<p>You are given an integer <code>n</code> and a 2D integer array <code>queries</code>.</p>

<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code>. Initially, there is a <strong>unidirectional</strong> road from city <code>i</code> to city <code>i + 1</code> for all <code>0 &lt;= i &lt; n - 1</code>.</p>

<p><code>queries[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents the addition of a new <strong>unidirectional</strong> road from city <code>u<sub>i</sub></code> to city <code>v<sub>i</sub></code>. After each query, you need to find the <strong>length</strong> of the <strong>shortest path</strong> from city <code>0</code> to city <code>n - 1</code>.</p>

<p>Return an array <code>answer</code> where for each <code>i</code> in the range <code>[0, queries.length - 1]</code>, <code>answer[i]</code> is the <em>length of the shortest path</em> from city <code>0</code> to city <code>n - 1</code> after processing the <strong>first </strong><code>i + 1</code> queries.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, queries = [[2,4],[0,2],[0,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[3,2,1]</span></p>

<p><strong>Explanation: </strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image8.jpg" style="width: 350px; height: 60px;" /></p>

<p>After the addition of the road from 2 to 4, the length of the shortest path from 0 to 4 is 3.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image9.jpg" style="width: 350px; height: 60px;" /></p>

<p>After the addition of the road from 0 to 2, the length of the shortest path from 0 to 4 is 2.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image10.jpg" style="width: 350px; height: 96px;" /></p>

<p>After the addition of the road from 0 to 4, the length of the shortest path from 0 to 4 is 1.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4, queries = [[0,3],[0,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,1]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image11.jpg" style="width: 300px; height: 70px;" /></p>

<p>After the addition of the road from 0 to 3, the length of the shortest path from 0 to 3 is 1.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image12.jpg" style="width: 300px; height: 70px;" /></p>

<p>After the addition of the road from 0 to 2, the length of the shortest path remains 1.</p>
</div>

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

<ul>
	<li><code>3 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= queries.length &lt;= 500</code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= queries[i][0] &lt; queries[i][1] &lt; n</code></li>
	<li><code>1 &lt; queries[i][1] - queries[i][0]</code></li>
	<li>There are no repeated roads among the queries.</li>
</ul>

class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:

        dp = [(n - i - 1) for i in range(n)]
        m = defaultdict(lambda: set())
        for i in range(n):
            m[i].add(i + 1)

        shortest = []
        for q in queries:
            s, e = q[0], q[1]
            m[s].add(e)
            dp[s] = min(dp[s], dp[e] + 1)
            for i in range(s - 1, -1, -1):
                for n in m[i]:
                    dp[i] = min(dp[n] + 1, dp[i], dp[i + 1] + 1, dp[s] + (s - i))
            shortest.append(dp[0])

        return shortest


        '''

                -------------
                |     |     |
             0  1  2  3  4  5
            [4, 2, 3, 2, 1, 0]


                        |-----|
                     _______
                     |     |  
                  0  1  2  3  4
            dp = [3, 2, 1, 1, 0]


            {
                source: [destinations]
            }


            0 to n - 1

            dist: (s - 0) + (road) + (n - e - 1)

            dp = [float('inf') for _ in range(n)]

            dp = [3, 2, 1, 1, 0] / [n - 1, n - 2, ..., n - n]

            dp[q[0]] = min(dp[q[0]], dp[q[1]] + 1)

            1. Initialize a DP array with distances [n - 1, n - 2, ..., n - n]
            2. For each query, update min distance for q[0] and q[1]
            3. Left shift all other values before q[0]
            4. Append DP[0] to result and return array
        '''
17100