KAVGMAT - Editorial

Author: Sayantan Jana
Tester: Aryan Choudhary
Editorialist: Sayantan Jana

Easy

PREREQUISITES:

Binary Search, 2 pointers technique

PROBLEM:

Given an integer K and a matrix A of dimensions N*M with integer elements, such that the elements are non decreasing along the row as well as column, find the number of square submatrices such the average of the elements in the square submatrix is \geq K.

QUICK EXPLANATION:

• For each cell (x,y) as the top left corner, we try to find the square submatrix with minimum length, such that average of the elements in the matrix is \geq K, i.e., we find minimum l such that f(x,y,l) = \frac{\sum_{i=x}^{i=x+l-1} \sum_{j=y}^{j=y+l-1} A_{i,j}}{l*l} \geq K. So if we get minimum l such that f(x,y,l) \geq K, we can count number of square submatrices with top left corner at (x,y) and average \geq K .
• f(x,y,l) can be computed in O(1) using prefix sum.
• Now there are 2 approaches to solve this problem.
• \textbf{Binary Search : } We find that the function f(x,y,l) increases with increase in l. So for each cell (x,y), we binary search for l and find the count number of square submatrices with top left corner at (x,y) and average \geq K .
• \textbf{2 pointers technique : } We also find that the function f(x,y,l) increases with increase in y. So fixing x, we find the minimum l_j for each (x,j), 1 \leq j \leq M such that f(x,j,l_j) \geq K in O(N+M) using 2 pointers technique.

EXPLANATION:

Notice that a square submatrix can be described by 3 variables, (x,y,l), (x,y) being the coordinates of the top left corner and l being the length of the square submatrix, i.e., the other 3 corners are (x+l-1,y), (x,y+l-1) and (x+l-1,y+l-1).

Similarly a general submatrix can be described by 4 variables, (x,y,l,r), (x,y) being the coordinates of the top left corner and l*r being the dimensions of the submatrix.

Let us define 2 terms :

• f(x,y,l) such that 1 \leq x \leq N, 1 \leq y \leq M, 1 \leq l \leq min(N-x+1,M-y+1) : average of the square submatrix with top left corner at (x,y) and length l. Similarly we can define f(x,y,l,r).
• r(x,y) : minimum length such that the square submatrix with top left corner at (x,y) has average \geq K, i.e., minimum l such that f(x,y,k) \geq K.

Monotonicity of f(x,y,l)

Consider the following submatrices :

• (x,y,l,l) : l*l submatrix with top left corner at (x,y).
• (x,y+l,l,1) : l*1 submatrix with top left corner at (x,y+l).
• (x+l,y,1,l) : 1*l submatrix with top left corner at (x+l,y).

Since the elements are increasing along the row, f(x,y+l,l,1) \geq f(x,y,l,l).
SInce the elements are increasing along the column, f(x+l,y,1,l) \geq f(x,y,l,l).

Also notice that the maximum of all the elements in the square submatrix (x,y,l) can be at most A_{x+l,y+l}. Hence we show that f(x,y,l+1) = \frac{l*l*f(x,y,l) + l*f(x+l,y,1,l) + l*f(x,y+l,l,1) + A_{x+l,y+l}}{(l+l)*(l+1)} \geq f(x,y,l).

We show that f(x,y,l) increases with increase in l. Similarly we can show that f(x,y,l) increase with increase in x as well as with increase in y, even independently.

O(N^2 M) solution

If we find prefix sums for the matrix A such that P_{i,j} tells the sum of elements in the submatrix (1,1,i-1,j-1), we can find f(x,y,l) in O(1) using matrix P.

So, fixing a cell as the top left corner, we want to count number of square submatrices with average \geq K . If we find r(x,y), the number of required submatrices would be min(N-(x+r(x,y)-1)+1,M-(y+r(x,y)-1)+1). We can find r(x,y) by iterating over 1 \leq l \leq min(N-x+1,M-y+1).

O(N M log(min(N,M)) solution

Instead of iterating over 1 \leq l \leq min(N-x+1,M-y+1), we can binary search on the length of submatrix, l, as we have shown before f(x,y,l) is increasing (not necessarily strictly though) over l.

O(N M) solution

As we have already talked about the monotonicity of f(), we can also find that r(x,y) is decreasing (not necessarily strictly) over y, i.e., r(x,y) \geq r(x,y+1).

Thus instead of fixing (x,y), if we just fix x, we can use 2 pointers technique to solve the problem. We solve as follows :

• We fix x and find r(x,1), we can either use binary search for the same or simply iterate on l, i.e., we keep left pointer at 1 and move the right pointer to get r(x,1).
• Now as we know r(x,1) \geq r(x,2), we move the left pointer by 1 and the right pointer by 1 stepwise as long as the submatrix has average \geq K to get r(x,2).
• Likewise step 2, we keep on moving the left pointer exactly by 1 as we increase y and have a gradual drop of the right pointer for getting corresponding r(x,y).

Fixing the row number x, we hence find r(x,y) for the entire row in O(N+M). Once we know r(x,y) for a cell, we have already discussed how to count number of square submatrices with average \geq K .

Note that in the above solution, left pointer denotes moving on y and right pointer on l, it isnâ€™t necessary that the value of left pointer is always lesser than the value of right pointer.

SOLUTIONS:

Setter's Solution - 2 pointer solution
#include<bits/stdc++.h>
using namespace std;

const int M = (1<<20)+5;
const int md = 1e9+7;

vector<int> mat[M];
vector<long long> prefix_sum[M];

int get_avg(int i,int j,int x,int y)
{
if(x<i || y<j) return -1;
return (prefix_sum[x][y]-prefix_sum[x][j-1]-prefix_sum[i-1][y]+prefix_sum[i-1][j-1])/(1ll*(x-i+1)*(y-j+1));
}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);

int testcases;
cin >> testcases;
while(testcases--)
{
int n,m,k;
cin >> n >> m >> k;
long long ans = 0;
for(int i=1;i<=n;++i)
{
mat[i].clear();
mat[i].push_back(0);
prefix_sum[i].clear();
prefix_sum[i].push_back(0);
}
mat[0].resize(m+1,0);
prefix_sum[0].resize(m+1,0);
for(int i=1;i<=n;++i)
{
for(int j=1;j<=m;++j)
{
int t;
cin >> t;
mat[i].push_back(t);
prefix_sum[i].push_back(prefix_sum[i-1][j] + prefix_sum[i][j-1] - prefix_sum[i-1][j-1] + mat[i][j]);
}
}
for(int i=1;i<=n;++i)
{
int j;
for(j=1;j<=m && i+j-1<=n;++j)
if(get_avg(i,1,i+j-1,j)>=k)
break;
int b = j;
if(b<=m && i+b-1<=n)
ans += min(m-b+1,n-i-b+2);
if(b>m)
b--;
for(j=2;j<=m;++j)
{
while(1)
{
if(get_avg(i,j,i+b-j,b)>=k)
b--;
else
{
b++;
break;
}
}
if(b<=m && i+b-j<=n)
ans += min(m-b+1,n-i-b+j+1);
if(b>m)
b--;
}
}
cout << ans << "\n";
}
}

Tester's Solution - Binary Search Solution
/* in the name of Anton */

/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

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);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

vi a(n);
for(int i=0;i<n-1;++i)
return a;
}

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end())         m.insert({x,cnt});
else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt)            m.erase(jt);
else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

void checkIfStrictlyIncreasing(const vector<vi> &a)
{
const int n=sz(a),m=sz(a[0]);
for(int i=0;i+1<n;++i)
for(int j=0;j<m;++j)
assert(a[i][j]<=a[i+1][j]);
for(int i=0;i<n;++i)
for(int j=0;j+1<m;++j)
assert(a[i][j]<=a[i][j+1]);
}

int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
lli nmsum=1e6;
while(T--)
{
nmsum-=n*m;
assert(nmsum>=0);
vector<vi> a(n);
for(auto &v:a)
checkIfStrictlyIncreasing(a);

vector<vi> b(n+1,vi(m+1));
fo(i,n) fo(j,m)     b[i+1][j+1]=a[i][j]-k;

repA(i,1,n)     repA(j,1,m)     b[i][j]+=b[i-1][j]+b[i][j-1]-b[i-1][j-1];
auto check=[&](lli r,lli c,lli len){
return b[r+len][c+len]+b[r][c]-b[r][c+len]-b[r+len][c]>=0;
};
lli ans=0;
for(int r=0;r<n;++r)
for(int c=0;c<m;++c)
{
const lli MX=min(n-r,m-c);
lli L=0,R=MX+1;
while(R-L>1)
{
lli M=(L+R)/2;
if(check(r,c,M))
R=M;
else
L=M;
}
// dbg(r,c,MX,L,R);
ans+=MX-L;
}
cout<<ans<<endl;
}   aryanc403();
return 0;
}

1 Like

can u please explain why did not i got TLE @shaanknight
https://www.codechef.com/viewsolution/44654102
i my point of view i iterate of all possible submatrix kk so total complexity will O(k(m*N))

4 Likes

I like your approach, but iterate over all the possible positions of the given matrix, you could have used binary search

1 Like

ya but i was wondering about why did not i get TLE.

Nice approach

can somebody please explain where I did mistake I followed some people solutions i am getting errorâ€¦

• whatâ€™s the difference if I take low<=high and low<high in while loop
• I added +1 for calculating mid if I didnâ€™t add I am getting division by 0 error. But in some codes they didnâ€™t add +1.