Number of Ways to Reach a Position After Exactly k Steps
<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>
Math; Dynamic Programming; Combinatorics
class Solution {
public:
    int numberOfWays(int startPos, int endPos, int k) {
        const int mod = 1e9 + 7;
        int f[k + 1][k + 1];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i, int j) -> int {
            if (i > j || j < 0) {
                return 0;
            }
            if (j == 0) {
                return i == 0 ? 1 : 0;
            }
            if (f[i][j] != -1) {
                return f[i][j];
            }
            f[i][j] = (dfs(i + 1, j - 1) + dfs(abs(i - 1), j - 1)) % mod;
            return f[i][j];
        };
        return dfs(abs(startPos - endPos), k);
    }
};
