In this HackerEarth Counting the number of intervals problem solution, You are given a sequence of N integers a1,a2,...,aN and an integer K. Your task is to count the number of intervals [l,r] such that al + ar + min(al,al+1,...,ar) <= K.

## HackerEarth Counting the number of intervals problem solution.

```#include <bits/stdc++.h>
using namespace std;

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int) 1e9 + 23111992;
const ll LINF = (ll) 1e18 + 23111992;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ull isqrt(ull k) {ull r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int myrand() {return abs((int) mt());}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";

const int maxn = 5e5 + 5;
int n;
long long k;
long long a[maxn];
int l[maxn];
int r[maxn];
vii g[maxn];

int fen[maxn];
void upd(int p, int val) {
p++;
for (; p < maxn; p += p & -p) {
fen[p] += val;
}
}
int query(int p) {
p++;
int res = 0;
for (; p > 0; p -= p & -p) {
res += fen[p];
}
return res;
}

void chemthan() {
cin >> n >> k;
assert(1 <= n && n <= 5e5);
assert(1 <= k && k <= 1e18);
vector<long long> dc;
FOR(i, 0, n) {
cin >> a[i];
assert(1 <= a[i] && a[i] <= 1e18);
dc.pb(a[i]);
}
sort(all(dc)), uni(dc);
FOR(i, 0, n) l[i] = r[i] = i;
FOR(i, 1, n) {
int ptr = i;
while (ptr && a[i] < a[ptr - 1]) ptr = l[ptr - 1];
l[i] = ptr;
}
FORd(i, n - 1, 0) {
int ptr = i;
while (ptr + 1 < n && a[i] <= a[ptr + 1]) ptr = r[ptr + 1];
r[i] = ptr;
}
FOR(i, 0, n) {
if (i - l[i] <= r[i] - i) {
FOR(j, l[i], i + 1) {
int d = upper_bound(all(dc), k - a[i] - a[j]) - dc.begin() - 1;
g[r[i]].pb(mp(d, 1));
if (i) {
g[i - 1].pb(mp(d, -1));
}
}
}
else {
FOR(j, i, r[i] + 1) {
int d = upper_bound(all(dc), k - a[i] - a[j]) - dc.begin() - 1;
g[i].pb(mp(d, 1));
if (l[i]) {
g[l[i] - 1].pb(mp(d, -1));
}
}
}
}
long long res = 0;
FOR(i, 0, n) {
int k = lower_bound(all(dc), a[i]) - dc.begin();
upd(k, 1);
for (pi ev : g[i]) {
res += query(ev.fi) * ev.se;
}
}
cout << res << "\n";
}

int main(int argc, char* argv[]) {
ios_base::sync_with_stdio(0), cin.tie(0);
if (argc > 1) {
assert(freopen(argv[1], "r", stdin));
}
if (argc > 2) {
assert(freopen(argv[2], "wb", stdout));
}
chemthan();
cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
return 0;
} ```

### Second solution

```#include<bits/stdc++.h>
#define rep(i,start,lim) for(lld i=start;i<lim;i++)
#define repd(i,start,lim) for(lld i=start;i>=lim;i--)
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define sz(a) (lld)((a).size())
#define all(c) (c).begin(),(c).end()
typedef long double ldb;
typedef long long lld;
const lld MOD = 1e9+7;
const lld INF = 1011111111;
const lld LLINF = 1000111000111000111LL;
const ldb EPS = 1e-10;
const ldb PI = 3.14159265358979323;
using namespace std;
lld powm(lld base,lld exp,lld mod=MOD) {lld ans=1;while(exp){if(exp&1) ans=(ans*base)%mod;exp>>=1,base=(base*base)%mod;}return ans;}
#define endl '\n'
#define fre freopen("1.in","r",stdin); freopen("1.out","w",stdout);
const lld N = 500005;
long long a[N],k;
vector<long long> comp;
int new_a[N];
#define LOGN 20
int n,log_base_2[N];
struct SparseTable{
int m[N][LOGN];
void pre() {
rep(i,2,N) log_base_2[i]=log_base_2[i>>1]+1;
rep(i,1,n+1) m[i][0]=i;
for(int j=1;(1<<j)<=n;j++) for(int i=1;i+(1<<j)-1<=n;i++) {
if(a[m[i][j-1]] > a[m[i+(1<<(j-1))][j-1]]) m[i][j] = m[i+(1<<(j-1))][j-1];
else m[i][j] = m[i][j-1];
}
}
int getmin(int l,int r) {
int tmp=log_base_2[r-l+1];
if(a[m[l][tmp]] > a[m[r-(1<<tmp)+1][tmp]]) return m[r-(1<<tmp)+1][tmp];
else return m[l][tmp];
}
} st;
int ft[N];
void update(int x,int val) {
for(;x<N;x+=(x&-x)) ft[x] += val;
}
int query(int x) {
int ret = 0;
for(;x>0;x-=(x&-x)) ret += ft[x];
return ret;
}

long long ans;
void go(int l,int r) {
if(r < l) return ;
int idx = st.getmin(l,r);
int lft = (idx-l+1), rgt = (r-idx+1);
if(lft <= rgt) {
go(l, idx-1);
rep(i,l,idx) update(new_a[i],-1);
go(idx+1, r);
update(new_a[idx],1);
rep(i,l,idx+1) {
if(k < a[idx] + a[i]) continue;
auto tmp = upper_bound(all(comp), k - a[idx] - a[i]);
if(tmp == comp.begin()) continue;
tmp--;
ans += query(tmp - comp.begin() + 1);
}
rep(i,l,idx) update(new_a[i],1);
}
else {
go(idx+1, r);
rep(i,idx+1,r+1) update(new_a[i],-1);
go(l, idx-1);
update(new_a[idx],1);
rep(i,idx,r+1) {
if(k < a[idx] + a[i]) continue;
auto tmp = upper_bound(all(comp), k - a[idx] - a[i]);
if(tmp == comp.begin()) continue;
tmp--;
ans += query(tmp - comp.begin() + 1);
}
rep(i,idx+1,r+1) update(new_a[i],1);
}
}
long long b[N];
int32_t main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin>>n>>k;
rep(i,1,n+1) {
cin>>a[i];
comp.push_back(a[i]);
}
sort(all(comp));
comp.erase(unique(all(comp)), comp.end());
rep(i,1,n+1) new_a[i] = lower_bound(all(comp), a[i]) - comp.begin() + 1;
st.pre();
go(1,n);
cout<<ans;
return 0;
}```