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:
- 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 is4 + 1 = 5
. - The path
0 -> 2 -> 3 -> 5
: The sum of weights is1 + 1 + 3 = 5
. - The path
0 -> 2 -> 3 -> 1 -> 5
: The sum of weights is1 + 1 + 2 + 1 = 5
.
- The path
Example 2:
- 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 weights1 + 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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