Learn
Practice
Interview
Resources
Newsletter
F
Toggle theme
0
F
0
Toggle menu
Animations
← Back to All Animations
Last Stone Weight II
Bookmark
Input
Example 1
Example 2
All same
Single stone
Custom
stones
=
[2, 7, 4, 1, 8, 1]
stones
2
7
4
1
8
1
smashing everything leaves |pileA - pileB|, so split the stones evenly
stones
2
7
4
1
8
1
smashing everything leaves |pileA - pileB|, so split the stones evenly
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+2
dp
T
0
F
1
F
2
F
3
F
4
F
5
F
6
F
7
F
8
F
9
F
10
F
11
j-2 = 4
j = 6
dp[6]
← dp[
4
] =
F
weight 6 stays unreachable
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+7
dp
T
0
F
1
T
2
F
3
F
4
F
5
F
6
F
7
F
8
T
9
F
10
F
11
j-7 = 2
j = 9
dp[9]
=
T
(2 + 7 = 9)
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+4
dp
T
0
F
1
T
2
F
3
F
4
F
5
F
6
T
7
F
8
T
9
F
10
T
11
j-4 = 4
j = 8
dp[8]
← dp[
4
] =
F
weight 8 stays unreachable
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+1
dp
T
0
F
1
T
2
F
3
T
4
F
5
T
6
T
7
F
8
T
9
F
10
T
11
j-1 = 9
j = 10
dp[10]
← dp[10 -
1
] = dp[
9
] =
T
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+1
dp
T
0
F
1
T
2
F
3
T
4
T
5
T
6
T
7
T
8
T
9
T
10
T
11
j-1 = 2
j = 3
dp[3]
← dp[3 -
1
] = dp[
2
] =
T
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+8
dp
T
0
T
1
T
2
T
3
T
4
T
5
T
6
T
7
T
8
T
9
T
10
T
11
j-8 = 2
j = 10
dp[10]
=
T
(2 + 8 = 10)
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+1
dp
T
0
T
1
T
2
T
3
T
4
T
5
T
6
T
7
T
8
T
9
T
10
T
11
j-1 = 9
j = 10
dp[10]
=
T
(9 + 1 = 10)
stones
2
7
4
1
8
1
sum =
23
target =
11
stone
+1
dp
T
0
T
1
T
2
T
3
T
4
T
5
T
6
T
7
T
8
T
9
T
10
T
11
j-1 = 3
j = 4
dp[4]
← dp[4 -
1
] = dp[
3
] =
T
stones
2
7
4
1
8
1
sum =
23
target =
11
dp
T
0
T
1
T
2
T
3
T
4
T
5
T
6
T
7
T
8
T
9
T
10
T
11
piles
11
vs
12
, last stone = 23 - 2×11 =
1
algo
master
.
io
Step:
Smashing all stones leaves |pileA - pileB|. Split the stones into two piles as evenly as possible.
0 / 93
Input
Example 1
Example 2
All same
Single stone
Custom
stones
=
[2, 7, 4, 1, 8, 1]
0 / 93
algo
master
.
io
Step:
Smashing all stones leaves |pileA - pileB|. Split the stones into two piles as evenly as possible.
