# 902ms Java Accepted solutiuon

• ``````public class Solution {
int dist( String x, String y ) {
int k = 0;
for ( int i = 0; i < x.length(); ++i )
if ( x.charAt(i) != y.charAt(i) && (++k) >= 2 )
return 2;
return k;
}
private final static int oo = (1<<29);
List<String> L;
int D,src,sink;
int []distance,d;
boolean []seen = new boolean[1<<20];
boolean [][]g;
void generate( int x, int costSoFar, int []idx, List<List<String>> lst ) {
if ( x == sink ) {
if ( costSoFar == D ) {
List<String> l = new ArrayList<>();
for ( int j = 0; j < costSoFar; ++j )
}
return ;
}
if ( costSoFar >= D || d[x] == +oo || distance[x] == +oo || d[x]+distance[x] != D ) return ;
for ( int y,i = 0; i < adj[x].length && distance[adj[x][i]] < +oo && d[x]+distance[adj[x][i]] <= D; ++i )
if ( !seen[y = adj[x][i]] && distance[y] < +oo && costSoFar+distance[y] <= D ) {
idx[costSoFar] = y;
seen[y] = true ;
generate(y,costSoFar+1,idx,lst);
seen[y] = false ;
}
}
public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
L = wordList;
if ( wordList.contains(beginWord) ) src = wordList.indexOf(beginWord);
else {
src = wordList.indexOf(beginWord);
}
if ( wordList.contains(endWord) ) sink = wordList.indexOf(endWord);
else return new ArrayList<>();
int i,j,k,n = wordList.size();
g = new boolean[n][n];
for ( i = 0; i < n; ++i )
for ( j = i+1; j < n; ++j )
g[i][j] = g[j][i] = dist(wordList.get(i),wordList.get(j))==1;
for ( i = 0; i < n; ++i ) {
for ( j = k = 0; k < n; ++k )
if ( g[i][k] ) ++j;
for ( k = 0, j = 0; j < n; ++j )
if ( g[i][j] )
}
d = new int[n];
for ( i = 0; i < n; d[i++] = +oo );
for ( q.add(src), d[src] = 0; !q.isEmpty(); )
for ( i = q.poll(), k = 0; k < adj[i].length; ++k )
if ( d[j=adj[i][k]] > d[i]+1 ) {
d[j] = d[i]+1;
}
if ( (D=d[sink]) == +oo )
return new ArrayList<>();
distance = new int[n];
for ( i = 0; i < n; ++i ) distance[i] = +oo;
for ( q.add(sink), distance[sink] = 0; !q.isEmpty(); )
for ( i = q.poll(), k = 0; k < adj[i].length; ++k )
if ( distance[j=adj[i][k]] > distance[i]+1 ) {
distance[j] = distance[i]+1;
}
for ( i = 0; i < n; ++i ) {
for ( boolean flag = true ; flag; )
for ( flag = false, k = 0; k < adj[i].length-1; ++k )
flag = true ;
}
}
List<List<String>> res = new ArrayList<>();
seen[src] = true ;
int []idx = new int[D];
generate(src,0,idx,res);
seen[src] = false ;
return res;
}
}
``````

I believe the problem is NP, so the optimizations in this problem can be pretty much ad hoc, and the center of gravity is certainly two-end BFS, and one cannot in principle do better than backtracking when generating the paths.

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