Number Of Ways To Divide A Long Corridor - Leetcode Solution

Problem Statement:

You are given a long corridor of length n. You want to divide it evenly into k parts such that each part has an odd length. Write a function to return the number of ways to divide the corridor.

Example:

Input: n = 6, k = 3 Output: 1 Explanation: You can only divide the corridor into [1,3,2] lengths.

Input: n = 5, k = 2 Output: 2 Explanation: You can divide the corridor into [1,4] or [3,2] lengths.

Approach:

We need to divide the corridor into k parts such that each part has an odd length. So the first observation is that the sum of k odd numbers is always even. And since the corridor has a length n, the sum of k odd numbers must be equal to n.

Let's assume that the k odd numbers are represented as: A1, A2, ..., Ak. Then we have:

A1 + A2 + ... + Ak = n

This is a classic problem of combinatorics that can be solved using dynamic programming. We can define a state dp(i,j) as the number of ways to divide the first i units of the corridor into j odd parts. The goal is to compute dp(n,k).

To compute dp(i,j), we need to consider the last odd part that we add. Let's assume that this last odd part has a length of L. Then we have:

L = 1, 3, 5, ..., i-1

The number of ways to add an odd part of length L is dp(i-L, j-1). The reason why we subtract L from i is that we want to avoid counting the same partitions multiple times.

The final formula for dp(i,j) is:

dp(i,j) = dp(i-1,j-1) + dp(i-3, j-1) + dp(i-5,j-1) + ... + dp(i-(2j-3), j-1)

We can use a bottom-up approach to solve the problem. We initialize dp(i,1) = 1 (there is only one way to divide the first i units of the corridor into one odd part), and dp(1,j) = 0 (you can't divide a length of 1 into an odd number of parts).

Code:

Here is the python code that implements the above approach:

def numWays(n: int, k: int) -> int: # Initialize dp array dp = [[0]*(k+1) for _ in range(n+1)] for i in range(1, n+1): dp[i][1] = 1

# Bottom-up approach to fill dp array
for i in range(1, n+1):
    for j in range(2, k+1):
        for L in range(1, i-(2*j-2), 2):
            dp[i][j] += dp[i-L][j-1]

return dp[n][k]

Time Complexity:

The time complexity of the above solution is O(n^2k). This is because we have three nested loops that run up to n, k and 2j-2 respectively. However, this can be optimized by noticing that the inner loop can be replaced by a constant-time operation.

We can define the sum S(j) as the sum of the first j odd numbers. This can be computed as:

S(j) = j^2

Therefore, we can replace the inner loop with a constant-time operation:

dp[i][j] += dp[i-S(j-1)][j-1]

The time complexity of the optimized solution is O(n*k). As the value of n can be as high as 10^9, we need to use the optimized solution to pass the time limit on leetcode.

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