This post will be a tutorial on 313B - Ilya and Queries, as well as some general python I/O optimizations that provided me with enough of a speedup to have my solution finish in the allotted time.

Solution

As an example, consider the input string #..###. From it, we create arrays

• S = [ # . . # # # ]
• A = [ 0 1 0 1 1 X ] = [ a_1 a_2 a_3 a_4 a_5 a_6 ]

where

• S is an array encoding the input string.
• A[i] = 1 if S[i] == S[i+1] and 0 otherwise (A[-1] is undefined).

Let f(l, r) be a function where f(l, r) = w_{l} + w_{l+1} + ... + w{r-1}. Clearly f corresponds to the quantity of interest in 313B - Ilya and Queries (convince yourself if you don't see it!).

A naive solution is to compute f(l, r) iteratively (i.e. by looping from l to r-1). However, the runtime of this algorithm is O(nm) = O(n^2) as 0 <= m <= n. Hence, in the worst case we will do (10^5)^2 = 10^10 operations as 0 <= n <= 10^5, which it too large a number for the allotted time. Hence we need to do better.

One idea is to pre-compute f(l, r) for 1 <= l < r <= n. However, it turns out this table would have 10^10 elements in it, so it won't do.

It turns out that instead of pre-computing every possible f(l, r) pair, we can pre-compute a smaller number of values which we can use very quickly to compute any f(l, r) we want. Consider the following function F(r), where F(r) = w_1 + w_2 + ... + w_{r-1}. Notice thatf(l, r)=F(l)=F(r-1). To see, this considerf(2, 5)`:

`f(2, 5)` =         `w_2` + `w_3` + `w_4` 
          = `w_1` + `w_2` + `w_3` + `w_4`
          - `w_1`
          = F(5)
          - F(2)

Hence all that remains is to compute F(r) for 0 <= r <= n. Note that this cuts the runtime down by a factor of m down to O(n*1) = O(n) as F(l) — F(r) are constant-time lookups. It turns out this is fast enough!

I/O Optimizations

Even with this optimization, my python solution was still exceeding the time limit. After profiling my code, I deduced that the way I was handling I/O was the bottleneck. I replaced

for _ in range(m):
    l, r = [int(num) for num in raw_input().split()]
    print A[r] - A[l]

with

lines = sys.stdin.readlines()
inputs = [[int(num)-1 for num in line.split()] for line in lines]
outputs = [str(SA[r]-SA[l]) for l, r in inputs]
print '\n'.join(outputs)

Eliminating the repeated calls to raw_input() and print gave me a substantial speedup which was fast enough to have my solution finish within the allotted time!