➤ Problem Link : 486B. OR in Matrix
✅ C++ Solution :
#include<bits/stdc++.h>
using namespace std;
int main()
{
int m,n;
cin>>m>>n;
bool B[m][n];
unordered_set<int> row,col;
row.clear();
col.clear();
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
cin>>B[i][j];
if(!B[i][j])
{
row.insert(i);
col.insert(j);
}
}
}
if(row.size()==m)
{
for(int j=0;j<n;j++)
col.insert(j);
}
else if(col.size()==n)
{
for(int i=0;i<m;i++)
row.insert(i);
}
bool flag=true;
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(B[i][j] && row.find(i)!=row.end() && col.find(j)!=col.end())
{
flag=false;
break;
}
}
}
if(!flag)
cout<<"NO";
else
{
bool A[m][n];
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
A[i][j]=-1;
}
for(auto it = row.begin();it!=row.end();it++)
{
for(int j=0;j<n;j++)
A[*it][j]=0;
}
for(auto it=col.begin();it!=col.end();it++)
{
for(int i=0;i<m;i++)
A[i][*it]=0;
}
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(A[i][j]==-1)
A[i][j]=1;
}
}
cout<<"YES\n";
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
cout<<A[i][j]<<" ";
cout<<"\n";
}
}
}
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I am a Member Technical at D.E. Shaw India Pvt. Ltd. CodingWithArt is a blog where you find tricks and solutions to challenging coding and development problems. I aim to build a universal tech platform for every enthusiast out there.