Submatrix Sum Queries
Last Updated :
18 Sep, 2023
Given a matrix of size M x N, there are large number of queries to find submatrix sums. Inputs to queries are left top and right bottom indexes of submatrix whose sum is to find out.
How to preprocess the matrix so that submatrix sum queries can be performed in O(1) time.
Example :
tli : Row number of top left of query submatrix
tlj : Column number of top left of query submatrix
rbi : Row number of bottom right of query submatrix
rbj : Column number of bottom right of query submatrix
Input: mat[M][N] = {{1, 2, 3, 4, 6},
{5, 3, 8, 1, 2},
{4, 6, 7, 5, 5},
{2, 4, 8, 9, 4} };
Query1: tli = 0, tlj = 0, rbi = 1, rbj = 1
Query2: tli = 2, tlj = 2, rbi = 3, rbj = 4
Query3: tli = 1, tlj = 2, rbi = 3, rbj = 3;
Output:
Query1: 11 // Sum between (0, 0) and (1, 1)
Query2: 38 // Sum between (2, 2) and (3, 4)
Query3: 38 // Sum between (1, 2) and (3, 3)
We strongly recommend you to minimize your browser and try this yourself first.
The idea is to first create an auxiliary matrix aux[M][N] such that aux[i][j] stores sum of elements in submatrix from (0,0) to (i,j). Once aux[][] is constructed, we can compute sum of submatrix between (tli, tlj) and (rbi, rbj) in O(1) time. We need to consider aux[rbi][rbj] and subtract all unnecessary elements. Below is complete expression to compute submatrix sum in O(1) time.
Sum between (tli, tlj) and (rbi, rbj) is,
aux[rbi][rbj] - aux[tli-1][rbj] -
aux[rbi][tlj-1] + aux[tli-1][tlj-1]
The submatrix aux[tli-1][tlj-1] is added because
elements of it are subtracted twice.
Illustration:
mat[M][N] = {{1, 2, 3, 4, 6},
{5, 3, 8, 1, 2},
{4, 6, 7, 5, 5},
{2, 4, 8, 9, 4} };
We first preprocess the matrix and build
following aux[M][N]
aux[M][N] = {1, 3, 6, 10, 16}
{6, 11, 22, 27, 35},
{10, 21, 39, 49, 62},
{12, 27, 53, 72, 89} }
Query : tli = 2, tlj = 2, rbi = 3, rbj = 4
Sum between (2, 2) and (3, 4) = 89 - 35 - 27 + 11
= 38
How to build aux[M][N]?
1. Copy first row of mat[][] to aux[][]
2. Do column wise sum of the matrix and store it.
3. Do the row wise sum of updated matrix aux[][] in step 2.
Below is the program based on above idea.
C++
#include<iostream>
using namespace std;
#define M 4
#define N 5
int preProcess( int mat[M][N], int aux[M][N])
{
for ( int i=0; i<N; i++)
aux[0][i] = mat[0][i];
for ( int i=1; i<M; i++)
for ( int j=0; j<N; j++)
aux[i][j] = mat[i][j] + aux[i-1][j];
for ( int i=0; i<M; i++)
for ( int j=1; j<N; j++)
aux[i][j] += aux[i][j-1];
}
int sumQuery( int aux[M][N], int tli, int tlj, int rbi,
int rbj)
{
int res = aux[rbi][rbj];
if (tli > 0)
res = res - aux[tli-1][rbj];
if (tlj > 0)
res = res - aux[rbi][tlj-1];
if (tli > 0 && tlj > 0)
res = res + aux[tli-1][tlj-1];
return res;
}
int main()
{
int mat[M][N] = {{1, 2, 3, 4, 6},
{5, 3, 8, 1, 2},
{4, 6, 7, 5, 5},
{2, 4, 8, 9, 4} };
int aux[M][N];
preProcess(mat, aux);
int tli = 2, tlj = 2, rbi = 3, rbj = 4;
cout << "\nQuery1: " << sumQuery(aux, tli, tlj, rbi, rbj);
tli = 0, tlj = 0, rbi = 1, rbj = 1;
cout << "\nQuery2: " << sumQuery(aux, tli, tlj, rbi, rbj);
tli = 1, tlj = 2, rbi = 3, rbj = 3;
cout << "\nQuery3: " << sumQuery(aux, tli, tlj, rbi, rbj);
return 0;
}
|
Java
import java.util.*;
import java.io.*;
class GFG {
static final int M = 4 ;
static final int N = 5 ;
static int preProcess( int mat[][], int aux[][])
{
for ( int i = 0 ; i < N; i++)
aux[ 0 ][i] = mat[ 0 ][i];
for ( int i = 1 ; i < M; i++)
for ( int j = 0 ; j < N; j++)
aux[i][j] = mat[i][j] +
aux[i- 1 ][j];
for ( int i = 0 ; i < M; i++)
for ( int j = 1 ; j < N; j++)
aux[i][j] += aux[i][j- 1 ];
return 0 ;
}
static int sumQuery( int aux[][], int tli,
int tlj, int rbi, int rbj)
{
int res = aux[rbi][rbj];
if (tli > 0 )
res = res - aux[tli- 1 ][rbj];
if (tlj > 0 )
res = res - aux[rbi][tlj- 1 ];
if (tli > 0 && tlj > 0 )
res = res + aux[tli- 1 ][tlj- 1 ];
return res;
}
public static void main (String[] args)
{
int mat[][] = {{ 1 , 2 , 3 , 4 , 6 },
{ 5 , 3 , 8 , 1 , 2 },
{ 4 , 6 , 7 , 5 , 5 },
{ 2 , 4 , 8 , 9 , 4 }};
int aux[][] = new int [M][N];
preProcess(mat, aux);
int tli = 2 , tlj = 2 , rbi = 3 , rbj = 4 ;
System.out.print( "\nQuery1: "
+ sumQuery(aux, tli, tlj, rbi, rbj));
tli = 0 ; tlj = 0 ; rbi = 1 ; rbj = 1 ;
System.out.print( "\nQuery2: "
+ sumQuery(aux, tli, tlj, rbi, rbj));
tli = 1 ; tlj = 2 ; rbi = 3 ; rbj = 3 ;
System.out.print( "\nQuery3: "
+ sumQuery(aux, tli, tlj, rbi, rbj));
}
}
|
Python3
M = 4
N = 5
def preProcess(mat, aux):
for i in range ( 0 , N, 1 ):
aux[ 0 ][i] = mat[ 0 ][i]
for i in range ( 1 , M, 1 ):
for j in range ( 0 , N, 1 ):
aux[i][j] = mat[i][j] + aux[i - 1 ][j]
for i in range ( 0 , M, 1 ):
for j in range ( 1 , N, 1 ):
aux[i][j] + = aux[i][j - 1 ]
def sumQuery(aux, tli, tlj, rbi, rbj):
res = aux[rbi][rbj]
if (tli > 0 ):
res = res - aux[tli - 1 ][rbj]
if (tlj > 0 ):
res = res - aux[rbi][tlj - 1 ]
if (tli > 0 and tlj > 0 ):
res = res + aux[tli - 1 ][tlj - 1 ]
return res
if __name__ = = '__main__' :
mat = [[ 1 , 2 , 3 , 4 , 6 ],
[ 5 , 3 , 8 , 1 , 2 ],
[ 4 , 6 , 7 , 5 , 5 ],
[ 2 , 4 , 8 , 9 , 4 ]]
aux = [[ 0 for i in range (N)]
for j in range (M)]
preProcess(mat, aux)
tli = 2
tlj = 2
rbi = 3
rbj = 4
print ( "Query1:" , sumQuery(aux, tli, tlj, rbi, rbj))
tli = 0
tlj = 0
rbi = 1
rbj = 1
print ( "Query2:" , sumQuery(aux, tli, tlj, rbi, rbj))
tli = 1
tlj = 2
rbi = 3
rbj = 3
print ( "Query3:" , sumQuery(aux, tli, tlj, rbi, rbj))
|
C#
using System;
class GFG
{
static int M = 4;
static int N = 5;
static int preProcess( int [,]mat, int [,]aux)
{
for ( int i = 0; i < N; i++)
aux[0,i] = mat[0,i];
for ( int i = 1; i < M; i++)
for ( int j = 0; j < N; j++)
aux[i,j] = mat[i,j] + aux[i-1,j];
for ( int i = 0; i < M; i++)
for ( int j = 1; j < N; j++)
aux[i,j] += aux[i,j-1];
return 0;
}
static int sumQuery( int [,]aux, int tli,
int tlj, int rbi, int rbj)
{
int res = aux[rbi,rbj];
if (tli > 0)
res = res - aux[tli-1,rbj];
if (tlj > 0)
res = res - aux[rbi,tlj-1];
if (tli > 0 && tlj > 0)
res = res + aux[tli-1,tlj-1];
return res;
}
public static void Main ()
{
int [,]mat = {{1, 2, 3, 4, 6},
{5, 3, 8, 1, 2},
{4, 6, 7, 5, 5},
{2, 4, 8, 9, 4}};
int [,]aux = new int [M,N];
preProcess(mat, aux);
int tli = 2, tlj = 2, rbi = 3, rbj = 4;
Console.Write( "\nQuery1: " +
sumQuery(aux, tli, tlj, rbi, rbj));
tli = 0; tlj = 0; rbi = 1; rbj = 1;
Console.Write( "\nQuery2: " +
sumQuery(aux, tli, tlj, rbi, rbj));
tli = 1; tlj = 2; rbi = 3; rbj = 3;
Console.Write( "\nQuery3: " +
sumQuery(aux, tli, tlj, rbi, rbj));
}
}
|
PHP
<?php
function preProcess(& $mat , & $aux )
{
$M = 4;
$N = 5;
for ( $i = 0; $i < $N ; $i ++)
$aux [0][ $i ] = $mat [0][ $i ];
for ( $i = 1; $i < $M ; $i ++)
for ( $j = 0; $j < $N ; $j ++)
$aux [ $i ][ $j ] = $mat [ $i ][ $j ] +
$aux [ $i - 1][ $j ];
for ( $i = 0; $i < $M ; $i ++)
for ( $j = 1; $j < $N ; $j ++)
$aux [ $i ][ $j ] += $aux [ $i ][ $j - 1];
}
function sumQuery(& $aux , $tli , $tlj , $rbi , $rbj )
{
$res = $aux [ $rbi ][ $rbj ];
if ( $tli > 0)
$res = $res - $aux [ $tli - 1][ $rbj ];
if ( $tlj > 0)
$res = $res - $aux [ $rbi ][ $tlj - 1];
if ( $tli > 0 && $tlj > 0)
$res = $res + $aux [ $tli - 1][ $tlj - 1];
return $res ;
}
$mat = array ( array (1, 2, 3, 4, 6),
array (5, 3, 8, 1, 2),
array (4, 6, 7, 5, 5),
array (2, 4, 8, 9, 4));
preProcess( $mat , $aux );
$tli = 2;
$tlj = 2;
$rbi = 3;
$rbj = 4;
echo ( "Query1: " );
echo (sumQuery( $aux , $tli , $tlj , $rbi , $rbj ));
$tli = 0;
$tlj = 0;
$rbi = 1;
$rbj = 1;
echo ( "\nQuery2: " );
echo (sumQuery( $aux , $tli , $tlj , $rbi , $rbj ));
$tli = 1;
$tlj = 2;
$rbi = 3;
$rbj = 3;
echo ( "\nQuery3: " );
echo (sumQuery( $aux , $tli , $tlj , $rbi , $rbj ));
?>
|
Javascript
<script>
var M = 4;
var N = 5;
function preProcess(mat, aux)
{
for ( var i = 0; i < N; i++)
aux[0,i] = mat[0,i];
for ( var i = 1; i < M; i++)
for ( var j = 0; j < N; j++)
aux[i][j] = mat[i][j] +
aux[i-1][j];
for ( var i = 0; i < M; i++)
for ( var j = 1; j < N; j++)
aux[i][j] += aux[i][j-1];
return 0;
}
function sumQuery( aux, tli, tlj, rbi, rbj)
{
var res = aux[rbi][rbj];
if (tli > 0)
res = res - aux[tli-1][rbj];
if (tlj > 0)
res = res - aux[rbi][tlj-1];
if (tli > 0 && tlj > 0)
res = res + aux[tli-1][tlj-1];
return res;
}
var mat= [[1, 2, 3, 4, 6],
[5, 3, 8, 1, 2],
[4, 6, 7, 5, 5],
[2, 4, 8, 9, 4]];
var aux = new Array(M,N);
preProcess(mat, aux);
var tli = 2, tlj = 2, rbi = 3, rbj = 4;
document.write( "\nQuery1: "
+ sumQuery(aux, tli, tlj, rbi, rbj)+ "<br>" );
tli = 0; tlj = 0; rbi = 1; rbj = 1;
document.write( "\nQuery2: "
+ sumQuery(aux, tli, tlj, rbi, rbj)+ "<br>" );
tli = 1; tlj = 2; rbi = 3; rbj = 3;
document.write( "\nQuery3: "
+ sumQuery(aux, tli, tlj, rbi, rbj));
}
}
</script>
|
Output
Query1: 38
Query2: 11
Query3: 38
Time complexity: O(M x N).
Auxiliary Space: O(M x N), since M x N extra space has been taken.
Source: https://www.geeksforgeeks.org/amazon-interview-experience-set-241-1-5-years-experience/
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