Last Updated: 2 Dec, 2020

Sub Query Sum

Easy

Asked in companies

Solve now

Problem statement

You have given a 2-dimensional array ‘ARR’ with ‘N’ rows and ‘M’ columns. The queries are given in a 2-dimensional array ‘Queries’ of size ‘K’, in which Queries[i] contains four integers denoting the left top and right bottom indices of the submatrix whose sum we need to find. Your task is to find the sum of the submatrix for each query given in the array ‘Queries’.

For example:

Consider ARR = [[1 , 2 , 3] , [3 , 4 , 1] , [2 , 1 , 2]] and Queries = [[0 , 0 , 1 , 2]], the submatrix with the left top index (0 , 0) and right bottom index (1 , 2) is  
                      [[1 , 2 , 3] , 
                       [3 , 4 , 1]]. 
The sum of the submatrix is 14. Hence, the answer is 14 in this case.

Input Format:

The first line of the input contains an integer, 'T,’ denoting the number of test cases.

The first line of each test case contains three space-separated integers, 'N', M’, and ‘K’, denoting the number of rows and the number of columns in the array 'ARR', and the number of rows in the array 'Queries' respectively.

The Next 'N' lines of each test case contain 'M' space-separated integers denoting the elements of the array 'ARR'.

The Next 'K' lines of each test case contain four space-separated integers denoting the elements of the array ‘Queries’.

Output Format:

For each test case, print ‘K’ space-separated integers - the sum of the submatrix for each query.

Print the output of each test case in a separate line.

Constraints :

 1 <= N <= 10 ^ 3     
 1 <= M <= 10 ^ 3
 1 <= K <= 10 ^ 3
 1 <= ARR[i][j] <= 10 ^ 6
 0 <= Queries[i][0] , Queries[i][2] < N
 0 <= Queries[i][1] , Queries[i][3] < M 

 Where 'T' denotes the number of test cases, 'N' and 'M' denotes the number of rows and the number of columns in the array ‘ARR’ respectively, ’K’ denotes the number of rows in the array ‘Queries’, 'ARR[i][j]' denotes the ’j-th’ element of  'i-th' row of the array 'ARR' and 'Queries[i]' contains four integers denoting the left top and right bottom indices of the submatrix.  

Time Limit: 1sec

Solve now

Approaches

01 Approach

A simple method is to traverse through the submatrix, and we will find the sum of the submatrix for each query.

Our approach will be to create an array answer to store the sum of the submatrix for each query. We will iterate queryNumber from 0 to K - 1, and we will maintain a variable sum to store the sum of the submatrix of the current query.

We iterate index from Queries[queryNumber][0] to Queries[queryNumber][2]. In each iteration, we will iterate pos from Queries[queryNumber][1] to Queries[queryNumber][3]. The element ARR[index][pos] is present in the submatrix. So, we will increment sum by ARR[index][pos].
The variable sum stores the sum of the submatrix, and we will insert the variable sum in the array answer.

In the end, we will return the array answer.

Algorithm:

Create an array answer of size K, which will store the sum of the submatrix for each query.
Iterate queryNumber from 0 to K - 1.
- Set the variable sum as 0. The variable sum stores the sum of the submatrix of the current query.
- Iterate index from Queries[queryNumber][0] to Queries[queryNumber][2].
  - Iterate pos from Queries[queryNumber][1] to Queries[queryNumber][2].
    - Increment sum by ARR[index][pos].
- Insert sum in the array answer.
Return the array answer.

Solve now

02 Approach

The basic idea is to create a matrix auxiliary with N rows and M columns in which auxiliary[r][c] will store the sum of the submatrix from (0 , 0) to (r , c). Using this idea, we can find the sum of the submatrix for each query in O(1) Space Complexity.

We can find the sum of the submatrix from (r1 , c1) to (r2 , c2) in the following way given below.

sum = auxiliary[r2][c2] - auxiliary[r2][c1 - 1] - auxiliary[r1 - 1][c2] + auxiliary[r1- 1][c1 - 1]

The element auxiliary[r2][c2] stores the sum of the submatrix from (0 , 0) to (r2 , c2), auxiliary[r1 - 1][c2] stores the sum of the submatrix from (0 , 0) to (r1 - 1 , c2), auxiliary[r2][c1 - 1] stores the sum of the submatrix from (0 , 0) to (r2 , c1 - 1), and auxiliary[r1 - 1][c1 - 1] stores the sum of the submatrix from (0 , 0) to (r1 - 1 , c1 - 1).

Our approach will be to create an array answer to store the sum of the submatrix for each query.

First, we will create the Auxiliary matrix. We will iterate the variable index from 1 to N - 1, and in each iteration, we will iterate pos from 0 to M - 1. We will increment auxiliary[index][pos] by ARR[index][pos] + auxiliary[index - 1][pos].
We will iterate the variable index from 0 to N - 1, and in each iteration, we will iterate pos from 1 to M - 1. We will increment auxiliary[index][pos] by auxiliary[index][pos - 1].
We have created the auxiliary matrix, which we will use to find the sum of the submatrix for each query. Now, we will find the sum of the submatrix for each query. We will iterate the variable queryNumber from 0 to K - 1, and we will apply conditions to use the formula discussed above to find the sum of the submatrix. We will insert the sum of the submatrix of the current query in the array answer.

In the end, we will return the array answer.

Algorithm:

Create an array answer of size K, which will store the sum of the submatrix for each query.
Create an array auxiliary of size N and columns of size M. We will use this array to find the sum of the submatrix for each query.
Copy the first row of the array ARR into the array auxiliary.
Iterate index from 1 to N - 1.
- Iterate pos from 0 to M - 1.
  - Increment auxiliary[index][pos] by ARR[index][pos] + auxiliary[index - 1][pos].
Iterate index from 0 to N - 1.
- Iterate pos from 1 to M - 1.
  - Increment auxiliary[index][pos] by auxiliary[index][pos - 1].
Iterate queryNumber from 0 to K - 1.
- Set the variable topRow as Queries[queryNumber][0], topColumn as Queries[queryNumber][1], bottomRow as Queries[queryNumber][2], bottomColumn as Queries[queryNumber][3].
- Set the variable sum as auxiliary[bottomRow][bottomColumn]. The variable sum stores the sum of the submatrix of the current query.
- Check if topRow is more than 0.
  - Decrement sum by auxiliary[topRow - 1][bottomColumn].
- Check if topColumn is more than 0.
  - Decrement sum by auxiliary[bottomRow][topColumn - 1].
- Check if topRow and topColumn are more than 0.
  - Increment sum by auxiliary[topRow-1][topColumn-1].
- Insert sum in the array answer.
Return the array answer.

Solve now