This https://codeforces.net/contest/1956/submission/256474898 solution consumes at least (2 * n + 1) * sizeof(int)
bytes, on the test #4 the memory usage must be at least 1_600_004
bytes instead of 56 KB
.
# | User | Rating |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3741 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3489 |
7 | Radewoosh | 3483 |
8 | Kevin114514 | 3442 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | atcoder_official | 162 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | nor | 150 |
This https://codeforces.net/contest/1956/submission/256474898 solution consumes at least (2 * n + 1) * sizeof(int)
bytes, on the test #4 the memory usage must be at least 1_600_004
bytes instead of 56 KB
.
https://codeforces.net/contest/1846/submission/215981788
The code (Ruby) below is a solution for https://leetcode.com/problems/uncrossed-lines/
Note: In Ruby Array#shift is O(1)
The trick is on the lines 3-4:
def max_uncrossed_lines a, b
# These two lines
c = a.to_set & b.to_set
a, b = *[a, b].map! { | v | v.filter { c === _1 } }
s = 0
while !a.empty? && a.first == b.first
a.shift
b.shift
s += 1
end
while !a.empty? && a.last == b.last
a.pop
b.pop
s += 1
end
r0, r1 = *Array.new(2) { [0] * (b.size + 1) }
for x in a
for y, j in b.each_with_index
r1[j + 1] = (x == y) ?
1 + r0[j] :
[r0[j + 1], r1[j]].max
end
r0, r1 = r1, r0
end
s + r0.last
end
Name |
---|