell us the minimum number inside the queue at any time, we are done. Thus our queue Q should have the following methods

• push: push an item in the queue
• pop: pop an item from the queue in FIFO order
• min: return the minimum value in the queue at this point

Let us implement Q using an internal double ended queue dQ, and an internal simple FIFO queue iQ.

About dQ

• dQ.size is the number of elements in dQ
• dQ.head is the first element in dQ
• dQ.tail is the last element in dQ
• dQ.push_front pushes the element at the head and changes head to this value
• dQ.pop_front pops the element at the head and shifts the head forward
• dQ.push_back pushes the element at the tail and changes the tail to this value
• dQ.pop_back pops the element at the tail and shifts the tail behind

About iQ

• iQ.size is the number of elements in iQ
• iQ.push pushes an element in iQ
• iQ.pop pops an element from iQ in FIFO order

Q.push is implemented as

function Q.push(item)
    while dQ.size && dQ.tail > item
        dQ.pop_back()
    dQ.push_back(item)
    iQ.push(item)

Q.pop is implemented as

function Q.pop()
    retVal = iQ.pop()
    if retVal == dQ.head then dQ.pop_front()
    return retVal

Q.min is implemented as

function Q.min()
    return dQ.head

Now, we make the following observations about Q

• iQ has all the elements inside Q
• dQ has a subset of elements, in Q, in increasing order
  • dQ.head has the smallest element in Q
  • followed by the next smallest element which was inserted after inserting the smallest element
  • so on, till dQ.tail, which has the last element inserted in Q
• If the current item to pop is the smallest item in Q, then we do a iQ.pop() as well as dQ.pop_front(). This ensures that we have the smallest value after the pop operation
• If the last item (or last few items) which was inserted are larger than the current item, then their values are not relevant towards finding the minimum value, since
  • They will be popped before popping the currently inserted item (FIFO order)
  • They cannot be the smallest value since the current item's value is smaller

You can work out several examples to see that this data structure works and returns the smallest value in Q at all times.

It remains to show that this is fast enough for this problem.

We will find the smallest values for each segment and store in an array Mi, by using Q as follows

Given: A[N]
Let Mi be an array of N-W+1 values
for i in 1 to W, inclusive
    Q.push(A[i])
Mi[1] = Q.min()
for i in W+1 to N, inclusive
    Q.push(A[i])
    Q.pop()
    Mi[i-W] = Q.min()

We see that Q.push is being called in a loop, and Q.push iterates over the length of dQ inside. But, if Q.push makes more than 1 comparison over dQ, it also pops items from dQ. Since each item inserted in dQ can only be popped once, there can be at most N pops through all the iterations.

That means that the sum of the number of comparisons made inside Q.push while it is being called in the loop from W+1 to N, will not be more than 2*N. The overall complexity of the algorithm to find the minimum value in each segment remains O(N).

Using the same ideas as above, you can build Ma[N], an array of the maximum values in each segment.

Following this, the result can be calculated as

result = 0
for i in 1 to N-W+1, inclusive
    if Ma[i] - Mi[i] + 1 == W then result = result + 1
print result

SETTERS SOLUTION

Can be found here

TESTERS SOLUTION

Can be found here