Submatrix Sum Queries

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++

// C++ program to compute submatrix query sum in O(1)
// time
#include<iostream>
using namespace std;
#define M 4
#define N 5

// Function to preprocess input mat[M][N].  This function
// mainly fills aux[M][N] such that aux[i][j] stores sum
// of elements from (0,0) to (i,j)
int preProcess(int mat[M][N], int aux[M][N])
{
   // Copy first row of mat[][] to aux[][]
   for (int i=0; i<N; i++)
      aux[0][i] = mat[0][i];

   // Do column wise sum
   for (int i=1; i<M; i++)
      for (int j=0; j<N; j++)
         aux[i][j] = mat[i][j] + aux[i-1][j];

   // Do row wise sum
   for (int i=0; i<M; i++)
      for (int j=1; j<N; j++)
         aux[i][j] += aux[i][j-1];
}

// A O(1) time function to compute sum of submatrix
// between (tli, tlj) and (rbi, rbj) using aux[][]
// which is built by the preprocess function
int sumQuery(int aux[M][N], int tli, int tlj, int rbi,
                                              int rbj)
{
    // result is now sum of elements between (0, 0) and
    // (rbi, rbj)
    int res = aux[rbi][rbj];

    // Remove elements between (0, 0) and (tli-1, rbj)
    if (tli > 0)
       res = res - aux[tli-1][rbj];

    // Remove elements between (0, 0) and (rbi, tlj-1)
    if (tlj > 0)
       res = res - aux[rbi][tlj-1];

    // Add aux[tli-1][tlj-1] as elements between (0, 0)
    // and (tli-1, tlj-1) are subtracted twice
    if (tli > 0 && tlj > 0)
       res = res + aux[tli-1][tlj-1];

    return res;
}

// Driver program
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

// Java program to compute submatrix query
// sum in O(1) time
import java.util.*;
import java.io.*;



class GFG {

    static final int M = 4;
    static final int N = 5;

    // Function to preprocess input mat[M][N].
    // This function mainly fills aux[M][N]
    // such that aux[i][j] stores sum of
    // elements from (0,0) to (i,j)
    static int preProcess(int mat[][], int aux[][])
    {

        // Copy first row of mat[][] to aux[][]
        for (int i = 0; i < N; i++)
            aux[0][i] = mat[0][i];

        // Do column wise sum
        for (int i = 1; i < M; i++)
            for (int j = 0; j < N; j++)
                aux[i][j] = mat[i][j] +
                                aux[i-1][j];

        // Do row wise sum
        for (int i = 0; i < M; i++)
            for (int j = 1; j < N; j++)
                aux[i][j] += aux[i][j-1];

        return 0;
    }

    // A O(1) time function to compute sum
    // of submatrix between (tli, tlj) and
    // (rbi, rbj) using aux[][] which is
    // built by the preprocess function
    static int sumQuery(int aux[][], int tli,
                    int tlj, int rbi, int rbj)
    {

        // result is now sum of elements
        // between (0, 0) and (rbi, rbj)
        int res = aux[rbi][rbj];

        // Remove elements between (0, 0)
        // and (tli-1, rbj)
        if (tli > 0)
            res = res - aux[tli-1][rbj];

        // Remove elements between (0, 0)
        // and (rbi, tlj-1)
        if (tlj > 0)
            res = res - aux[rbi][tlj-1];

        // Add aux[tli-1][tlj-1] as elements
        // between (0, 0) and (tli-1, tlj-1)
        // are subtracted twice
        if (tli > 0 && tlj > 0)
            res = res + aux[tli-1][tlj-1];

        return res;
    }

    // Driver code
    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));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python 3 program to compute submatrix
# query sum in O(1) time

M = 4
N = 5

# Function to preprocess input mat[M][N].
# This function mainly fills aux[M][N]
# such that aux[i][j] stores sum
# of elements from (0,0) to (i,j)
def preProcess(mat, aux):

    # Copy first row of mat[][] to aux[][]
    for i in range(0, N, 1):
        aux[0][i] = mat[0][i]

    # Do column wise sum
    for i in range(1, M, 1):
        for j in range(0, N, 1):
            aux[i][j] = mat[i][j] + aux[i - 1][j]

    # Do row wise sum
    for i in range(0, M, 1):
        for j in range(1, N, 1):
            aux[i][j] += aux[i][j - 1]

# A O(1) time function to compute sum of submatrix
# between (tli, tlj) and (rbi, rbj) using aux[][]
# which is built by the preprocess function
def sumQuery(aux, tli, tlj, rbi, rbj):

    # result is now sum of elements
    # between (0, 0) and (rbi, rbj)
    res = aux[rbi][rbj]

    # Remove elements between (0, 0)
    # and (tli-1, rbj)
    if (tli > 0):
        res = res - aux[tli - 1][rbj]

    # Remove elements between (0, 0)
    # and (rbi, tlj-1)
    if (tlj > 0):
        res = res - aux[rbi][tlj - 1]

    # Add aux[tli-1][tlj-1] as elements
    # between (0, 0) and (tli-1, tlj-1)
    # are subtracted twice
    if (tli > 0 and tlj > 0):
        res = res + aux[tli - 1][tlj - 1]

    return res

# Driver Code
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))

# This code is contributed by
# Shashank_Sharma

C#

// C# program to compute submatrix
// query sum in O(1) time
using System;

class GFG
{
    static int M = 4;
    static int N = 5;

    // Function to preprocess input mat[M][N].
    // This function mainly fills aux[M][N]
    // such that aux[i][j] stores sum of
    // elements from (0,0) to (i,j)
    static int preProcess(int [,]mat, int [,]aux)
    {
        // Copy first row of mat[][] to aux[][]
        for (int i = 0; i < N; i++)
            aux[0,i] = mat[0,i];

        // Do column wise sum
        for (int i = 1; i < M; i++)
            for (int j = 0; j < N; j++)
                aux[i,j] = mat[i,j] + aux[i-1,j];

        // Do row wise sum
        for (int i = 0; i < M; i++)
            for (int j = 1; j < N; j++)
                aux[i,j] += aux[i,j-1];

        return 0;
    }

    // A O(1) time function to compute sum
    // of submatrix between (tli, tlj) and
    // (rbi, rbj) using aux[][] which is
    // built by the preprocess function
    static int sumQuery(int [,]aux, int tli,
                        int tlj, int rbi, int rbj)
    {
        // result is now sum of elements
        // between (0, 0) and (rbi, rbj)
        int res = aux[rbi,rbj];

        // Remove elements between (0, 0)
        // and (tli-1, rbj)
        if (tli > 0)
            res = res - aux[tli-1,rbj];

        // Remove elements between (0, 0)
        // and (rbi, tlj-1)
        if (tlj > 0)
            res = res - aux[rbi,tlj-1];

        // Add aux[tli-1][tlj-1] as elements
        // between (0, 0) and (tli-1, tlj-1)
        // are subtracted twice
        if (tli > 0 && tlj > 0)
            res = res + aux[tli-1,tlj-1];

        return res;
    }

    // Driver code
    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));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to compute submatrix
// query sum in O(1) time

// Function to preprocess input mat[M][N].
// This function mainly fills aux[M][N]
// such that aux[i][j] stores sum
// of elements from (0,0) to (i,j)
function preProcess(&$mat, &$aux)
{
    $M = 4;
    $N = 5;

    // Copy first row of mat[][] to aux[][]
    for ($i = 0; $i < $N; $i++)
        $aux[0][$i] = $mat[0][$i];

    // Do column wise sum
    for ($i = 1; $i < $M; $i++)
        for ($j = 0; $j < $N; $j++)
            $aux[$i][$j] = $mat[$i][$j] +
                           $aux[$i - 1][$j];

    // Do row wise sum
    for ($i = 0; $i < $M; $i++)
        for ($j = 1; $j < $N; $j++)
            $aux[$i][$j] += $aux[$i][$j - 1];
}

// A O(1) time function to compute sum of
// submatrix between (tli, tlj) and
// (rbi, rbj) using aux[][] which is built
// by the preprocess function
function sumQuery(&$aux, $tli, $tlj, $rbi,$rbj)
{
    // result is now sum of elements
    // between (0, 0) and (rbi, rbj)
    $res = $aux[$rbi][$rbj];

    // Remove elements between (0, 0)
    // and (tli-1, rbj)
    if ($tli > 0)
    $res = $res - $aux[$tli - 1][$rbj];

    // Remove elements between (0, 0)
    // and (rbi, tlj-1)
    if ($tlj > 0)
    $res = $res - $aux[$rbi][$tlj - 1];

    // Add aux[tli-1][tlj-1] as elements between (0, 0)
    // and (tli-1, tlj-1) are subtracted twice
    if ($tli > 0 && $tlj > 0)
    $res = $res + $aux[$tli - 1][$tlj - 1];

    return $res;
}

// Driver Code
$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));

// This code is contributed by Shivi_Aggarwal
?>

Javascript

<script>
// Javascript program to compute submatrix query
// sum in O(1) time


    var M = 4;
    var N = 5;

    // Function to preprocess input mat[M][N].
    // This function mainly fills aux[M][N]
    // such that aux[i][j] stores sum of
    // elements from (0,0) to (i,j)
    function preProcess(mat, aux)
    {

        // Copy first row of mat[][] to aux[][]
        for (var i = 0; i < N; i++)
            aux[0,i] = mat[0,i];

        // Do column wise sum
        for (var i = 1; i < M; i++)
            for (var j = 0; j < N; j++)
                aux[i][j] = mat[i][j] +
                                aux[i-1][j];

        // Do row wise sum
        for (var i = 0; i < M; i++)
            for (var j = 1; j < N; j++)
                aux[i][j] += aux[i][j-1];

        return 0;
    }

    // A O(1) time function to compute sum
    // of submatrix between (tli, tlj) and
    // (rbi, rbj) using aux[][] which is
    // built by the preprocess function
    function sumQuery( aux, tli, tlj, rbi, rbj)
    {

        // result is now sum of elements
        // between (0, 0) and (rbi, rbj)
        var res = aux[rbi][rbj];

        // Remove elements between (0, 0)
        // and (tli-1, rbj)
        if (tli > 0)
            res = res - aux[tli-1][rbj];

        // Remove elements between (0, 0)
        // and (rbi, tlj-1)
        if (tlj > 0)
            res = res - aux[rbi][tlj-1];

        // Add aux[tli-1][tlj-1] as elements
        // between (0, 0) and (tli-1, tlj-1)
        // are subtracted twice
        if (tli > 0 && tlj > 0)
            res = res + aux[tli-1][tlj-1];

        return res;
    }

    // Driver code

        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));
    }
}

// This code is contributed by shivanisinghss2110
</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.

This article is contributed by Aarti_Rathi and Shivam Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Source: https://www.geeksforgeeks.org/amazon-interview-experience-set-241-1-5-years-experience/