Two Pointers Technique & Binary Search For Beginners

Maximum sum 2D:

Problem: Given 2D array with some +ve & -ve integers. You have to compute the sum of submatrix(subrectangle-may be square) with maximum sum.

Link: here

 IDEA:

   1.) Fix one column lft .

   2.) Fix one column rgt from lft to rgtmost corner.

   3.) For every fixed lft & rgt column, traverse below rowwise & calculate cumulative sum.

        Note: We have to precalculate rowise sum earlier.

   4.) If sum becomes negative then, make it 0 & update maximum sum on every step.

So, overall complexity is O(n^3).

Source code:

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define M 1000000

#define S(a) scanf("%d",&a)
#define P(a) printf("%d\n",a)
#define S2(a,b) scanf("%lld%lld",&a,&b)
#define clr(a) memset(a,0,sizeof(a))
#define ff(a) memset(a,false,sizeof(a))
#define pb(x) push_back(x)
#define all(v) v.begin(),v.end()
#define fr(i,a,n) for(i=a;i<=n;i++)
using namespace std;
typedef long long ll;
int i,j,k,n,m,lft,rgt,row,mx,sm,ans,rsum[105][105],ar[105][105];
int main()
{
    while(S(n)==1)
    {
        fr(i,1,n)fr(j,1,n)
        {
            S(ar[i][j]);
            rsum[i][j]=ar[i][j]+rsum[i][j-1];
        }
        mx=INT_MIN;
        fr(lft,1,n)fr(rgt,lft,n)///moving from lft col to corner in each step!
        {
            sm=0;
            fr(row,1,n)
            {
                sm+=rsum[row][rgt]-rsum[row][lft-1];
                if(sm<0)sm=0;
                mx=max(mx,sm);
            }
        }
        P(mx);
    }
    return 0;
}