Cube_Summation.py

"""
Cube Summation
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Link: https://www.hackerrank.com/contests/wissen-coding-challenge-2021/challenges/cube-summation
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Define a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinates (1,1,1) and the last block is defined by the coordinates (n,n,n). There are two types of queries.

UPDATE x y z W
Update the value of block (x,y,z) to W.

QUERY x1 y1 z1 x2 y2 z2
Calculate the sum of the values of blocks whose x coordinate is between x1 and x2 (inclusive), y coordinate between y1 and y2 (inclusive) and z coordinate between z1 and z2 (inclusive).

Function Description

Complete the cubeSum function in the editor below.

cubeSum has the following parameters: - *int n: the dimensions of the 3-d matrix - string operations[m]: the operations to perform

Returns
- int[]: the results of each QUERY operation

Input Format
The first line contains an integer T, the number of test-cases. T testcases follow.

For each test case, the first line contains two space-separated integers, n and m. n defines the n x n x n matrix.
m defines the number of operations.
The next m lines will contain an operation either of these forms:

 1. UPDATE x y z W
 2. QUERY  x1 y1 z1 x2 y2 z2 

Sample Input

2
4 5
UPDATE 2 2 2 4
QUERY 1 1 1 3 3 3
UPDATE 1 1 1 23
QUERY 2 2 2 4 4 4
QUERY 1 1 1 3 3 3
2 4
UPDATE 2 2 2 1
QUERY 1 1 1 1 1 1
QUERY 1 1 1 2 2 2
QUERY 2 2 2 2 2 2

Sample Output

4
4
27
0
1
1

Explanation
In the first test case, there is a cube of 4 * 4 * 4 and there are 5 queries. Initially all the cells (1,1,1) to (4,4,4) are 0.
UPDATE 2 2 2 4 makes the cell (2,2,2) = 4
QUERY 1 1 1 3 3 3. As (2,2,2) is updated to 4 and the rest are all 0. The answer to this query is 4.
UPDATE 1 1 1 23. updates the cell (1,1,1) to 23. QUERY 2 2 2 4 4 4. Only the cell (1,1,1) and (2,2,2) are non-zero and (1,1,1) is not between (2,2,2) and (4,4,4). So, the answer is 4.
QUERY 1 1 1 3 3 3. 2 cells are non-zero and their sum is 23+4 = 27.
"""

(Python Solution)

#!/bin/python3

import math
import os
import random
import re
import sys

source = sys.stdin

def update(A,x,y,z,val):
    for a in A:
        if (a[0] == x) and (a[1] == y) and (a[2] == z) :
            a[3] = val
            return
    A.append([x,y,z,val])

def query(A,x1,y1,z1,x2,y2,z2):
    total = 0
    for a in A:
        if ((a[0] >= x1) and (a[0] <= x2) 
        and (a[1] >= y1) and (a[1] <= y2) 
        and (a[2] >= z1) and (a[2] <= z2)):
            total += a[3] 
    return total

A = []
T = int(source.readline())

for t in range(T):
    [N,M] = [int(s) for s in source.readline().split()]

    A.clear()
    for _ in range(M):
        c = source.readline().split()
        args = [int(s) for s in c[1:]]
        if c[0]=='UPDATE':
            update(A, *args)
        else:
            val = query(A,*args)
            print(val)