PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4
Setter: TheScrasse
Tester: Manan Grover, Lavish Gupta
Editorialist: Devendra Singh
DIFFICULTY:
To be calculated
PREREQUISITES:
None
PROBLEM:
You are given an array A of length N.
Determine the count of subarrays of A which contain their length as an element.
Formally, count the number of pairs (L, R) (1\le L\le R\le N) such that: (R-L+1) \in \{A_L, A_{L+1}, \ldots, A_R\}.
EXPLANATION:
Let us iterate over the elements of array A and for each element A_i count the number of pairs (L, R) (1\le L\le i\le R\le N) such that: (R-L+1) = A_i. The number of subarrays of a particular length which contain the element A_i can be found using the following equations:
Left limit for the start of such subarrays = max( 1,\: i - A_i +1)
Right limit for start of such subarrays = min(i, N - A_i+1)
Therefore the number of such subarrays is equal to Right limit-Left limit+1. But this does not account for the overcounting of subarrays. Same subarray could be counted more than once for two same values of array A. For index i = 1\:\: to\:\: N , to avoid overcounting, we will not count the subarrays which have been counted before due to some index j such that A_j=A_i.
Let last_{A_i} represent the index of last occurrence of A_i (initially 0), then the new left limits and right limits are:
left limit = max(last_{A_i} + 1, i - A_i + 1),
right limit = min(i, N - A_i+1)
Therefore the number of such subarrays is equal to Right limit-Left limit+1.
Sum of such subarrays for each index i =\: 1\: to\: N is the answer to the problem
TIME COMPLEXITY:
O(N) for each test case.
SOLUTION:
Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
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;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
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){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int sumN=0;
void solve()
{
int n=readInt(1,200000,'\n');
sumN+=n;
assert(sumN<=500000);
int A[n+1]={0};
for(int i=1;i<=n;i++)
{
if(i==n)
A[i]=readInt(1,n,'\n');
else
A[i]=readInt(1,n,' ');
adj[A[i]].pb(i);
}
int prev[n+1]={0};
ll ans=0;
for(int i=1;i<=n;i++)
{
ll exceed=i+A[i]-1-n;
exceed=max(exceed,(ll)0);
ll total=i-prev[A[i]];
total=min(total,(ll)A[i]);
total-=exceed;
total=max(total,(ll)0);
ans+=total;
prev[A[i]]=i;
}
cout<<ans<<'\n';
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
int T=readInt(1,1000,'\n');
//cin>>T;
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
const int N = 2e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
int last[N];
void sol(void)
{
int n;
cin >> n;
for (int i = 1; i <= n; i++)
last[i] = -1;
vll v(n);
for (int i = 0; i < n; i++)
cin >> v[i];
ll ans = 0;
for (int i = 0; i < n; i++)
{
ll l = max(last[v[i]] + 1, i - v[i] + 1), r = min(i, n - v[i]);
last[v[i]] = i;
if (r >= l)
ans += (r - l + 1);
}
cout << ans << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}
