PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Setter: Venkata Nikhil Medam
Tester: Rahul Dugar
Editorialist: Taranpreet Singh
DIFFICULTY
Simple
PREREQUISITES
None
PROBLEM
Given a matrix of integers divided into N rows and M integers, and Q queries on matrix, each query replacing an element in the matrix with a specific value, determine after each query whether the matrix is good.
A matrix is considered good if, for each diagonal from top left to the bottom right direction, all elements have the same value.
QUICK EXPLANATION
Maintain a multiset of values for each diagonal. A matrix is good if all multiset contains one distinct value each.
EXPLANATION
Let’s consider each diagonal as a separate array.
For example, for matrix
1 2 3 4
5 6 7 8
9 10 11 12
We get the following arrays
9
5 10
1 6 11
2 7 12
3 8
4
For the matrix to be good, all elements in each array should be equal. So we can simply represent these elements using a multiset, which allows fast insertion, removal, and checking if all values in the multiset are the same or not.
Constant memory solution
The above solution was boring, nothing interesting, but there exists a way elegant solution to solve this problem in time O(N*M+Q) and memory complexity O(1) excluding input array.
Click here for solution
Let’s denote count as the number of adjacent pairs in the above arrays such that the values differ. Compute it for the given matrix.
Now, it is easy to recalculate this count after each update, since at most two pairs are affected. The matrix would be good if and only if the count is 0 after the update. The proof of why this work is left as an exercise.
TIME COMPLEXITY
Time complexity O(N*M*log(min(N,M))+Q*log(N+M))
Memory complexity O(N*M)
SOLUTIONS
Setter's Solution
// Setter: Nikhil_Medam
#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
int32_t main()
{
IOS;
int t;
cin >> t;
while(t--)
{
int n, m;
cin >> n >> m;
vector<vector<int>> a(n, vector<int>(m));
int count = 0;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
cin >> a[i][j];
}
}
for(int i = 1; i < n; i++)
{
for(int j = 1; j < m; j++)
{
count += (a[i][j] != a[i - 1][j - 1]);
}
}
int q, x, y, v;
cin >> q;
while(q--)
{
cin >> x >> y >> v;
x--, y--;
if(x - 1 >= 0 and y - 1 >= 0)
count -= (a[x][y] != a[x - 1][y - 1]);
if(x + 1 < n and y + 1 < m)
count -= (a[x][y] != a[x + 1][y + 1]);
a[x][y] = v;
if(x - 1 >= 0 and y - 1 >= 0)
count += (a[x][y] != a[x - 1][y - 1]);
if(x + 1 < n and y + 1 < m)
count += (a[x][y] != a[x + 1][y + 1]);
cout << (count == 0 ? "YES" : "NO") << endl;
}
}
return 0;
}
Tester's Solution
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#ifndef rd
#define trace(...)
#define endl '\n'
#endif
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double f80;
#define double long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define sz(x) ((long long)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int infi=0x3f3f3f3f;
const int mod=998244353;
//const int mod=1000000007;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0,lim-1);
return uid(rang);
}
int powm(int a, int b) {
int res=1;
while(b) {
if(b&1)
res=(res*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return res;
}
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;
}
assert(l<=x&&x<=r);
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);HRSCPMTR
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 a[505][505];
int sum_nm=0,sum_q=0;
multiset<int> holo[1005];
void solve() {
int n=readIntSp(1,500),m=readIntLn(1,500);
fr(i,1,n) {
rep(j,1,m)
a[i][j]=readIntSp(-1000'000'000,1000'000'000);
a[i][m]=readIntLn(-1000'000'000,1000'000'000);
}
int q=readIntLn(1,200'000);
sum_nm+=n*m;
sum_q+=q;
// assert(sum_nm<=500000&&sum_q<=200000);
fr(i,0,1000)
holo[i].clear();
fr(i,1,n)
fr(j,1,m)
holo[i-j+500].insert(a[i][j]);
int vald=0;
fr(i,500-m+1,500+n-1)
if((*holo[i].begin())==(*holo[i].rbegin()))
vald++;
trace(vald,n+m-1);
while(q--) {
int x=readIntSp(1,n),y=readIntSp(1,m),v=readIntLn(-1000'000'000,1000'000'000);
if((*holo[x-y+500].begin())==(*holo[x-y+500].rbegin()))
vald--;
holo[x-y+500].erase(holo[x-y+500].lower_bound(a[x][y]));
a[x][y]=v;
holo[x-y+500].insert(a[x][y]);
if((*holo[x-y+500].begin())==(*holo[x-y+500].rbegin()))
vald++;
if(vald==n+m-1) {
cout<<"YES"<<endl;
} else
cout<<"NO"<<endl;
}
}
signed main() {
ios_base::sync_with_stdio(0),cin.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
cout<<fixed<<setprecision(7);
int t=readIntLn(1,100);
fr(i,1,t)
solve();
assert(getchar()==EOF);
#ifdef rd
cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class HRSCPMTR{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int N = ni(), M = ni();
int[][] mat = new int[N][M];
for(int i = 0; i< N; i++)for(int j = 0; j< M; j++)mat[i][j] = ni();
int count = 0;
for(int i = 1; i< N; i++)for(int j = 1; j< M; j++)if(mat[i][j] != mat[i-1][j-1])count++;
int Q = ni();
for(int q = 0; q< Q; q++){
int r = ni()-1, c = ni()-1, v = ni();
if(Math.min(r, c) > 0 && mat[r][c] != mat[r-1][c-1])count--;
if(r+1 < N && c+1 < M && mat[r][c] != mat[r+1][c+1])count--;
mat[r][c] = v;
if(Math.min(r, c) > 0 && mat[r][c] != mat[r-1][c-1])count++;
if(r+1 < N && c+1 < M && mat[r][c] != mat[r+1][c+1])count++;
pn(count==0?"YES":"NO");
}
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new HRSCPMTR().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
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