Given N rectangular cuboids in a three-dimensional space the task is to find for each cuboid the number of other cuboids that share a face with it In other words we need to count the number of cuboids

To solve this problem, we can use a brute-force approach where we iterate through each pair of cuboids and check if they share a face or not. However, this approach would be inefficient and would result in a time complexity of O(N^2), which is not feasible given the constraints.

Instead, we can use a more efficient approach by counting the number of cuboids that intersect each face of a given cuboid. To do this, we can create six arrays to store the number of intersecting cuboids for each face of a cuboid. Let's call these arrays "countX1", "countX2", "countY1", "countY2", "countZ1", and "countZ2".

To calculate the number of intersecting cuboids for each face, we can iterate through all the cuboids and check if their faces intersect with the face of the current cuboid. If a face intersects with another cuboid, we increment the corresponding count array by 1.

Here's the step-by-step algorithm to solve this problem:

1. Initialize the count arrays "countX1", "countX2", "countY1", "countY2", "countZ1", and "countZ2" with zeros.

2. Iterate through each cuboid in the input.

3. For each cuboid, calculate the minimum and maximum values for each coordinate axis: minX, maxX, minY, maxY, minZ, and maxZ. These values represent the ranges of each face of the cuboid.

4. Iterate through all the cuboids again.

5. For each cuboid, calculate the minimum and maximum values for each coordinate axis: minX2, maxX2, minY2, maxY2, minZ2, and maxZ2.

6. Check if the face of the current cuboid intersects with the face of the other cuboid. To do this, compare the ranges of each axis and check if there is an overlap. For example, if maxX of the current cuboid is greater than or equal to minX2 of the other cuboid, and minX of the current cuboid is less than or equal to maxX2 of the other cuboid, then the faces intersect along the X-axis.

7. If a face intersects with another cuboid, increment the corresponding count array by 1. For example, if the X-axis face of the current cuboid intersects with another cuboid, increment countX1 and countX2 by 1.

8. Repeat steps 4-7 for all the cuboids.

9. The final count of intersecting cuboids for each cuboid is the sum of the corresponding count arrays. For example, the count of intersecting cuboids for the first cuboid is countX1[0] + countX2[0] + countY1[0] + countY2[0] + countZ1[0] + countZ2[0].

10. Return the counts for all the cuboids.

This approach has a time complexity of O(N), where N is the number of cuboids. This is because we iterate through all the cuboids twice, and the operations inside the loops have constant time complexity.

Here's the Python code implementation of the above algorithm:

def count_intersecting_cuboids(cuboids):
    n = len(cuboids)
    countX1 = [0] * n
    countX2 = [0] * n
    countY1 = [0] * n
    countY2 = [0] * n
    countZ1 = [0] * n
    countZ2 = [0] * n

    for i in range(n):
        minX, maxX, minY, maxY, minZ, maxZ = cuboids[i]
        for j in range(n):
            if i == j:
                continue
            minX2, maxX2, minY2, maxY2, minZ2, maxZ2 = cuboids[j]
            if maxX >= minX2 and minX <= maxX2:
                countX1[i] += 1
                countX2[j] += 1
            if maxY >= minY2 and minY <= maxY2:
                countY1[i] += 1
                countY2[j] += 1
            if maxZ >= minZ2 and minZ <= maxZ2:
                countZ1[i] += 1
                countZ2[j] += 1

    return [countX1[i] + countX2[i] + countY1[i] + countY2[i] + countZ1[i] + countZ2[i] for i in range(n)]

You can test the code with sample inputs to verify its correctness