 # ARRGAME - Editorial

Setter: Rithvik Chatterjee
Tester: Yash Chandnani
Editorialist: Ishmeet Singh Saggu

Easy

Games

# PROBLEM:

You are given a strip A of length N where A[i]=0 represent a free cell and A[i]=1 represent a blocked cell. Two players Nayeon and Tzuyu play this game and alternate turns.

• Nayeon plays first.
• In the first turn player can go in any free-cell and after that, the player can only go in cells which are free and are adjacent to the current cell.
• If the player visits a free-cell then it is marked as a blocked cell.
• If a player is unable to move to a free cell during her turn, this player loses the game.

Print “Yes” if Nayeon wins else print “No”.

# QUICK EXPLANATION:

• In array A we can note that it is in the form of continuous segments of 0's and 1's that is it is in the form of 11110000011111000011110111.... Let us refer to each continuous segment of 0's as zeroSegment.
• When all cells are blocked then Nayeon loses.
• When we have only 1 zeroSegment then Nayeon will win only if its length is odd.
• When there are more than 1 zeroSegment then, let us refer to length of largest zeroSegment as L and the length of the second-largest segment as L'. Then Nayeon will win if following conditions are true.
• L is odd.
• L' < \frac{L+1}{2}

# EXPLANATION:

When we observe the array A we can note that it is in the form of continuous segments of 0's and 1's that is it is in the form of 11110000011111000011110111.... Let us to each continuous segment of 0's as zeroSegment. Note that once a player enters a zeroSegment(that enters any free cell belonging to that zeroSegment) it cannot go into any other zeroSegment as there will be blocked cells(or no cell) at the ends of the zeroSegment.

Let us handle the corner test case when there is no zeroSegment, i.e. all cells are blocked then Nayeon cannot make the move hence answer will be No.

Now let us solve the easier version when array A contains only one zeroSegment.

Claim - Optimal move for Nayeon will be to enter the middle cell of the zeroSegment. And if the length L of zeroSegment is odd then Nayeon will win and if its length is even then Tzuyu will win. Note here I am referring the middle cell as the cell which when blocked divides the zeroSegment into two equal-sized zeroSegment of smaller size, so when L is even there is no middle element and when L is odd there is a middle element.

Let us see why choosing the middle cell is the optimal strategy. Consider the length of zeroSegment as length L. Now consider we enter the ith free cell, then marking it as blocked cell after the move there will be 2 zeroSegment. One will be of i-1 free cell to the left and the other will be of L-i cells to the right. When i is not the middle cell or L is even, then the length i-1 and L-i will be unequal. Let us consider that right segment with L-i free cell is longer, then Tzuyu will choose the i+1 box in next turn and there will be no way for Nayeon to enter the free cell or the right segment as i+1 cell is blocked now and the only option for the Nayeon will be to move to the left and enter cells i, i-1, i-2,\dots1 whereas for Tzuyu the only option will be to move to the right and enter cells i+1, i+2, i+3, \dots L, but as we know that the right segment is bigger than left so Nayeon will be out of moves first and will lose. Similarly, you can prove it when the left segment is bigger than the right segment.

So the only case when Nayeon will win is when L is odd and she chooses the middle element as it will divide the zeroSegment into 2 equal segments.

if(numOfZeroSegment == 1) {
if(segmentLen % 2) cout << "Yes\n"; // Only in odd length strip Nayeon can win.
else cout << "No\n";
}


Now when there are more than 1 zeroSegment.

All the condition shown above is true the only thing is that Nayeon will choose the middle element(if possible) of the largest zeroSegment. The only extra condition in this problem is there is a possibility that Nayeon might lose even if the length L of the largest segment is odd. That possibility is :
Suppose L of the largest segment is odd(in all other cases Nayeon will lose as shown above). And the length of second-largest segment L' \geq \frac{L+1}{2}. Because when she chooses the middle element she can make total \frac{L+1}{2} moves (as if Nayeon doesn't choose the middle element thinking that she can make more moves but then she will still lose, as Tzuyu will follow the moves described above) and Tzuyu will enter the leftmost cell of the second-largest zeroSegement will have total moves \geq \frac{L+1}{2}(number of moves of Nayeon) hence Tzuyu will win.

if(numOfZeroSegment > 1) {
if((segmentLenMax % 2) && (segmentLenSecondMax < ((segmentLenMax+1)/2))) {
cout << "Yes\n"; // Nayeon wins
}
else cout << "No\n";
}


# TIME COMPLEXITY:

• O(N) in total to find the length of all the zeroSegment.
• In case when the number of zeroSegment > 1. The maximum length and second maximum can be found in 2 for loops of O(N) (you can refer to the editorialist’s code).
• Total time complexity per test case is O(N)

# SOLUTIONS:

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

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define rep(i, n)    for(int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define fill(a)  memset(a, 0, sizeof (a))
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
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 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){
}
}
}
void pre(){

}
int a;
ll sum = 0;
void solve(){
rep(i,n-1){
}
assert(a==1);
assert(a[n-1]==1);
sum+=n;
vi v;
int cnt  =0;
rep(i,n){
if(a[i]==1) v.pb(cnt),cnt=0;
else cnt++;
}
sort(all(v));
reverse(all(v));
if(sz(v)==1){
if(v%2==1) cout<<"Yes\n";
else cout<<"No\n";
}
else{
if(v%2==1&&(v+1)/2>v) cout<<"Yes\n";
else cout<<"No\n";
}
}

int main() {
cin.sync_with_stdio(0); cin.tie(0);
cin.exceptions(cin.failbit);
pre();
rep(i,n) solve();
assert(getchar()==EOF);
assert(sum<=1e6);
return 0;
}

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

void solveTestCase() {
int N;
cin >> N;
vector<int> A(N), zeroSegmentLen;
for(int i = 0; i < N; i ++) {
cin >> A[i];
}
for(int i = 0; i < N; i ++) {
if(A[i]) continue;
int ptr = i;
while((ptr+1 < N) && (A[ptr+1] == 0)) {
ptr ++;
}
zeroSegmentLen.push_back(ptr-i+1);
i = ptr;
}
if(zeroSegmentLen.size() == 0) { // when all cells are blocked.
cout << "No\n";
}
else if(zeroSegmentLen.size() == 1) {
if(zeroSegmentLen % 2) cout << "Yes\n"; // only in odd length strip Nayeon can win.
else cout << "No\n";
}
else {
int maxLen = 0, id = -1;
int M = zeroSegmentLen.size();
for(int i = 0; i < M; i ++) {
if(maxLen < zeroSegmentLen[i]) {
maxLen = zeroSegmentLen[i];
id = i;
}
}
swap(zeroSegmentLen, zeroSegmentLen[id]);
maxLen = 0, id = -1;
for(int i = 1; i < M; i ++) {
if(maxLen < zeroSegmentLen[i]) {
maxLen = zeroSegmentLen[i];
id = i;
}
}
swap(zeroSegmentLen, zeroSegmentLen[id]);
// By doing above operation we get maximum at index 0 and second maximum at index 1.
if((zeroSegmentLen % 2) && (zeroSegmentLen <= ((zeroSegmentLen-1)/2))) {
cout << "Yes\n";
}
else cout << "No\n";
}
}

int main() {
ios_base::sync_with_stdio(0); // fast IO
cin.tie(0);
cout.tie(0);

int testCase;
cin >> testCase;
for(int i = 1; i <= testCase; i ++) {
solveTestCase();
}

}


## Video Editorial

Feel free to share your approach. In case of any doubt or anything is unclear please ask it in the comment section. Any suggestions are welcomed. 18 Likes

Wasted 2 hours on this Nice problem.

16 Likes

nice editorial

2 Likes

https://www.codechef.com/viewsolution/37271813

All test cases are giving correct output. can someone find the mistake please

nicely written

Really wasted 2.4 hour and gets demotivated.

8 Likes

invested*

14 Likes

N=11
Arr=[1,0,0,0,0,0,1,0,0,0,1]
Output should be ‘No’. Your code prints ‘Yes’

I think you might have understood it by now, I exactly did this prblm the same way you did. Anyways I’m gonna tell you the wrong part of your logic…

if there are more than one zero segments, since both players play optimally the second person might not only have to stick with even length zero segments. (which you and I didn’t think of sadly )

since the 1st player has to make the first move she definetley has to pick the largest odd length zero segment (if it exists) and occupy n/2th position. The second player now has to check if any of the remaing zero segments (odd or even) have length greater than half the length of the segment occupied by 1st player. Hope I didn’t confuse you.

4 Likes

Nice editorial

Btw if anyone need a Video Editorial in Hindi - Link

8 Likes

I unnecessarily wasted more 30 minutes until I released that output was in the format ["Yes" & "No"] not ["YES" & "NO"]!! 4 Likes

https://www.codechef.com/viewsolution/37266535

PLEASE TELL ME WHATS IS WRONG!    really nice problem and logic. (unfortunately got 50 though ).

2 Likes

I did manage to solve it, but I could’ve tried ELOMAX which looked like cancer brute force (so I didn’t have time to implement).
I could’ve had a much better rank with 250 points rather than 200.

1 Like

https://www.codechef.com/viewsolution/37245176
can anyone tell me what I have done wrong

my approach

I commited the same mistake only to demotivate myself I’m idiot …

Feeling lucky.  I thought that if there is only one zerosegment we wil win if it’s length is not equal to 2.

but some bug got me AC.
Because I was also considering segments with length zero.

suppose this test case
10001000001
i guess your code will give output Yes but answer is no.