Dynamic Range Minimum Queries
Medium C++
Key Idea: Solution implementation
Solution
#include <bits/stdc++.h>
using namespace std;
//author: von_Braun
#define ll long long
#define lli long long int
#define pb push_back
#define rep(var, start, num) for(ulli var = start; var <start + num; var++)
#define all(x) x.begin(), x.end()
#define ulli unsigned long long int
#define ull unsigned long long
bool sortbysec(const pair<ll,ll> &a,const pair<ll,ll> &b) { return (a.second < b.second); }
int query(int l, int r,int rl, int rr, int pos, vector<int> &tree) {
// cout<<"query for - \n";
// cout<<"current range "<<l<<" "<<r<<endl;
// cout<<"req range "<<rl<<" "<<rr<<endl;
int mid = (l+r)/2;
if (l==rl && r==rr) {
return tree[pos];
}
if (rr <= mid) {
return query(l,mid,rl,rr,2*pos+1,tree);
} else if (rl >= 1+mid) {
return query(mid+1,r,rl,rr,2*pos+2,tree);
} else {
return min(query(l,mid,rl,mid,2*pos+1,tree), query(mid+1,r,mid+1,rr,2*pos+2,tree));
}
}
void build(int l, int r, int pos, vector<int> &main, vector<int> &tree) {
if (l==r) {
tree[pos] = main[l];
return;
}
int mid = (l+r)/2;
build(l,mid,2*pos+1,main,tree);
build(mid+1,r,2*pos+2,main,tree);
tree[pos] = min(tree[2*pos+1],tree[2*pos+2]);
}
void update(int l, int r, int newidx, int pos, vector<int> &main, vector<int> &tree) {
if (l==r && r == newidx) {
tree[pos] = main[l];
return;
}
int mid = (l+r)/2;
if (newidx<=mid) {
update(l,mid,newidx,2*pos+1,main,tree);
}
else {
update(mid+1,r,newidx,2*pos+2,main,tree);
}
tree[pos] = min(tree[2*pos+1],tree[2*pos+2]);
}
void solve() {
int n,q;
int t;
cin>>n>>q;
vector<int> main(n);
vector<int> tree(4*n);
rep(i,0,n) {cin>>main[i];}
build(0,n-1,0,main,tree);
// for(int i=0;i<4*n;i++) {cout<<tree[i]<<" ";} cout<<endl;
rep(i,0,q) {
cin>>t;
if (t==1) {
int k,u; cin>>k>>u;
main[k-1]=u;
update(0,n-1,k-1,0,main,tree);
} else {
int a,b;
cin>>a>>b;
cout<<query(0,n-1,a-1,b-1,0,tree)<<endl;
}
}
}
int main() {
//add quotes incase input output file
//freopen(input.txt,r,stdin);
//freopen(output.txt,w,stdout);
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int tc = 1;
// cin >> tc;
for (int t = 1; t <= tc; t++) {
solve();
}
}