"select a.playername as player1, b.playername as player2,
abs(a.level-b.level) as level_disparity
from players a
cross join players b
on a.playername < b.playername
where abs(a.level-b.level) <=5
order by level_disparity
`"
Gowtami K. - "select a.playername as player1, b.playername as player2,
abs(a.level-b.level) as level_disparity
from players a
cross join players b
on a.playername < b.playername
where abs(a.level-b.level) <=5
order by level_disparity
`"See full answer
"def traprainwater(height: List[int]) -> int:
n = len(height)
totalwaterlevel = 0
for i in range(n):
j = i+1
while j = n:
break
rows = j - i -1
intrwaterlevel = min(height[j], height[i]) * rows
for k in range(i+1, j):
intrwaterlevel -= height[k]
totalwaterlevel += intrwaterlevel
i = j
return totalwaterlevel"
Manoj R. - "def traprainwater(height: List[int]) -> int:
n = len(height)
totalwaterlevel = 0
for i in range(n):
j = i+1
while j = n:
break
rows = j - i -1
intrwaterlevel = min(height[j], height[i]) * rows
for k in range(i+1, j):
intrwaterlevel -= height[k]
totalwaterlevel += intrwaterlevel
i = j
return totalwaterlevel"See full answer
"The question is incomplete --- the code only passes if you return the data frame sorted by BOTH department name AND rank. While in the problem description, it mentions to only rank by department name: "The results should be ordered by department name."
Not a big difference I know, but students shouldn't need to look into the solution to get the necessary knowledge to answer the question."
Chao peter Y. - "The question is incomplete --- the code only passes if you return the data frame sorted by BOTH department name AND rank. While in the problem description, it mentions to only rank by department name: "The results should be ordered by department name."
Not a big difference I know, but students shouldn't need to look into the solution to get the necessary knowledge to answer the question."See full answer
"Make current as root.
2 while current is not null,
if p and q are less than current,
go left.
If p and q are greater than current,
go right.
else return current.
return null"
Vaibhav D. - "Make current as root.
2 while current is not null,
if p and q are less than current,
go left.
If p and q are greater than current,
go right.
else return current.
return null"See full answer
"Leetcode 347: Heap + Hashtable
Follow up question: create heap with the length of K instead of N (more time complexity but less space )"
Chen J. - "Leetcode 347: Heap + Hashtable
Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer
"
import pandas as pd
from datetime import datetime
def findfastestlike(log: pd.DataFrame) -> pd.DataFrame:
log=log.sortvalues(['userid','timestamp'])
#get the prev event, time by user
log['prevevent'] = log.groupby('userid')['event'].shift(1)
log['prevtimestamp'] = log.groupby('userid')['timestamp'].shift(1)
True only on rows where the previous event was a login
and the current event is a like
log['loginlike'] = (log['prevevent'] == 'log"
Sean L. - "
import pandas as pd
from datetime import datetime
def findfastestlike(log: pd.DataFrame) -> pd.DataFrame:
log=log.sortvalues(['userid','timestamp'])
#get the prev event, time by user
log['prevevent'] = log.groupby('userid')['event'].shift(1)
log['prevtimestamp'] = log.groupby('userid')['timestamp'].shift(1)
True only on rows where the previous event was a login
and the current event is a like
log['loginlike'] = (log['prevevent'] == 'log"See full answer
"def countuniqueoutfits(totalpants: int, uniquepants: int,
totalshirts: int, uniqueshirts: int,
totalhats: int, uniquehats: int) -> int:
"""
Number of unique outfits can simply be defined by
(uniquepantschoose1uniqueshirtschoose1uniquehatschoose_1)
(uniquepantschoose1*uniqueshirtschoose1) # Not wearing a hat
nchoosek is n
"""
res = (uniquepants*uniqueshirtsuniquehats) + (uniquepantsunique_shirts)
return res
print(countuniqueoutfits(2, 1, 1, 1, 3, 2))"
Sai R. - "def countuniqueoutfits(totalpants: int, uniquepants: int,
totalshirts: int, uniqueshirts: int,
totalhats: int, uniquehats: int) -> int:
"""
Number of unique outfits can simply be defined by
(uniquepantschoose1uniqueshirtschoose1uniquehatschoose_1)
(uniquepantschoose1*uniqueshirtschoose1) # Not wearing a hat
nchoosek is n
"""
res = (uniquepants*uniqueshirtsuniquehats) + (uniquepantsunique_shirts)
return res
print(countuniqueoutfits(2, 1, 1, 1, 3, 2))"See full answer
"#include
#include
#include
using namespace std;
void printComs(int prev, int start, int end, int target)
{
if (start >= end) return;
while (start target)
{
end--;
}
else
{
st"
Iris F. - "#include
#include
#include
using namespace std;
void printComs(int prev, int start, int end, int target)
{
if (start >= end) return;
while (start target)
{
end--;
}
else
{
st"See full answer