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Word Break
Bookmark
Input
Standard
Reuse word
No solution
Custom
s
=
leetcode
,
wordDict
=
[leet, code]
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
F
0
F
1
F
2
F
3
F
4
F
5
F
6
F
7
F
8
can "leetcode" be split entirely into dictionary words?
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
F
0
F
1
F
2
F
3
F
4
F
5
F
6
F
7
F
8
can "leetcode" be split entirely into dictionary words?
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
F
4
F
5
F
6
F
7
F
8
i = 2
i = 2
does some dict word end a valid split of
"le"
?
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
F
4
F
5
F
6
F
7
F
8
i = 3
j = 0
dp[0] =
T
but
"lee"
is not a dict word
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
"leet"
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
F
8
i = 4
j = 0
dp[0] =
T
and
"leet"
is a word:
dp[4]
=
T
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
F
8
i = 5
j = 3
split at
j = 3
: dp[3] =
F
, no valid break ends there
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
F
8
i = 6
j = 2
split at
j = 2
: dp[2] =
F
, no valid break ends there
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
F
8
i = 7
i = 7
does some dict word end a valid split of
"leetcod"
?
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
F
8
i = 7
j = 4
dp[4] =
T
but
"cod"
is not a dict word
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
F
8
i = 8
j = 0
dp[0] =
T
but
"leetcode"
is not a dict word
s =
"leetcode"
dict
leet
code
s
l
e
e
t
c
o
d
e
dp
T
0
F
1
F
2
F
3
T
4
F
5
F
6
F
7
T
8
"leetcode" =
leet
+
code
algo
master
.
io
Step:
Can "leetcode" be split entirely into dictionary words?
0 / 46
Input
Standard
Reuse word
No solution
Custom
s
=
leetcode
,
wordDict
=
[leet, code]
0 / 46
algo
master
.
io
Step:
Can "leetcode" be split entirely into dictionary words?
