# POSSPEW - Editorial

Tester: Samarth Gupta
Editorialist: Taranpreet Singh

Easy

None

# PROBLEM

Given a circular array A of length N with non-negative integers and an integer K, the following process is repeated every second

• For each i such that A_i \gt 0, elements adjacent to i-th element get incremented by 1.

Find the sum of the array A after K seconds.

# QUICK EXPLANATION

• Once an element becomes non-zero, it stays non-zero.
• Each non-zero element contributes 2 to the sum every second. If A_p becomes non-zero at time T, then it contributes 2*max(0, K-T) to the overall sum.
• For each element, we can calculate the first time it becomes non-zero by finding the nearest non-zero values in the given circular array.
• Only edge case is when all elements are 0. The sum remains 0 in this case.

# EXPLANATION

Instead of trying to maintain the actual array, let’s only keep relevant information. We only care about the sum of the array after K seconds, so let’s try computing that directly. The sum of the final array would be the sum of the initial array plus the increments from operations.

In each operation, for each non-zero A_i, one is added to adjacent elements on both sides, which increases the sum of the array by 2. This happens for each non-zero value in the array.

So, if before an operation, there are x non-zero values in an array, the sum is increased by 2*x by that operation.

Also, we can see that once an element becomes non-zero, it only increases, and thus, keeps contributing to the overall sum.

### Minimum time an element becomes non-zero

Now, the problem reduces to computing T_i, where T_i is the minimum time when A_i becomes non-zero. For non-zero elements in initial array, T_i = 0.

Let’s assume A_p is the nearest non-zero value for A_i. Then it takes |p-i| seconds, as, after each second, the element adjacent to the nearest non-zero element becomes positive, reducing |p-i| by 1.

Hence, for each element, find the position of the nearest non-zero element in both directions, and take the minimum distance to both elements.

It can be easily done via ordered sets in O(N*log(N)).

But we can use DP as well, using recurrence T_{i+1} = min(T_{i+1}, T_i +1) (looping i from 1 to N twice to handle circular array) and T_{i-1} = min(T_{i-1}, T_i +1) (looping i from N to 1 twice to handle circular array).

Hence, this way we are able to compute T_i, so we can compute sum of final array as \displaystyle \sum_{i = 1}^N A_i + 2*max(0, K - T_i)

### Bonus

Solve this problem for non-circular array A.

# TIME COMPLEXITY

The time complexity is O(N) or O(N*log(N)) depending upon implementation.

# SOLUTIONS

Setter's Solution
#include<bits/stdc++.h>
using namespace std;
#define int long long

void solve() {
int n; cin >> n;
assert(n >= 3);
int k; cin >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
int sum = accumulate(a.begin(), a.end(), 0LL);
if (sum == 0) {
cout << 0 << '\n';
return;
}
vector<int> time(n);
int L = -n;
for (int i = n - 1; i >= 0; i--) {
if (a[i] > 0) {
L = -(n - i);
break;
}
}
for (int i = 0; i < n; i++) {
if (a[i] > 0) {
L = i;
}
time[i] = i - L;
}
int R = 2 * n;
for (int i = 0; i < n; i++) {
if (a[i] > 0) {
R = n + i;
break;
}
}
for (int i = n - 1; i >= 0; i--) {
if (a[i] > 0) {
R = i;
}
time[i] = min(time[i], R - i);
}
int ans = sum;
for (int i = 0; i < n; i++) ans += 2 * max(0LL, k - time[i]);
cout << ans << '\n';
}

signed main() {
ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
int t; cin >> t;
while (t--) {
solve();
}
return 0;
}

Tester's Solution
#include <bits/stdc++.h>
using namespace std;

long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
int sum = 0;
while(t--){
int n, k;
//cerr << "done" << endl;
sum += n;
assert(sum <= 1e6);
vector<int> vec(n);
queue<int> q;
vector<int> vis_time(n, 1e9);
for(int i = 0; i < n ; i++){
if(i == n - 1)
else
if(vec[i] > 0)
q.push(i), vis_time[i] = 0;
}
while(!q.empty()){
int idx = q.front();
q.pop();
if(vis_time[(idx+1)%n] == 1e9)
q.push((idx+1)%n), vis_time[(idx+1)%n] = vis_time[idx] + 1;
if(vis_time[(idx-1+n)%n] == 1e9)
q.push((idx-1+n)%n), vis_time[(idx-1+n)%n] = vis_time[idx] + 1;
}
long long ans = 0;
for(int i = 0; i < n ; i++){
ans += (vec[i] + max(k - vis_time[(i+1)%n], 0) + max(k - vis_time[(i-1+n)%n], 0));
}
cout << ans << '\n';
}
return 0;
}

Editorialist's Solution
import java.util.*;
import java.io.*;
class POSSPEW{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int N = ni(), K = ni();
int[] A = new int[N];
for(int i = 0; i< N; i++)A[i] = ni();
int[] minTime = new int[N];
int INF = 3*N;
Arrays.fill(minTime, INF);
boolean allzero = true;
for(int i = 0; i< N; i++)if(A[i] > 0){
minTime[i] = 0;
allzero = false;
}
for(int i = 0; i< 2*N; i++)
minTime[(i+1)%N] = Math.min(minTime[(i+1)%N], minTime[i%N]+1);
for(int i = 2*N-1; i>= 0; i--)
minTime[(i+N-1)%N] = Math.min(minTime[(i+N-1)%N], minTime[i%N]+1);

if(allzero){
pn(0);
return;
}
long sum = 0;
for(int x:A)sum += x;
for(int i = 0; i< N; i++)sum += 2*Math.max(0, K-minTime[i]);
pn(sum);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
void run() throws Exception{
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new POSSPEW().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}

StringTokenizer st;
}

}

String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
}catch (IOException  e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}

String nextLine() throws Exception{
String str = "";
try{
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}


Feel free to share your approach. Suggestions are welcomed as always.

11 Likes

what’s wrong with my solution https://www.codechef.com/viewsolution/51174003

1 Like

My Approach :
First make all elements Non - Zero (if at least one element is zero) . Then, if the sum of array elements is s and each element of array is non-zero. Answer will become s + 2nk where k is number of operations and n is number of elements.
P.S. This approach should have given TLE on the basis of the contraints given in the question. Maybe Test Cases were not proper.

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define ll long long
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
const unsigned int M = 1000000007;
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> tree;
bool checkzeroes(vector<ll> data){
for(auto elem : data){
if(elem == 0)
return true;
}
return false;
}
void solve()
{
int t;
ll n, k;
cin>>t;
while(t--){
cin>>n>>k;
bool check_non_zero = false;
vector<ll> data(n);
for(auto &elem : data)
cin>>elem;

for(auto elem : data){
if(elem != 0)
{
check_non_zero = true;
break;
}
}
if(!check_non_zero){
cout<<"0\n";
continue;
}

while(checkzeroes(data) and k > 0){
vector<bool> pts(n,false);
for(int i = 0; i < n; i ++)
if(data[i] != 0)
pts[i] = true;

for(int i = 0; i < n ; i++ ){
if(pts[i]){
if(i == 0)
data[n-1] ++ , data[i+1] ++ ;
else if(i == n-1)
data[0] ++, data[i-1] ++ ;
else
data[i-1] ++ , data[i+1] ++ ;
}
}
k--;
}
ll sum = accumulate(data.begin(),data.end(),0);
sum += (n*k*2);

cout<<sum<<"\n";

}

}
int main()
{
ios_base::sync_with_stdio(false);
cout.tie(NULL);
cin.tie(NULL);
solve();
return 0;
}

6 Likes

the simple and easy-to-understand solution,
solved using multisource BFS concept, queue with visited arrays.

https://www.codechef.com/viewsolution/51125036

10 Likes

I have solved it using prefix and suffix concept in O(n) . I calculated total zeroes before and after the element , from this we can simple know the time when it will become positive .
https://www.codechef.com/viewsolution/51126119

9 Likes

Isn’t this solution https://www.codechef.com/viewsolution/51122029 have time complexity O(n^2)?

Assume a case of n=10^5, k=10^9 and only one element in the array is non-zero, in this case, shouldn’t the above solution result in TLE??

7 Likes

Exactly that’s what I have thought this method is O(n^2)? Otherwise question will be just implementation. This should result in TLE

2 Likes

No this wont give TLE. It will take very few iterations to make all elements of array to non zero

1 Like

Test it locally, when N=1e4, the code is running for 5e7 iterations, so it should definitely TLE for full constraints.

1 Like

I think only 2 numbers (according to my test case) which are zero get converted into non-zeros in each iteration. It approximately takes n/2 iterations (according to my test case) to make all elements of the array non-zero.

2 Likes

the sum you have to use long long I guess, int will cause overflow

1 Like

Yes atmost n/2 operations is possible. This should give TLE.

2 Likes

I also did this but I am getting wrong answer. Could you tell what is difference between my and your code .
My submission : Solution: 51176594 | CodeChef

1 Like

Hi!
https://www.codechef.com/viewsolution/51164974
I have done the same approach. Can you please tell me why it is showing TLE in Task 1?

1 Like

Great Editorial ! Thanks

1 Like

Can anyone please explain why this solution got SIGSEGV and where it is going wrong.

https://www.codechef.com/viewsolution/51174797

1 Like

Testcase
1
3 2
0 0 0

1 Like

Your solution may overflow. arr[i] <= 10e6. Use Long Long array

1 Like

Sorry Bro ! I found that my solution should also have given TLE. Maybe the test cases were not proper. Here is a Sample Code. For 10e3, it takes 500 iterations. It means complexity goes to O(n^2) and it should give TLE.

n = 10**3
a = [0 for i in range(n)]
a[n//2] = 1

counter = 0
while 0 in a :
check = [False for i in range(len(a))]
for i in range(len(a)):
if a[i] != 0:
check[i] = True
for i in range(len(a)):
if check[i]:
if i == 0:
a[len(a)-1] += 1
a[i + 1] += 1
elif i == len(a)-1:
a[len(a)-1] += 1
a[i-1] += 1
else:
a[i-1]+= 1
a[i+1] += 1

counter += 1

print(counter)

1 Like