<p>You are given two <strong>positive</strong> integers <code>startPos</code> and <code>endPos</code>. Initially, you are standing at position <code>startPos</code> on an <strong>infinite</strong> number line. With one step, you can move either one position to the left, or one position to the right.</p>

<p>Given a positive integer <code>k</code>, return <em>the number of <strong>different</strong> ways to reach the position </em><code>endPos</code><em> starting from </em><code>startPos</code><em>, such that you perform <strong>exactly</strong> </em><code>k</code><em> steps</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two ways are considered different if the order of the steps made is not exactly the same.</p>

<p><strong>Note</strong> that the number line includes negative integers.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> startPos = 1, endPos = 2, k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -&gt; 2 -&gt; 3 -&gt; 2.
- 1 -&gt; 2 -&gt; 1 -&gt; 2.
- 1 -&gt; 0 -&gt; 1 -&gt; 2.
It can be proven that no other way is possible, so we return 3.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> startPos = 2, endPos = 5, k = 10
<strong>Output:</strong> 0
<strong>Explanation:</strong> It is impossible to reach position 5 from position 2 in exactly 10 steps.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= startPos, endPos, k &lt;= 1000</code></li>
</ul>

# Problem 2400 - Number of Ways to Reach a Position After Exactly k Steps

class Solution:
    @cache
    def numberOfWays(self, startPos: int, endPos: int, k: int) -> int:
        # What is the distance between start and end?
        distance = abs(endPos - startPos)

        if distance > k:
            return 0
        if (k - distance) % 2 == 1:
            return 0
        
        # How many times we can go to left, how many in the opposite direction?
        oneDirection = distance + (k - distance) // 2 
        anotherDirection = (k - distance) // 2

        # In how many ways we can order these steps?
        # permutations with repetition (think it as the number of
        #  ways to assign m >= k indices from 1 to m to a pool of k items)
        # C_k_oneDirection = k!/(oneDirection! * anotherDirection!)

        def nCk(n, k):
            kMod = 1_000_000_007

            # dp[i] := C(n so far, i)
            # (is the line in an imaginary matrix filled
            #  until the main diagonal)
            dp = [0] * (k + 1)
            dp[0] = 1

            for i in range(n): # Calculate n times.
                # Use the previous line in the imaginary matrix
                for j in range(min(i+1, k) - 1, -1, -1):
                    dp[j + 1] += dp[j]
                    dp[j + 1] %= kMod
            return dp[k]

        return nCk(k, oneDirection)

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