PROBLEM LINK:
Contest Division 1
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Contest Division 3
Contest Division 4
Setter: Flame Storm
Tester: Harris Leung
Editorialist: Aman Dwivedi
DIFFICULTY:
Easy
PREREQUISITES:
Observation
PROBLEM:
There is an array A with N elements. Each element of A has a fixed polarity: either north or south.
Chef is allowed to perform some (possibly zero) operations on the array A. In one operation, Chef can:
 Pick some subarray of the array A, such that, the first and last elements of the subarray have different polarities, and, rearrange the elements in this subarray any way he wants.
Note that the polarity of each element remains unchanged after an operation.
Find the minimum number of operations required to sort the array in nondecreasing order, or state that it is impossible.
EXPLANATION:
Letâ€™s first see when it is not possible to sort the array at all. Consider this situation and think about whether it is possible to achieve a sorted array or not.
 Given an unsorted array in which every element has the same polarity either north or south, then it wonâ€™t be possible to achieve a sorted array.
Itâ€™s easy to visualise cause you canâ€™t take any subarray and rearrange elements, hence the array will remain as it was given to us.
In all other cases, it is possible to sort the array and the maximum operations in the worst case will be only 2.
Wait, how is that possible only 2?

Letâ€™s say the polarity of the first and last element is the same i.e North. Now as there exists at least one element which will have different polarity. We can bring that element to the end or start of the array in one operation.

Now the start and end of the array will have different parity and hence you can rearrange the whole array as you like. So why not make the array sorted in this operation.

Hence the maximum number of operations required in the worst case will be 2.
So we are left with finding when it will require only a single operation. Letâ€™s see:

Identify the sorted prefix and suffix of the given array. Letâ€™s say the prefix starts from index 0 and ends at index p. Similarly, say suffix starts from index s and ends at index n1.

Hence the subarray A[p+1, s1] is the unsorted part of the given array. Now if there exists an element in prefix P[0,p+1] whose polarity is different to that of any elements in suffix S [s1,n1]. Then we can consider that subarray and rearrange the elements to get the sorted array.

In such cases, only 1 operation would be required.
TIME COMPLEXITY:
O(N \log N) per test case
SOLUTIONS:
Setter
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
int n;
cin >> n;
int a[n + 7];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
string s;
cin >> s;
if (is_sorted(a, a + n)) {cout << 0 << '\n'; return;}
bool hasN = false, hasS = false;
for (char c : s) {
if (c == 'N') {hasN = true;}
if (c == 'S') {hasS = true;}
}
if (!hasN  !hasS) {cout << 1 << '\n'; return;}
if (s[0] != s[n  1]) {cout << 1 << '\n'; return;}
int sorted[n + 7];
for (int i = 0; i < n; i++) {
sorted[i] = a[i];
}
sort(sorted, sorted + n);
bool seen = false;
for (int i = 0; i < n; i++) {
if (sorted[i] == a[i]) {
if (s[i] != s[0]) {seen = true;}
}
else if (i == 0  sorted[i  1] == a[i  1]) {
if (s[i] != s[0]) {seen = true;}
break;
}
}
for (int i = n  1; i >= 0; i) {
if (sorted[i] == a[i]) {
if (s[i] != s[n  1]) {seen = true;}
}
else if (i == n  1  sorted[i + 1] == a[i + 1]) {
if (s[i] != s[n  1]) {seen = true;}
break;
}
}
if (seen) {cout << 1 << '\n';}
else {cout << 2 << '\n';}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Tester
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
const int N=2e5+1;
int n;
ll a[N],b[N];
char c[N];
bool check(){
for(int i=2; i<=n ;i++){
if(b[i1]>b[i]) return false;
}
return true;
}
void solve(){
cin >> n;
bool hvn=false,hvs=false;
for(int i=1; i<=n ;i++) cin >> a[i];
for(int i=1; i<=n ;i++){
cin >> c[i];
hvn=c[i]=='N';
hvs=c[i]=='S';
}
for(int i=1; i<=n ;i++) b[i]=a[i];
if(check()){
cout << "0\n";
return;
}
if(!hvn  !hvs){
cout << "1\n";
return;
}
if(c[1]!=c[n]){
cout << "1\n";
return;
}
for(int i=1; i<=n ;i++) b[i]=a[i];
int ptr=1;
while(c[ptr]==c[1]) ptr++;
sort(b+ptr,b+n+1);
if(check()){
cout << "1\n";
return;
}
for(int i=1; i<=n ;i++) b[i]=a[i];
ptr=n;
while(c[ptr]==c[n]) ptr;
sort(b+1,b+ptr+1);
if(check()){
cout << "1\n";
return;
}
cout << "2\n";
}
int main(){
ios::sync_with_stdio(false);cin.tie(0);
int t;cin >> t;
while(t) solve();
//solve();
}