# Tile Stacking Problem

A stable tower of height n is a tower consisting of exactly n tiles of unit height stacked vertically in such a way, that no bigger tile is placed on a smaller tile. An example is shown below:

We have infinite number of tiles of sizes 1, 2, …, m. The task is calculate the number of different stable tower of height n that can be built from these tiles, with a restriction that you can use at most k tiles of each size in the tower.

Note: Two tower of height n are different if and only if there exists a height h (1 <= h <= n), such that the towers have tiles of different sizes at height h.

Examples:

```Input : n = 3, m = 3, k = 1.
Output : 1
Possible sequences: { 1, 2, 3}.

Input : n = 3, m = 3, k = 1.
Output : 7
{1, 1, 2}, {1, 1, 3}, {1, 2, 2},
{1, 2, 3}, {1, 3, 3}, {2, 2, 3},
{2, 3, 3}.```

We basically need to count number of decreasing sequences of length n using numbers from 1 to m where every number can be used at most k times. We can recursively compute count for n using count for n-1.

The idea is to use Dynamic Programming. Declare a 2D array dp[][], where each state dp[i][j] denotes the number of decreasing sequences of length i using numbers from j to m. We need to take care of the fact that a number can be used a most k times. This can be done by considering 1 to k occurrences of a number. Hence our recurrence relation becomes:

Also, we can use the fact that for a fixed j we are using the consecutive values of previous k values of i. Hence, we can maintain a prefix sum array for each state. Now we have got rid of the k factor for each state.

Below is the C++ implemantation of this approach:

```// CPP program to find number of ways to make stable
// tower of given height.
#include <bits/stdc++.h>
using namespace std;
#define N 100

int possibleWays(int n, int m, int k)
{
int dp[N][N];
int presum[N][N];
memset(dp, 0, sizeof dp);
memset(presum, 0, sizeof presum);

// Initialing 0th row to 0.
for (int i = 1; i < n + 1; i++) {
dp[0][i] = 0;
presum[0][i] = 1;
}

// Initialing 0th column to 0.
for (int i = 0; i < m + 1; i++)
presum[i][0] = dp[i][0] = 1;

// For each row from 1 to m
for (int i = 1; i < m + 1; i++) {

// For each column from 1 to n.
for (int j = 1; j < n + 1; j++) {

// Initialing dp[i][j] to presum of (i - 1, j).
dp[i][j] = presum[i - 1][j];
if (j > k) {
dp[i][j] -= presum[i - 1][j - k - 1];
}
}

// Calculating presum for each i, 1 <= i <= n.
for (int j = 1; j < n + 1; j++)
presum[i][j] = dp[i][j] + presum[i][j - 1];
}

return dp[m][n];
}

// Driver Program
int main()
{
int n = 3, m = 3, k = 2;
cout << possibleWays(n, m, k) << endl;
return 0;
}
```

Output:

```7
```

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/tile-stacking-problem/
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