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**2D Range minimum Query in <O(n*m*logn*logm),O(1)>**

I solved THIS problem from "codechef June16 long challenge" by extending the " 1D RMQ sparse table method" to 2D version which finds maximum value in a submatrix in O(1) after O(n*m*logn*logm) preprocessing time.

*Problem: Given an array[n] and Q queries. Each query asks for minimum in range x to y where 0=<x<=y<n*

*Preprocess : O(n*logn)*

```
For(i=0;i<n;i++)
table[ 0 ][ i ]=array[ i ] //building base
For(j=1;j<=log(n);j++)
For(i=0;i+2^(j-1)<n;i++)
table[ j ][ i ] = min ( table[ j-1 ][ i ], table[ j-1 ][ i+2^(j-1) ] ) //use previously computed
```

we created a sparse table[ 1+logn ][ n ]

**table[ j ][ i ] contains the minimum value in range i to i-1 + 2^j inclusive**

Each range is a power of 2. First we build base of the table. Then for each range we split it into 2 equal parts for which value is already calculated. This way table is build in O(n*logn).

*Query(x,y) : O(1)*

```
len=y-x+1
k=log2(len)
return min(table[ k ][ x ],table[ k ][ y+1-2^k ]) // ranges may overlap
```

Query is then simply splitting the range in 2 parts each of which is a power of 2 and then using the pre-computed value from table. The 2 partitioned Ranges may overlap.

I found THIS video which explains it in detail

This was 1D RMQ. Now lets extend this concept to solve 2D RMQ

*Problem:Given a matrix[n][m] and Q queries. Each query asks for minimum in submatrix[ (x1,y1), (x2,y2) ]*

*x1,y1 is the top left-most point and x2,y2 is bottom right-most point of the submatrix.Assume 0-based indexing*

*Preprocess : O(n*m*logn*logm)*

- we create a table[ 1+logn ][ n ][ 1+logm ][ m ]
- Each box of the table[ 1+logn ][ n ] is a sparse table of size [ 1+logm ][ m ]
- Let us see what table[ jr ][ ir ][ jc ][ ic ] actually contains:

It contains the minimum element from column ic to ic-1+2^jc of all rows from ir to ir-1+2^jr

In other words, it contain the**minimum element in the submatrix [ (ir,ic), (ir-1+2^jr , ic-1+2^jc) ]**

where submatrix [ (x1,y1),(x2,y2) ] denotes the submatrix with x1,y1 as its top left-most and x2,y2 as its bottom right-most point. - Now you can easily conclude that, table[ 0 ][ ir ][ jc ][ ic ] is nothing but the 1D RMQ table if we take our array as row ir

Now,lets see how to code it

At First,for each row ir, we compute table[ 0 ][ ir ][ jc ][ ic ] ,the same as we did in 1D version.

Taking each row as an array and then computing its sparse table in same fashion as in 1D RMQ.

```
For(ir=0;ir<n;ir++)
{
For(ic=0;ic<m;ic++)
table[ 0 ][ ir ][ 0 ][ ic ] = Matrix[ ir ][ ic ];
For(jc=1;jc<=log2(m);jc++)
For(ic=0;ic+2^(jc-1)<m;ic++)
table[0 ][ir ][jc ][ic ] = min(table[0 ][ir ][jc-1 ][ic ],table[0 ][ir ][ jc-1 ][ ic+2^(jc-1) ])
}
```

The above step is nothing but computing the sparse table for each row.

The complexity for one row is O(m*logm) and so for all rows O(n*m*logm).

Now,we will use this base to build the entire table.

```
For(jr=1;jr<=log(n);jr++)
For(ir=0;ir<n;ir++)
For(jc=0;jc<=log(m);jc++)
For(ic=0;ic<m;ic++)
table[jr ][ir ][jc ][ic ] = min(table[jr-1 ][ir ][jc ][ic ],table[jr-1 ][ir+2^(jr-1) ][jc ][ic ])
```

Its not very difficult to understand this,just go through each line of the above code and try to visualise.

Clearly,the above step is O(n*m*logn*logm)

*Query(x1,y1,x2,y2) : O(1)*

We have to find the minimum in submatrix [(x1,y1),(x2,y2)]

```
lenx=x2-x1+1
kx=log2(lenx)
leny=y2-y1+1
ky=log2(leny)
```

First we will query on n

we have to query from row x1 to row x2.

But we have only computed for the range which are powers of 2 so we will divide in into two range

each of which is a power of 2.

x1 to x1-1+2^kx .....(say range R1)

x2+1-2^kx to x2 .....(say range R2)

The partition may overlap but that's not the problem here as we have to find the min/max only.

Now for each of the above two ranges R1 and R2, we have to query in column y1 to y2.

For this we will do the same range partition as we did above, i.e.

y1 to y1-1+2^ky ......(say range C1)

y2+1-2^ky to y2 ......(say range C2)

so query is simply 3 lines of code,

```
min_R1 = min ( table[kx ][x1 ][ky ][y1 ] , table[kx ][x1 ][ky ][ y2+1-2^ky ] )
min_R2 = min ( table[kx ][x2+1-2^kx ][ky ][y1 ], table[kx ][x2+1-2^kx ][ky ][y2+1-2^ky ] )
return min ( min_R1, min_R2 )
```

Done.

We can also use 2D segment tree to solve 2D RMQ in <O(n*m), O(logn*logm)>. It has a special benefit that we can also do update in O(logn*logm). We will see that in next blog.

1D RMQ Video

1D RMQ problem //use fats I/O

2D RMQ problems

problem1 // 2D segtree gives TLE in last subtask

problem2

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

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