[中文版 | Chinese version](https://cp-wiki.vercel.app/tutorial/kick-start/2020H/)↵
↵
## Problem A — [Retype](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043adc7)↵
↵
We only have two options, thus↵
↵
$$↵
ans=K-1+\min(N + 1, K - S + N - S + 1)↵
$$↵
↵
Time complexity is $O(1)$.↵
↵
<spoiler summary="Code (C++)">↵
↵
~~~~~↵
#include <algorithm>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
long long N, K, S;↵
cin >> N >> K >> S;↵
cout << K - 1 + min(N + 1, K - S + N - S + 1) << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
## Problem B — [Boring Numbers](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043b0c6)↵
↵
All problems such that require counting of numbers within range $[L,R]$ can be transformed into solving for $[0,R]$ and $[0,L]$ separately, and taking their difference as the final answer.↵
↵
Now suppose $X$ has $D$ digits and we want to count boring numbers within $[0,X]$.↵
↵
First, let's consider all numbers with $d<D$ digits. For $d$ digits, we can generate $5^d$ boring nubmers since we have $5$ options for each position (the most significant nubmer must be add so it cannot be $0$). So all numbers with $d<D$ digits make a contribution of $\sum_{i<D}5^i$.↵
↵
Then we consider numbers with $D$ digits and are no larger than $X$.↵
↵
Start from the most significant digit, and suppose that we are at the $i$-th digit now.↵
↵
- If $X[i]$ does not satisfy the requirement of parity, we just need to count the digits that are smaller than $X[i]$ and can satisfy the parity (we can precalculate such numbers in $b[X[i]]$), then add to the total number $b[X[i]]\cdot5^{D-i}$. Since for these $b[X[i]]$ numbers, the following $D-i$ digits can be chose arbitrarily. In this case, we can stop right here.↵
- Otherwise, we first count the digits that are smaller than $X[i]$ and can satisfy the parity (we can precalculate such numbers in $a[X[i]]$) and add to the total number $a[X[i]]\cdot5^{D-i}$. Then we are going to count boring numbers that have exactly same $i$ digits as $X$ and continue our processing. Note that if $i=D$, we need to add $1$ to the total number, since this means $X$ itself is a boring number.↵
↵
Time complexity is $O(\log R)$ if we exclude the precalculations.↵
↵
<spoiler summary="Code (C++)">↵
↵
~~~~~↵
#include <algorithm>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
typedef long long ll;↵
ll five[20], pre[20];↵
int a[10] = {0, 0, 1, 1, 2, 2, 3, 3, 4, 4};↵
int b[10] = {0, 1, 1, 2, 2, 3, 3, 4, 4, 5};↵
↵
class Solution {↵
ll count(ll x) {↵
string s = to_string(x);↵
int n = s.size();↵
ll ans = pre[n - 1];↵
for (int i = 1; i <= n; ++i) {↵
int c = s[i - 1] - '0';↵
if (c % 2 != i % 2) {↵
ans += five[n - i] * b[c];↵
break;↵
} else {↵
ans += five[n - i] * a[c];↵
if (i == n)↵
ans++;↵
}↵
}↵
return ans;↵
}↵
↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
ll L, R;↵
cin >> L >> R;↵
cout << count(R) - count(L - 1) << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
five[0] = 1;↵
for (int i = 1; i < 20; ++i)↵
five[i] = five[i - 1] * 5;↵
pre[0] = 0;↵
for (int i = 1; i < 20; ++i)↵
pre[i] = pre[i - 1] + five[i];↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
## Problem C — [Rugby](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043b027)↵
↵
Apparently, we can solve for $x$ and $y$ independently.↵
↵
First consider $y$. Since all the people will be in the same row, this becomes a classical problem in which we just need to take the median of $Y_i$ as the meeting place.↵
↵
Then we consider $x$. It is obvious that once we determine the starting point $x$, the optimal movement is determined. The leftmost person will go to the leftmost cell, and the rest follow.↵
↵
Thus we can solve this problem via ternary search. In order to prove the correctness, we need to prove that $dist(x)$ has only one extreme point, which is also its minimum point. (If we consider integer points, there might be two, but the two must be $x$ and $x+1$).↵
↵
Obviously, when $x+N-1\leq\min(X_i)$, $dist(x)$ decreases with $x$. While when $x\geq\max(X_i)$, $dist(x)$ increases with $x$.↵
↵
We then observe that, when we move the starting point from $x$ to $x+1$, there will be $k(x)$ people who will move $1$ less, and $N-k(x)$ people who will move $1$ more. So $dist(x+1)-dist(x)=N-2\cdot k(x)$. During the process where $x$ moves from $-\infty$ to $\infty$, $k(x)$ goes to $0$ from $N$, and will never increase. So $dist(x+1)-dist(x)$ will increase from $-N$ to $N$ and will never increase. So $dist(x)$ will take its extreme value (also its minimum) at the minimum $x$ that makes $dist(x+1)-dist(x)\geq0$.↵
↵
The final time complexity is $O(N\log N+N\log MAX)$, in which $MAX$ is our search range.↵
↵
<spoiler summary="Code (C++), Ternary search">↵
↵
~~~~~↵
#include <algorithm>↵
#include <climits>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
typedef long long ll;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
int N;↵
cin >> N;↵
vector<int> X(N), Y(N);↵
for (int i = 0; i < N; ++i)↵
cin >> X[i] >> Y[i];↵
sort(Y.begin(), Y.end());↵
ll ylo = 0;↵
for (int yi : Y)↵
ylo += abs(yi - Y[N / 2]);↵
sort(X.begin(), X.end());↵
ll l = -2e9, r = 2e9;↵
ll xlo = LLONG_MAX;↵
auto dist = [&](ll start) {↵
ll ret = 0;↵
int idx = 0;↵
for (int xi : X) {↵
ret += abs(start + idx - xi);↵
idx++;↵
}↵
xlo = min(xlo, ret);↵
return ret;↵
};↵
while (l <= r) {↵
ll ml = l + (r - l) / 3, mr = r - (r - l) / 3;↵
ll dl = dist(ml), dr = dist(mr);↵
if (dl <= dr)↵
r = mr - 1;↵
if (dl >= dr)↵
l = ml + 1;↵
}↵
cout << ylo + xlo << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
<spoiler summary="Code (C++), Binary search">↵
↵
~~~~~↵
#include <algorithm>↵
#include <climits>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
typedef long long ll;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
int N;↵
cin >> N;↵
vector<int> X(N), Y(N);↵
for (int i = 0; i < N; ++i)↵
cin >> X[i] >> Y[i];↵
sort(Y.begin(), Y.end());↵
ll ylo = 0;↵
for (int yi : Y)↵
ylo += abs(yi - Y[N / 2]);↵
sort(X.begin(), X.end());↵
ll l = -2e9, r = 2e9;↵
ll xlo = LLONG_MAX;↵
auto dist = [&](ll start) {↵
ll ret = 0;↵
int idx = 0;↵
for (int xi : X) {↵
ret += abs(start + idx - xi);↵
idx++;↵
}↵
xlo = min(xlo, ret);↵
return ret;↵
};↵
while (l <= r) {↵
ll mid = (l + r) / 2;↵
ll delta = dist(mid + 1) - dist(mid);↵
if (delta >= 0)↵
r = mid - 1;↵
else↵
l = mid + 1;↵
}↵
cout << ylo + xlo << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
## Problem D — [Friends](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043aee7)↵
↵
If we build a graph with the strings, we will have too many edges.↵
↵
So instead we can build a graph with different letters (in this problem we have $C=26$ letters). We will save this graph in an adjacent matrix.↵
↵
The initial setting is $d[i][j]=\infty$ and $d[i][i]=0$. Then we enumerate on all $N$ strings. If two different letters $a$ and $b$ both occur in the same string $S$, we set $d[a][b]=d[b][a]=1$. The meaning is that, if we have a string $S'$ with $a$ and another string $S''$ with $b$, we can build a chain $S'\to S\to S''$ which has $1$ middle point.↵
↵
Then we do Floyd-Warshall on this adjacent matrix. Now $d[i][j]$ means the minimum middle points that are needed to build a chain from $i$ to $j$.↵
↵
For each query, we enumerate on letters in $S[X_i]$ and $S[Y_i]$, and the final answer will be↵
↵
$$↵
\min_{p\in S[X_i],q\in S[Y_i]}d[p][q] + 2↵
$$↵
↵
If the answer is $\infty$, we just output $-1$.↵
↵
The total time complexity is $O((N+Q)L^2+C^3)$, in which $C=26$ is the size of the alphabet.↵
↵
↵
<spoiler summary="Code (C++)">↵
↵
~~~~~↵
#include <algorithm>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
const int INF = 0x3f3f3f3f;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
int N, Q;↵
cin >> N >> Q;↵
vector<string> S(N + 1);↵
for (int i = 1; i <= N; ++i)↵
cin >> S[i];↵
vector<vector<int>> d(26, vector<int>(26, INF));↵
for (string s : S)↵
for (char c1 : s)↵
for (char c2 : s)↵
if (c1 != c2)↵
d[c1 - 'A'][c2 - 'A'] = 1;↵
for (int k = 0; k < 26; ++k)↵
for (int i = 0; i < 26; ++i) {↵
if (i == k)↵
continue;↵
for (int j = 0; j < 26; ++j) {↵
if (j == i || j == k)↵
continue;↵
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);↵
}↵
}↵
↵
for (int i = 1; i <= Q; ++i) {↵
int X, Y;↵
cin >> X >> Y;↵
int ans = INF;↵
bool found = false;↵
for (char c1 : S[X]) {↵
for (char c2 : S[Y]) {↵
if (c1 == c2) {↵
cout << 2 << " ";↵
found = true;↵
break;↵
}↵
ans = min(ans, d[c1 - 'A'][c2 - 'A'] + 2);↵
}↵
if (found)↵
break;↵
}↵
if (!found)↵
cout << (ans == INF ? -1 : ans) << " ";↵
}↵
cout << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
↵
## Problem A — [Retype](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043adc7)↵
↵
We only have two options, thus↵
↵
$$↵
ans=K-1+\min(N + 1, K - S + N - S + 1)↵
$$↵
↵
Time complexity is $O(1)$.↵
↵
<spoiler summary="Code (C++)">↵
↵
~~~~~↵
#include <algorithm>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
long long N, K, S;↵
cin >> N >> K >> S;↵
cout << K - 1 + min(N + 1, K - S + N - S + 1) << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
## Problem B — [Boring Numbers](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043b0c6)↵
↵
All problems such that require counting of numbers within range $[L,R]$ can be transformed into solving for $[0,R]$ and $[0,L]$ separately, and taking their difference as the final answer.↵
↵
Now suppose $X$ has $D$ digits and we want to count boring numbers within $[0,X]$.↵
↵
First, let's consider all numbers with $d<D$ digits. For $d$ digits, we can generate $5^d$ boring nubmers since we have $5$ options for each position (the most significant nubmer must be add so it cannot be $0$). So all numbers with $d<D$ digits make a contribution of $\sum_{i<D}5^i$.↵
↵
Then we consider numbers with $D$ digits and are no larger than $X$.↵
↵
Start from the most significant digit, and suppose that we are at the $i$-th digit now.↵
↵
- If $X[i]$ does not satisfy the requirement of parity, we just need to count the digits that are smaller than $X[i]$ and can satisfy the parity (we can precalculate such numbers in $b[X[i]]$), then add to the total number $b[X[i]]\cdot5^{D-i}$. Since for these $b[X[i]]$ numbers, the following $D-i$ digits can be chose arbitrarily. In this case, we can stop right here.↵
- Otherwise, we first count the digits that are smaller than $X[i]$ and can satisfy the parity (we can precalculate such numbers in $a[X[i]]$) and add to the total number $a[X[i]]\cdot5^{D-i}$. Then we are going to count boring numbers that have exactly same $i$ digits as $X$ and continue our processing. Note that if $i=D$, we need to add $1$ to the total number, since this means $X$ itself is a boring number.↵
↵
Time complexity is $O(\log R)$ if we exclude the precalculations.↵
↵
<spoiler summary="Code (C++)">↵
↵
~~~~~↵
#include <algorithm>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
typedef long long ll;↵
ll five[20], pre[20];↵
int a[10] = {0, 0, 1, 1, 2, 2, 3, 3, 4, 4};↵
int b[10] = {0, 1, 1, 2, 2, 3, 3, 4, 4, 5};↵
↵
class Solution {↵
ll count(ll x) {↵
string s = to_string(x);↵
int n = s.size();↵
ll ans = pre[n - 1];↵
for (int i = 1; i <= n; ++i) {↵
int c = s[i - 1] - '0';↵
if (c % 2 != i % 2) {↵
ans += five[n - i] * b[c];↵
break;↵
} else {↵
ans += five[n - i] * a[c];↵
if (i == n)↵
ans++;↵
}↵
}↵
return ans;↵
}↵
↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
ll L, R;↵
cin >> L >> R;↵
cout << count(R) - count(L - 1) << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
five[0] = 1;↵
for (int i = 1; i < 20; ++i)↵
five[i] = five[i - 1] * 5;↵
pre[0] = 0;↵
for (int i = 1; i < 20; ++i)↵
pre[i] = pre[i - 1] + five[i];↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
## Problem C — [Rugby](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043b027)↵
↵
Apparently, we can solve for $x$ and $y$ independently.↵
↵
First consider $y$. Since all the people will be in the same row, this becomes a classical problem in which we just need to take the median of $Y_i$ as the meeting place.↵
↵
Then we consider $x$. It is obvious that once we determine the starting point $x$, the optimal movement is determined. The leftmost person will go to the leftmost cell, and the rest follow.↵
↵
Thus we can solve this problem via ternary search. In order to prove the correctness, we need to prove that $dist(x)$ has only one extreme point, which is also its minimum point. (If we consider integer points, there might be two, but the two must be $x$ and $x+1$).↵
↵
Obviously, when $x+N-1\leq\min(X_i)$, $dist(x)$ decreases with $x$. While when $x\geq\max(X_i)$, $dist(x)$ increases with $x$.↵
↵
We then observe that, when we move the starting point from $x$ to $x+1$, there will be $k(x)$ people who will move $1$ less, and $N-k(x)$ people who will move $1$ more. So $dist(x+1)-dist(x)=N-2\cdot k(x)$. During the process where $x$ moves from $-\infty$ to $\infty$, $k(x)$ goes to $0$ from $N$, and will never increase. So $dist(x+1)-dist(x)$ will increase from $-N$ to $N$ and will never increase. So $dist(x)$ will take its extreme value (also its minimum) at the minimum $x$ that makes $dist(x+1)-dist(x)\geq0$.↵
↵
The final time complexity is $O(N\log N+N\log MAX)$, in which $MAX$ is our search range.↵
↵
<spoiler summary="Code (C++), Ternary search">↵
↵
~~~~~↵
#include <algorithm>↵
#include <climits>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
typedef long long ll;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
int N;↵
cin >> N;↵
vector<int> X(N), Y(N);↵
for (int i = 0; i < N; ++i)↵
cin >> X[i] >> Y[i];↵
sort(Y.begin(), Y.end());↵
ll ylo = 0;↵
for (int yi : Y)↵
ylo += abs(yi - Y[N / 2]);↵
sort(X.begin(), X.end());↵
ll l = -2e9, r = 2e9;↵
ll xlo = LLONG_MAX;↵
auto dist = [&](ll start) {↵
ll ret = 0;↵
int idx = 0;↵
for (int xi : X) {↵
ret += abs(start + idx - xi);↵
idx++;↵
}↵
xlo = min(xlo, ret);↵
return ret;↵
};↵
while (l <= r) {↵
ll ml = l + (r - l) / 3, mr = r - (r - l) / 3;↵
ll dl = dist(ml), dr = dist(mr);↵
if (dl <= dr)↵
r = mr - 1;↵
if (dl >= dr)↵
l = ml + 1;↵
}↵
cout << ylo + xlo << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
<spoiler summary="Code (C++), Binary search">↵
↵
~~~~~↵
#include <algorithm>↵
#include <climits>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
typedef long long ll;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
int N;↵
cin >> N;↵
vector<int> X(N), Y(N);↵
for (int i = 0; i < N; ++i)↵
cin >> X[i] >> Y[i];↵
sort(Y.begin(), Y.end());↵
ll ylo = 0;↵
for (int yi : Y)↵
ylo += abs(yi - Y[N / 2]);↵
sort(X.begin(), X.end());↵
ll l = -2e9, r = 2e9;↵
ll xlo = LLONG_MAX;↵
auto dist = [&](ll start) {↵
ll ret = 0;↵
int idx = 0;↵
for (int xi : X) {↵
ret += abs(start + idx - xi);↵
idx++;↵
}↵
xlo = min(xlo, ret);↵
return ret;↵
};↵
while (l <= r) {↵
ll mid = (l + r) / 2;↵
ll delta = dist(mid + 1) - dist(mid);↵
if (delta >= 0)↵
r = mid - 1;↵
else↵
l = mid + 1;↵
}↵
cout << ylo + xlo << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
## Problem D — [Friends](https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff49/000000000043aee7)↵
↵
If we build a graph with the strings, we will have too many edges.↵
↵
So instead we can build a graph with different letters (in this problem we have $C=26$ letters). We will save this graph in an adjacent matrix.↵
↵
The initial setting is $d[i][j]=\infty$ and $d[i][i]=0$. Then we enumerate on all $N$ strings. If two different letters $a$ and $b$ both occur in the same string $S$, we set $d[a][b]=d[b][a]=1$. The meaning is that, if we have a string $S'$ with $a$ and another string $S''$ with $b$, we can build a chain $S'\to S\to S''$ which has $1$ middle point.↵
↵
Then we do Floyd-Warshall on this adjacent matrix. Now $d[i][j]$ means the minimum middle points that are needed to build a chain from $i$ to $j$.↵
↵
For each query, we enumerate on letters in $S[X_i]$ and $S[Y_i]$, and the final answer will be↵
↵
$$↵
\min_{p\in S[X_i],q\in S[Y_i]}d[p][q] + 2↵
$$↵
↵
If the answer is $\infty$, we just output $-1$.↵
↵
The total time complexity is $O((N+Q)L^2+C^3)$, in which $C=26$ is the size of the alphabet.↵
↵
↵
<spoiler summary="Code (C++)">↵
↵
~~~~~↵
#include <algorithm>↵
#include <iostream>↵
#include <vector>↵
↵
using namespace std;↵
const int INF = 0x3f3f3f3f;↵
↵
class Solution {↵
public:↵
void solve(int case_num) {↵
cout << "Case #" << case_num << ": ";↵
int N, Q;↵
cin >> N >> Q;↵
vector<string> S(N + 1);↵
for (int i = 1; i <= N; ++i)↵
cin >> S[i];↵
vector<vector<int>> d(26, vector<int>(26, INF));↵
for (string s : S)↵
for (char c1 : s)↵
for (char c2 : s)↵
if (c1 != c2)↵
d[c1 - 'A'][c2 - 'A'] = 1;↵
for (int k = 0; k < 26; ++k)↵
for (int i = 0; i < 26; ++i) {↵
if (i == k)↵
continue;↵
for (int j = 0; j < 26; ++j) {↵
if (j == i || j == k)↵
continue;↵
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);↵
}↵
}↵
↵
for (int i = 1; i <= Q; ++i) {↵
int X, Y;↵
cin >> X >> Y;↵
int ans = INF;↵
bool found = false;↵
for (char c1 : S[X]) {↵
for (char c2 : S[Y]) {↵
if (c1 == c2) {↵
cout << 2 << " ";↵
found = true;↵
break;↵
}↵
ans = min(ans, d[c1 - 'A'][c2 - 'A'] + 2);↵
}↵
if (found)↵
break;↵
}↵
if (!found)↵
cout << (ans == INF ? -1 : ans) << " ";↵
}↵
cout << endl;↵
}↵
};↵
↵
int main() {↵
int t;↵
cin >> t;↵
for (int i = 1; i <= t; ++i) {↵
Solution solution = Solution();↵
solution.solve(i);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
