Summary Ranges - Leetcode 228 - Arrays & Strings (Python)

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[GregHogg]

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[davidbuckingham]

Don't often comment on here but feel I have to say a huge thank you for making all these videos. Your explanations are great and although i find many of the problems difficult, I usually understand after watching two or three times. You're probably saving me from early onset dementia - I've watched quite a few other presenters on you tube explaining this stuff but you are the best by quite a margin in my opinion.

[zynthrix]

hey, just wanna let you know your free dsa leetcode roadmap is helpful af. I'm a 2nd year cs student and my dsa professor last semester couldn't teach for the sake of his life, and I'm learning more from here than anywhere else. Big Thanks!

[sebastianibarraherrera]

awsome explination, definetly not seen this aproach in college as a second year
thanks so much Greg
!!

[ZaneFalcao]

My solution checks the difference between the next number and current number

class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
ans = []
n = len(nums)
i = 0

while i < n:
start = nums[i]
while i + 1 < n and nums[i + 1] - nums[i] == 1:
i += 1
if start != nums[i]:

else:
ans.append(str(start))
i += 1

return ans

[christianjt]

I really liked this solution, thanks a lot for sharing this valuable content.

[kidyoshi]

The trick is to start at a value, then loop again to see how many times a number is the increment of the start.

[teamtitus]

def summaryRanges(self, nums: list) -> list:
# empty list
if not nums:
return []

# initialisation
a = nums[0]
b = nums[0]
ans = []

for i in range(1, len(nums)):
# tracks the end of interval
if nums[i] == b + 1:
b = nums[i]
continue

# empty interval or not
if a == b:
ans.append(str(a))
else:
ans.append(str(a) + "->" + str(b))

# next interval
a = nums[i]
b = nums[i]

# last interval
if a == b:
ans.append(str(a))
else:
ans.append(str(a) + "->" + str(b))

return ans

# time -> O(n)
# space -> O(n)

[mohamedthaiebu]

Thank you for providing this solution.😇😇

[Rutik]

My appraoch -

class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
lst = []
l = 0
length = 0
for r in range(len(nums)):
if nums[r] != nums[l] + length:
if l == r-1:
lst.append(str(nums[l]))
else:
lst.append(str(nums[l]) + "->" + str(nums[r-1]))
length = 1
l = r
else:
length+=1

if l<len(nums)-1:
lst.append(str(nums[l]) + "->" + str(nums[r]))
elif l==len(nums)-1:
lst.append(str(nums[l]))

return lst

[Ivan-Shyriaiev]

omg so easy solution, i thought about using nested for loops

[spongsquad]

What would you rate my solution? I used two pointers method

def summaryRanges(self, nums: List[int]) -> List[str]:
results = []
nums.append(-999)
lp = nums[0]

for i in range(1, len(nums)):
if nums[i-1] != nums[i]-1:
if(lp == nums[i-1]):
string = str(lp)
else:
string = str(lp)+"->"+str(nums[i-1])
results.append(string)
lp = nums[i]

return results

The nums.append(-999) is a bit of a hack but it passed all the test cases so i guess its fine

[atTien-kopg]

GREG HOGG + NEETCODE = 200 LEETCODE PROBLEMS SOLVED