WOUT - Editorial

aug15
dynamic-programming
easy
editorial
prefix-sum
subarray

#1

PROBLEM LINK:

Contest
Practice

Author: Vitalij Kozhukhivskij
Tester: Hiroto Sekido
Editorialist: Kevin Atienza

DIFFICULTY:

Easy

PREREQUISITES:

Dynamic programming, subarray sums, prefix sums, segment trees

PROBLEM:

There is a grid of N imes N cells. The N blocks of each column are made up of soil, except for a contiguous sequence of cells: from the $l_i$th cell to the $h_i$th cell (starting from the bottom, 0-indexing). Cells can be cleared of soil.

We need to create a subgrid of length N and height H containing no soil. What is the minimum number of cells needed to be cleared of soil?

QUICK EXPLANATION:

There are only N-H+1 possible N imes H subgrids. The answer is the the minimum number of soil cells among all such subgrids. We can preprocess our grid so we can compute the sum of each subgrid in constant time.

To preprocess:

  • We need to know the number of soil cells in each row (denoted by r_i for 0 \le i < N). r_i is equal to the number of $j$s such that 0 \le j < N and i < l_j or i > h_j. This is also equal to the following expression:
N - \#\{j : 0 \le j < N, i \le h_j\} + \#\{j : 0 \le j < N, i < l_j\}

Thus the $r_i$s can be computed quickly by considering the $l_j$s and $h_j$s in sorted order.

  • We also need to know the number of soil cells in each prefix of rows, i.e. let s_i be the sum r_0 + r_1 + \cdots + r_{i-1}.

Then the number of soil cells in each N imes H subgrid can be computed as s_i - s_{i-H} for H \le i \le N.

EXPLANATION:

Clearly, we need to find the N imes H subgrid with the minimum number of soil cells in it. There are only N-H+1 such subgrids: the following illustrates the case N = 7 and H = 3 (there are N-H+1 = 5 subgrids):

.......          #######   .......   .......   .......   .......
.......          #######   #######   .......   .......   .......
.......          #######   #######   #######   .......   .......
....... -------> .......   #######   #######   #######   .......
.......          .......   .......   #######   #######   #######
.......          .......   .......   .......   #######   #######
.......          .......   .......   .......   .......   #######

We can simply construct the N imes N grid, and compute the number of soil cells in these subgrids. Since there are N-H+1 subgrids and each one has NH cells to check, this algorithm runs in O((N-H+1)NH) time. The worst case is when H is around half of N (which makes the running time O(N^3)), so unfortunately this algorithm is only good for the first subtask. For the second subtask, you can’t even store the whole grid due to the memory requirements!

To answer the second subtask, we need a way to sum up these subgrids without constructing the whole grid. The first thing we notice is that the only information we need from its row is the number of soil cells in it, i.e. we don’t need to know their positions in the row. Let’s say the $i$th row (0 \le i < N) contains r_i soil cells. Then the number of soil cells in each of the N-H+1 subgrids are the following:

  • r_0 + r_1 + r_2 + \cdots + r_{H-1}
  • r_1 + r_2 + r_3 + \cdots + r_H
  • r_2 + r_3 + r_4 + \cdots + r_{H+1}
  • r_3 + r_4 + r_5 + \cdots + r_{H+2}
  • \ldots
  • r_{N-H} + r_{N-H+1} + r_{N-H+3} + \cdots + r_{N-1}

The smallest of these is the answer! Thus, it would be very helpful if we can compute the sequence r_0, r_1, \ldots, r_{N-1} quickly, without constructing the whole grid!

To do so, we use the following observation: r_i is equal to the number of $j$s such that 0 \le j < N and i < l_j or i > h_j. Now, counting all such $j$s this way is still not fast enough, so we do some manipulations first. For a statement \phi(j), let C_{\phi(j)} be the number of $j$s such that 0 \le j < N and \phi(j) is true. To familiarize yourself with this notation, we give a few basic facts (we invite you to verify each one):

  • C_{ ext{true}} = N
  • C_{ ext{false}} = 0
  • C_{\phi(j)} + C_{ ext{not }\phi(j)} = C_{ ext{true}} = N
  • C_{\phi(j)} = C_{ ext{true}} - C_{ ext{not }\phi(j)} = N - C_{ ext{not }\phi(j)}
  • C_{\phi_1(j) ext{ or } \phi_2(j)} = C_{\phi_1(j)} + C_{\phi_2(j)} - C_{\phi_1(j) ext{ and } \phi_2(j)}
  • C_{c < f(j)} + C_{c = f(j)} + C_{c > f(j)} = N (trichotomy)
  • C_{c < f(j)} + C_{c = f(j)} = C_{c \le f(j)}
  • If f_1(j) \le f_2(j) for all j, then C_{f_1(j) \le c \le f_2(j)} = C_{c \le f_2(j)} - C_{c < f_1(j)}

Now, back to r_i. We have the following (using some of the facts above:

\begin{align*} r_i &= C_{i < l_j ext{ or } i > h_j} \\\ &= N - C_{ ext{not } (i < l_j ext{ or } i > h_j)} \\\ &= N - C_{i \ge l_j ext{ and } i \le h_j} \\\ &= N - C_{l_j \le i \le h_j} \\\ &= N - (C_{i \le h_j} - C_{i < l_j}) \end{align*}

The last one is true because l_j \le h_j. Now, we have:

r_i = N - C_{i \le h_j} + C_{i < l_j}

To compute the $r_i$s, we just need to compute the quantity - C_{i \le h_j} + C_{i < l_j} for all 0 \le i < N.

The key to this is to notice that r_i can be computed by a simple adjustment from r_{i-1}! In other words, we can just calculate the difference r_i - r_{i-1}, and if we have already computed r_{i-1}, then we can calculate r_i by adding this difference. In more detail, let’s try to compute r_i - r_{i-1}:

\begin{align*} r_i - r_{i-1} &= (N - C_{i \le h_j} + C_{i < l_j}) - (N - C_{i-1 \le h_j} + C_{i-1 < l_j}) \\\ &= N - C_{i \le h_j} + C_{i \le l_j-1} - N + C_{i \le h_j+1} - C_{i \le l_j} \\\ &= C_{i \le h_j+1} - C_{i \le h_j} + C_{i \le l_j-1} - C_{i \le l_j} \\\ &= (C_{i \le h_j+1} - C_{i \le h_j}) + (C_{i \le l_j-1} - C_{i \le l_j}) \\\ &= C_{i = h_j+1} - C_{i = l_j} \end{align*}

But we can compute the values C_{i = h_j+1} and C_{i = l_j} for 0 \le i < N quickly, via a linear pass of all pairs (l_j,h_j) for 0 \le j < N! The following pseudocode does it:

# arrays are initialized with zeroes
Ch[0...N]   # Ch* will contain C[i = h_j + 1]
Cl[0...N]   # Ch* will contain C[i = l_j]
for j = 0...N-1:
    Ch[h_j + 1] += 1
    Cl[l_j] += 1

Now that all the C_{i = h_j+1} and C_{i = l_j} are computed, we can now compute all the $r_i$s using the following recurrence:

r_{-1} = N
r_i = r_{i-1} + C_{i = h_j+1} - C_{i = l_j}

The following pseudocode does it:

# array is initialized with zeroes
r[0...N]
curr = N
for i in 0...N-1:
    curr += Ch* - Cl*
    r* = curr

Clearly, these pseudocodes run in O(N) time!

Finally, to compute the answer, we need to know the following sums:

  • r_0 + r_1 + r_2 + \cdots + r_{H-1}
  • r_1 + r_2 + r_3 + \cdots + r_H
  • r_2 + r_3 + r_4 + \cdots + r_{H+1}
  • r_3 + r_4 + r_5 + \cdots + r_{H+2}
  • \ldots
  • r_{N-H} + r_{N-H+1} + r_{N-H+3} + \cdots + r_{N-1}

and then compute the minimum among them. But this is easy! Notice that r_i + r_{i+1} + \cdots + r_j is simply (r_0 + \cdots + r_j) - (r_0 + \cdots + r_{i-1}), so we can first try computing the prefix sums. Let s_i be the sum r_0 + r_1 + \cdots + r_{i-1}. Then r_i + r_{i+1} + \cdots + r_j is simply s_{j+1} - s_i. The $s_i$s can be computed in O(N) too, because s_i = s_{i-1} + r_{i-1}, with the base case s_0 = 0. Afterwards, the sums we need are simply:

  • s_H - s_0
  • s_{H+1} - s_1
  • s_{H+2} - s_2
  • \ldots
  • s_N - s_{N-H}

Since all the steps of this algorithm runs is O(N), the answer can thus be computed in O(N) time in total!

The following is a sample implementation in Python:

for cas in xrange(input()):
    n, h = map(int, raw_input().strip().split())
    row = [0]*(n+2)
    for i in xrange(n):
        a, b = map(int, raw_input().strip().split())
        row[a+1] -= 1
        row[b+2] += 1
    row[0] = n
    for i in xrange(n): row[i+1] += row*
    for i in xrange(n): row[i+1] += row*
    print min(row* - row[i-h] for i in xrange(h,n+1))

A few things to notice about this implementation:

  • The pairs (l_j, h_j) are never stored in an array: they are obtained from the input on the fly, processed, and thrown away immediately.
  • Instead of having two arrays for Ch* and Cl*, we only use a single array containing Ch* - Cl*.
  • We reuse the same array row to contain the values Ch* - Cl*, r_i and s_i. Furthermore, r_i is stored in index i+1 of row.

Time Complexity:

O(N)

AUTHOR’S AND TESTER’S SOLUTIONS:

setter
tester


#2

WAYOUT was a regular BIT/segment tree problem. To make the users think about O(n) solution rather than O(nlogn), the time limit could have been made strict, maybe 0.5 seconds or N<=10^6


#3

WAYOUT was a regular BIT/segment tree problem. To make the users think about O(n) solution rather than O(nlogn), the time limit could have been made strict, maybe 0.5 seconds or n<=10^6. BIT solution link https://www.codechef.com/viewsolution/7723896.


#4

Can someone please explain why my code gives WA on some of the test cases of subtask 2 WOUT


#5

Because everything is in int and ans can come out to be in long.


#6

@admin

but if long long int is required only by 2 or 3 variable why we declare all variable long long int…it slow down solution…because that can not handle at time in 64-bit machine it require double time compare int calculation because long long int 128-bit long variable cpu stored it into two-place and do calculation which simply increase the time required by machine to variable manipulation and calculation…
I think it is not good advice…you should only be aware which variable might get value out of range of int[-2^16 2^16-1]
integer range is quite long no need too worry much just take care of range only some cases…
Happy coding…


#7

why editorialists use too much complex thing which really hard to understand why cann’t they use some images or any other resources to easily visualization of things…

this question was quite easy who knows about BIT/segment tree data structure(I also know…): P
but moment I saw explanation I got stuck and think is this question that much difficult??

is there a way where user also can make editorial too??.1.

Happy coding


#8

Awesome solution! I dunno why the other users are complaining. Even I used a BIT solution to solve this. Thanks for showing me a faster way!


#9

Segment Tree solution of WayOut:-
https://www.codechef.com/viewsolution/7856183


#10

I used the same algorithm with Big(O) complexity but still I got TLE , I still don’t understand
Can anyone help me ? i am new to coding contests I solved a lot of practice problems although
https://www.codechef.com/viewsolution/7804230
https://www.codechef.com/viewsolution/7806640


#11

I didn’t understand the question during the contest. But reading the problem statement in the editorial made me understand. I solved without reading the editorial any further. Should have paid more attention in the contest. Btw why are you using BIT or segment tree? Simple prefix sum works too.

My solution


#12

What do we mean by this statement

Let C[ϕ(j)] be the number of js such that 0≤j<N and ϕ(j) is true.


#13

what do we mean by this statement

Let C[ϕ(j)] be the number of js such that 0≤j<N and ϕ(j) is true.


#14

Why my segment tree solution has not passed and given TLE please explain

https://www.codechef.com/viewsolution/7686894


#15

@nil96 your solution didn’t passed as you need to optimize it by avoiding use of long long int. Use int instead and at last you can use long long int.


#16

sorry can some one explain, how
C[i=hj+1] and C[i=lj] are computed in O(N) ? Aren’t they supposed to be computed for every row ? In which case it would cost around O(N^2)?
I understood the rest of the prefix sum and how it is O(N) though.
I am in learning stage, any explanation would be helpful


#17

Hi ,
I used the following approach for solving the problem without using bit/segment trees.

https://www.codechef.com/viewsolution/7726107

Hope this helps!!


#18

I am getting wrong answer in 2 sections of subtask 2. Could somebody look and point out what me be wrong.
Here is a link to my solution: https://www.codechef.com/submit/complete/11254238


#19

Why is r-1=N? Couldn’t get that.


#20

this is not a dp question too a far and large extent … why wrong tag ??