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MD ARIFUL HAQUE
MD ARIFUL HAQUE

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3123. Find Edges in Shortest Paths

3123. Find Edges in Shortest Paths

Hard

You are given an undirected weighted graph of n nodes numbered from 0 to n - 1. The graph consists of m edges represented by a 2D array edges, where edges[i] = [ai, bi, wi] indicates that there is an edge between nodes ai and bi with weight wi.

Consider all the shortest paths from node 0 to node n - 1 in the graph. You need to find a boolean array answer where answer[i] is true if the edge edges[i] is part of at least one shortest path. Otherwise, answer[i] is false.

Return the array answer.

Note that the graph may not be connected.

Example 1:

graph35drawio-1

  • Input: n = 6, edges = [[0,1,4],[0,2,1],[1,3,2],[1,4,3],[1,5,1],[2,3,1],[3,5,3],[4,5,2]]
  • Output: [true,true,true,false,true,true,true,false]
  • Explanation: The following are all the shortest paths between nodes 0 and 5:
    • The path 0 -> 1 -> 5: The sum of weights is 4 + 1 = 5.
    • The path 0 -> 2 -> 3 -> 5: The sum of weights is 1 + 1 + 3 = 5.
    • The path 0 -> 2 -> 3 -> 1 -> 5: The sum of weights is 1 + 1 + 2 + 1 = 5.

Example 2:

graphhhh

  • Input: n = 4, edges = [[2,0,1],[0,1,1],[0,3,4],[3,2,2]]
  • Output: [true,false,false,true]
  • Explanation: There is one shortest path between nodes 0 and 3, which is the path 0 -> 2 -> 3 with the sum of weights 1 + 2 = 3.

Constraints:

  • 2 <= n <= 5 * 104
  • m == edges.length
  • 1 <= m <= min(5 * 104, n * (n - 1) / 2)
  • 0 <= ai, bi < n
  • ai != bi
  • 1 <= wi <= 105
  • There are no repeated edges.

Solution:

class Solution {

    /**
     * @param String $s
     * @param Integer $k
     * @return Integer
     */
    public function longestIdealString($s, $k) {
        // dp[$i] := the longest subsequence that ends in ('a' + $i)
        $dp = array_fill(0, 26, 0);

        for ($i = 0; $i < strlen($s); $i++) {
            $c = $s[$i];
            $charIndex = ord($c) - ord('a');
            $dp[$charIndex] = 1 + $this->getMaxReachable($dp, $charIndex, $k);
        }

        return max($dp);
    }

    private function getMaxReachable($dp, $i, $k) {
        $first = max(0, $i - $k);
        $last = min(25, $i + $k);
        $maxReachable = 0;
        for ($j = $first; $j <= $last; $j++) {
            $maxReachable = max($maxReachable, $dp[$j]);
        }
        return $maxReachable;
    }
}
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