# SLAYS - Editorial

Author: kugo
Tester: yash_daga
Editorialist: iceknight1093

TBD

# PREREQUISITES:

Dynamic programming, range-max queries

# PROBLEM:

You’re given an array S. For Q queries of (L, R), answer the following:

• For a fixed value of D, you start from L with a score of D and move towards R. At each index, S_i \leq D must hold. Further, if S_i = D, reduce D by 1.
Find the smallest value of D that allows you to reach R from L.

# EXPLANATION:

Let M = \max(S_L, S_{L+1}, \ldots, S_R).
Clearly, the answer to query (L, R) must be at least M, since we will encounter an M along our way and need to be \geq M when we do.

Further, it’s also easy to see that D = M+1 always allows us to reach R, since it’s strictly larger than anything in this range and hence will never reduce.
So, our task reduces to checking whether D = M allows us to reach R or not.

Let’s see how this process would go.

• Let i_1 \geq L be the first index such that S_{i_1} = M. We will definitely reach this index, after which we’d be left with D = M-1.
• From here, we move right till we
• Reach R, at which point we’re done
• Reach an occurrence of something \gt D, at which point we can’t continue
• Reach another occurrence of D, at which point we move to D-1; and then this process repeats.

This is easy to simulate in \mathcal{O}(N), but we need to do better.

Notice that once we reach the first position where S_i = M, the process is rather uniquely defined, because we can pretend we start there with D = M.

So, let’s find for each index i, the furthest we can travel if we start from i with value S_i.
Let this be dp_i.

dp_i can be computed as follows:

• Let j \gt i be the first index such that S_j \geq S_i
• Let k \gt i be the first index such that S_k = S_i - 1.
• Then, dp_i = \min(j-1, dp_k), because we can’t cross index j no matter what; and if we reach k we’re essentially starting from there instead, so we can’t go beyond dp_k either.

dp_i can be precomputed quite easily for all indices by iterating i from N down to 1. The only things that need to be taken care of are quickly computing indices j and k.

• Computing k is easy. Since we’re iterating in reverse anyway, simply keep a record (using, say, a map) of the last occurrence of every element. Let this be \text{last}[x]; then k = \text{last}[S_i - 1].
Note that after processing index i, you should set \text{last}[S_i] = i.
• Computing j needs a bit more effort, but in the case of this specific problem, can be actually done easier.
Note that rather than finding the next element that’s \geq A_i, it suffices to find the next element that’s equal to S_i, i.e, \text{last}[S_i] itself. Do you see why?

At any rate, by maintaining the \text{last} map, all the dp values can be computed quickly.

After this, for a query (L, R) we need to:

• Find the maximum M on the range [L, R], which is a standard data structure task (use a segment tree/sparse table/whatever).
• Find the leftmost occurrence of this maximum, which is also not too hard: for example, the segment tree/sparse table merge function can be modified slightly to return this, by maintaining pairs of (value, index).
• Let i be the index we found. Then, if dp_i \geq R the answer is M, otherwise the answer is M+1.

# TIME COMPLEXITY

\mathcal{O}((N + Q)\log N) per test case.

# CODE:

Setter's code (C++)
#include "bits/stdc++.h"
using namespace std;

typedef long long           lol;
typedef std::pair<int,int>  pii;
#define pb                  push_back
#define ub                  upper_bound
#define lb                  lower_bound
#define fo(i,l,r,d)         for (auto i=(l); (d)<0?i>(r):((d)>0?i<(r):0); i+=(d))
#define all(x)              x.begin(), x.end()
#define ff                  first
#define ss                  second

std::mt19937 rng (std::chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename A, typename B> std::ostream& operator<< (std::ostream &cout, const std::pair<A, B> &p) { return cout << p.first << ' ' << p.second; } template <typename A, size_t n> std::ostream& operator<< (std::ostream &cout, const std::array<A, n> &v) { for (int i = 0; i < n - 1; ++i) cout << v[i] << ' '; return (n ? cout << v.back(): cout << '\n'); } template <typename A> std::ostream& operator<< (std::ostream &cout, const std::vector<A> &v) { for (int i = 0; i < v.size() - 1; ++i) cout << v[i] << ' '; return (v.size() ? cout << v.back(): cout << '\n'); }
template <typename A, typename B> std::istream& operator>> (std::istream &cin, std::pair<A, B> &p) { cin >> p.first; return cin >> p.second; } template <typename A, size_t n> std::istream& operator>> (std::istream &cin, std::array<A, n> &v) { assert(n); for (int i = 0; i < n - 1; i++) cin >> v[i]; return cin >> v.back(); } template <typename A> std::istream& operator>> (std::istream &cin, std::vector<A> &v) { assert(v.size()); for (int i = 0; i < v.size() - 1; i++) cin >> v[i]; return cin >> v.back(); }
template <typename A, typename B> auto amax (A &a, const B b){ if (b > a) a = b ; return a; }
template <typename A, typename B> auto amin (A &a, const B b){ if (b < a) a = b ; return a; }

template <
class Node,
class Calc,
bool kNearestPowOf2 = false
>
class Segtree {
public:

explicit Segtree (const int n, const Node id, const Calc& F)
: sz(n), N(kNearestPowOf2 ? 1 << 32 - __builtin_clz(std::max(1, sz - 1)) : sz), a(N << 1, id), id(id), F(F)
{

}

explicit Segtree (const std::vector<Node>& x, const Node id, const Calc& F)
: sz(x.size()), N(kNearestPowOf2 ? 1 << 32 - __builtin_clz(std::max(1, sz - 1)) : sz), id(id), F(F)
{
a.resize(N << 1, id);
std::copy(x.begin(), x.end(), a.begin() + N);

for (int i = N; --i; )
a[i] = F(a[i << 1], a[i << 1 | 1]);
}

void set (int i, const Node x) {
// assert(0 <= i and i < sz);
for (a[i += N] = x; i >>= 1; )
a[i] = F(a[i << 1], a[i << 1 | 1]);
}

Node qu (int l, int r) const {
// assert(0 <= l and l <= r and r <= sz);

Node x = id, y = id;
for (l += N, r += N; l < r; l >>= 1, r >>= 1) {
if (l & 1) x = F(x, a[l++]);
if (r & 1) y = F(a[--r], y);
}

return F(x, y);
}

// First j in [l, N] such that pred(F[l, j)) is FALSE, if pred is monotonic
template<class Predicate>
int max_right (int l, const Predicate& pred) const {
// assert(0 <= l and l <= N and pred(id));
if (l == N) return l;

Node prev = id, t = id;
l += N;

do {
l >>= __builtin_ctz(l);
if (!pred(F(prev, a[l]))) {
while (l < N)
if (pred(t = F(prev, a[l <<= 1])))
prev = t, l++;
return l - N;
}

prev = F(prev, a[l++]);
} while ((l & -l) != l);

return N;
}

// First j in [0, r] such that pred(F[j, r)) is TRUE, if pred is monotonic
template<class Predicate>
int min_left (int r, const Predicate& pred) const {
// assert(r > -1 and r <= N and pred(id));
if(r == 0) return r;

Node last = id, t = id;
r += N;

do {
r--, r >>= __builtin_ctz(~r);
if (r == 0) r = 1;
if (!pred(F(a[r], last))){
while (r < N)
if (pred(t = F(a[(r <<= 1) += 1], last)))
last = t, r--;
return r + 1 - N;
}

last = F(a[r], last);
} while((r & -r) != r);

return 0;
}

private:

const int sz;
const int N;
std::vector<Node> a;
const Node id;
const Calc F;
};

void darling (const int kase) {

int n, q; cin >> n >> q;
vector a(n, 0); cin >> a;

a.pb(-1);

stack<pair<int, int>> s;
vector geq(n, -1);

s.push(pair(0, a[0]));

for (int i = 1; i < n; i++) {
while (s.size() and s.top().ss < a[i])
s.pop();
if (s.size())
geq[i] = s.top().ff;
s.push(pair(i, a[i]));
}

vector dp(n, n);

fo(i,n-1,-1,-1) {
int j = geq[i];
if (a[j] == a[i])
dp[j] = i;
else if (a[j] == a[i] + 1)
amin(dp[j], dp[i]);
}

// cout << a << '\n' << dp << '\n';

vector it(n, 0);
iota(all(it), 0);

Segtree mx(it, n, [&](int x, int y){
if (a[x] > a[y])
return x;
else if (a[x] < a[y])
return y;
else
return min(x, y);
});

while (q--) {
int l, r; cin >> l >> r, l--;

auto j = mx.qu(l, r);

if (dp[j] < r)
cout << a[j] + 1 << '\n';
else
cout << a[j] << '\n';
}

}

int main () {
ios_base::sync_with_stdio(0), cin.tie(0);

int t; cin >> t, assert(t >= 0);
for (int i = 0; t--; )
darling(++i);

}

Tester's code (C++)
#include <bits/stdc++.h>
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
#define pii pair<int, int>
#define ll long long
#define ff first
#define ss second
#define rep(i,x,y) for(int i=x; i<y; i++)
using namespace std;
const long long N=500005, INF=2000000000000000000;

int a[N];
pii st[4*N];
void build(int v, int l, int r)
{
if(l==r)
{
st[v]={a[l], -l};
return;
}
int m=(l+r)/2;
build(v*2, l, m);
build((v*2)+1, m+1, r);
st[v]=max(st[v*2], st[(v*2)+1]);
//cout<<l<<" "<<r<<" "<<t[v].o<<" "<<t[v].c<<" "<<t[v].ans<<"\n";
}
pii query(int v, int tl, int tr, int l, int r)
{
if(l>r)
return {-INF, 0};
if(tl==l&&tr==r)
return st[v];
int tm=(tl+tr)/2;
return max(query((2*v), tl, tm, l, min(tm, r)), query((2*v)+1, tm+1, tr, max(tm+1, l), r));
}

int32_t main()
{
IOS;
int t;
cin>>t;
while(t--)
{
int n, q;
cin>>n>>q;
rep(i,0,n)
cin>>a[i];
build(1, 0, n-1);
int dp[n];
stack <int> s;
for(int i=n-1;i>=0;i--)
{
dp[i]=n;
while(!s.empty() && a[s.top()]<a[i]-1)
s.pop();
if(!s.empty() && a[s.top()]==a[i]-1)
{
dp[i]=min(dp[i], dp[s.top()]);
s.pop();
}
if(!s.empty())
dp[i]=min(dp[i], s.top());
s.push(i);
}
while(q--)
{
int l, r;
cin>>l>>r;
l--, r--;
pii p=query(1, 0, n-1, l, r);
if(dp[-p.ss]<=r)
cout<<p.ff+1<<"\n";
else
cout<<p.ff<<"\n";
}
}
}

Editorialist's code (Python)
class RangeQuery:
def __init__(self, data, func=max):
self.func = func
self._data = _data = [list(data)]
i, n = 1, len(_data[0])
while 2 * i <= n:
prev = _data[-1]
_data.append([func(prev[j], prev[j + i]) for j in range(n - 2 * i + 1)])
i <<= 1

def query(self, start, stop):
"""func of data[start, stop)"""
depth = (stop - start).bit_length() - 1
return self.func(self._data[depth][start], self._data[depth][stop - (1 << depth)])

def __getitem__(self, idx):
return self._data[0][idx]

for _ in range(int(input())):
n, q = map(int, input().split())
s = list(map(int, input().split()))
queries = [ [] for _ in range(n)]
for i in range(q):
l, r = map(int, input().split())
queries[l-1].append((r-1, i))

RMQ = RangeQuery(s)
ans = [0]*q
dp = [0]*n
pos = {}
for i in reversed(range(n)):
dp[i] = n-1
if s[i] in pos: dp[i] = min(dp[i], pos[s[i]] - 1)
if s[i]-1 in pos: dp[i] = min(dp[i], dp[pos[s[i] - 1]])

pos[s[i]] = i

for r, id in queries[i]:
mx = RMQ.query(i, r+1)
ans[id] = mx+1

where = pos[mx]
if dp[where] >= r: ans[id] = mx
print(*ans, sep = '\n')

1 Like

This problem can be done using Segment trees which I think far easier than this approach
See the code of Just_a_looser1 who have done it using segment trees
https://www.codechef.com/viewsolution/93391916

Can you Please make a video, probably a one explaining with handwritten on a paper, so that we can understand clearly. It’s very difficult to read line by line after failing to solve the problem and understanding what is, and debug the approaches.

By Making diagrams and everything