post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1942319/Python-One-Liner! | class Solution:
def reversePrefix(self, word, ch):
i = word.find(ch)+1
return word[:i][::-1] + word[i:] if i > 0 else word | reverse-prefix-of-word | Python - One-Liner! | domthedeveloper | 0 | 34 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,900 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1942319/Python-One-Liner! | class Solution:
def reversePrefix(self, word, ch):
return (lambda i : word[:i][::-1] + word[i:] if i > 0 else word)(word.find(ch)+1) | reverse-prefix-of-word | Python - One-Liner! | domthedeveloper | 0 | 34 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,901 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1906789/Python-Solution-or-List-Slicing-and-Reversing-or-Over-90-Faster | class Solution:
def reversePrefix(self, s: str, ch: str) -> str:
pos = s.find(ch)
return s[:pos+1][::-1] + s[pos+1:] | reverse-prefix-of-word | Python Solution | List Slicing and Reversing | Over 90% Faster | Gautam_ProMax | 0 | 29 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,902 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1881304/Python-dollarolution | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
if ch not in word:
return word
x = word.index(ch)+1
return (word[:x][::-1] + word[x:]) | reverse-prefix-of-word | Python $olution | AakRay | 0 | 25 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,903 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1860631/Python-solution-35-ms | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
try:
ind = word.index(ch)+1
except ValueError:
return word
sub = word[0:ind]
return sub[::-1] + word[ind:] | reverse-prefix-of-word | Python solution 35 ms | alishak1999 | 0 | 35 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,904 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1808637/Python-Easy-and-Fast-95 | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
if ch not in word:
return word
i=word.index(ch)
a=word[:i+1]
b=word[i+1:]
a=a[::-1]
return a+b | reverse-prefix-of-word | Python Easy and Fast 95% | adityabaner | 0 | 44 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,905 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1758348/Python-Easy-Sol-oror-Beats-90 | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
try:
ri = word.index(ch)
except ValueError:
return word
if ri == len(word)-1:
return word[::-1]
else:
rr = word[:ri+1]
rem = word[ri+1:]
... | reverse-prefix-of-word | Python Easy Sol || Beats 90% | dos_77 | 0 | 47 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,906 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1700190/Python3-accepted-solution | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
if(ch not in word): return word
return word[:word.index(ch)+1][::-1] + word[word.index(ch)+1:] | reverse-prefix-of-word | Python3 accepted solution | sreeleetcode19 | 0 | 37 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,907 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1683421/Python-one-line-solution-logic-explained | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
return word[0:word.find(ch)+1][::-1] + word[word.find(ch) + 1:] | reverse-prefix-of-word | Python one line solution, logic explained | snagsbybalin | 0 | 32 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,908 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1555618/Easy-Python3-solution | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
index: int = word.find(ch)
return word if index == -1 else (word[:index + 1][::-1] + word[index + 1:]) | reverse-prefix-of-word | Easy Python3 solution | sirenescx | 0 | 54 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,909 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1534163/Python-Runtime-97%2B-Faster(24ms)-Memory-98%2B-Solution | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
try:
index = word.index(ch)
except:
return word
word = list(word)
part = word[:index+1]
word[:index+1] = part[::-1]
return "".join(word) | reverse-prefix-of-word | Python Runtime 97%+ Faster(24ms) Memory 98%+ Solution | aaffriya | 0 | 91 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,910 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1511325/Python-1-line | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
return word if ch not in word else ch + word.split(ch)[0][::-1] + word[word.find(ch)+1:] | reverse-prefix-of-word | Python 1-line | SmittyWerbenjagermanjensen | 0 | 47 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,911 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1461241/Python3-2-line | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
k = word.find(ch)
return word[:k+1][::-1] + word[k+1:] | reverse-prefix-of-word | [Python3] 2-line | ye15 | 0 | 33 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,912 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1461135/Python3-Fastest-Solution-Faster-Than-100-Really-Easy-To-Understand | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
idx = -1
for i in range(len(word)):
if word[i] == ch:
idx = i
break
return word[:idx + 1][::-1] + word[idx + 1:] if idx != -1 else word | reverse-prefix-of-word | Python3 Fastest Solution, Faster Than 100%, Really Easy To Understand | Hejita | 0 | 44 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,913 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1459918/Python-the-standard-algorithmic-O(N)-solution | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
for i, c in enumerate(word):
if c == ch:
w = list(word)
for j in range((i+1)//2):
w[j], w[i-j] = w[i-j], w[j]
return "".join(w)
return ... | reverse-prefix-of-word | Python, the standard algorithmic O(N) solution | blue_sky5 | 0 | 36 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,914 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1459211/Intuitive-approach-by-index | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
try:
i = word.index(ch)
i += len(ch) - 1
return ''.join(word[:i+1][::-1]) + word[i+1:]
except:
return word | reverse-prefix-of-word | Intuitive approach by index | puremonkey2001 | 0 | 24 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,915 |
https://leetcode.com/problems/reverse-prefix-of-word/discuss/1487225/Python-3oror-Faster-than-91-oror-28ms-oror-simple-slicing | class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
idx = word.find(ch)
if idx==-1:
return word
else:
res = list(word)
first = res[idx::-1]
last = res[idx+1:]
ans = first+last
return ''.join(ans) | reverse-prefix-of-word | Python 3|| Faster than 91% || 28ms || simple slicing | ana_2kacer | -1 | 39 | reverse prefix of word | 2,000 | 0.778 | Easy | 27,916 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/2818774/O(n)-solution-of-Combined-Dictionary-and-Pre-sum-in-Python | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
preSum = []
for rec in rectangles:
preSum.append(rec[1]/rec[0])
dic1 = {}
for i in range(len(preSum)-1, -1, -1):
if preSum[i] not in dic1.keys():
... | number-of-pairs-of-interchangeable-rectangles | O(n) solution of Combined Dictionary and Pre sum in Python | DNST | 0 | 3 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,917 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/2766675/Number-of-pairs-of-interchangeables-Rectangles | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
count=0
d=dict()
for i in range(len(rectangles)):
ratio=rectangles[i][0]/rectangles[i][1]
if ratio not in d:
d[ratio]=1
else:
count+=d[... | number-of-pairs-of-interchangeable-rectangles | Number of pairs of interchangeables Rectangles | shivansh2001sri | 0 | 6 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,918 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/2705014/Python-Easy-O(N)-Solution-with-Counter | class Solution:
def interchangeableRectangles(self, rectangles: [[int]]) -> int:
counter = Counter([i / j for i, j in rectangles])
ans = 0
for i, j in rectangles:
ans += (counter.get(i / j) - 1)
counter.update({i / j: - 1})
return ans | number-of-pairs-of-interchangeable-rectangles | Python Easy O(N) Solution with Counter | alan_ahmet | 0 | 13 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,919 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/2545410/4146-TEST-CASES-PASSED. | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
for i in range(0,len(rectangles)):
r=rectangles[i][1]/rectangles[i][0]
rectangles[i]=r
count=0
for j in range(0,len(rectangles)-1):
Slice=rectangles[j+1:]
c=Slice.count(rectangles[j])
count=count+c
return c... | number-of-pairs-of-interchangeable-rectangles | 41/46 TEST CASES PASSED. | keertika27 | 0 | 11 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,920 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/2452570/Python-3-clean-Solution | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
hashmap , pairs = {} , 0
for i in range(len(rectangles)):
key = rectangles[i][0] / rectangles[i][1]
if key in hashmap:
... | number-of-pairs-of-interchangeable-rectangles | Python 3 clean Solution | abhisheksanwal745 | 0 | 15 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,921 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1984358/Python-one-line | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
c = Counter(w / h for w, h in rectangles)
return sum(v*(v-1)//2 for v in c.values()) | number-of-pairs-of-interchangeable-rectangles | Python one line | SmittyWerbenjagermanjensen | 0 | 62 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,922 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1984358/Python-one-line | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
return sum(v*(v-1)//2 for v in Counter(w / h for w, h in rectangles).values()) | number-of-pairs-of-interchangeable-rectangles | Python one line | SmittyWerbenjagermanjensen | 0 | 62 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,923 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1461244/Python3-2-line | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
freq = Counter(w/h for w, h in rectangles)
return sum(v*(v-1)//2 for v in freq.values()) | number-of-pairs-of-interchangeable-rectangles | [Python3] 2-line | ye15 | 0 | 55 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,924 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1461244/Python3-2-line | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
ans = 0
freq = defaultdict(int)
for w, h in rectangles:
ans += freq[w/h]
freq[w/h] += 1
return ans | number-of-pairs-of-interchangeable-rectangles | [Python3] 2-line | ye15 | 0 | 55 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,925 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1459988/Easy-Python-Solution-using-dictionary | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
diction = defaultdict(int)
n = len(rectangles)
count = 0
for i in range(0, n):
diction[rectangles[i][0]/rectangles[i][1]] += 1
x = list(diction.values())
m = len(x)
... | number-of-pairs-of-interchangeable-rectangles | Easy Python Solution using dictionary | Zikai_Lian | 0 | 42 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,926 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1458677/JavaPython3-Concise-Solution | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
m = len(rectangles)
n = len(rectangles[0])
res = 0
d = {}
for i in rectangles:
ratio = i[0]/i[1]
if ratio in d:
res += d[ratio]
... | number-of-pairs-of-interchangeable-rectangles | [Java/Python3] Concise Solution | abhijeetmallick29 | 0 | 24 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,927 |
https://leetcode.com/problems/number-of-pairs-of-interchangeable-rectangles/discuss/1458322/Python3-solution-Faster-than-100-less-than-100 | class Solution:
def interchangeableRectangles(self, rectangles: List[List[int]]) -> int:
counts = {}
for i in range(len(rectangles)):
rectangles[i] = rectangles[i][0]/rectangles[i][1]
if rectangles[i] in counts:
counts[rectangles[i]] += 1
else:
... | number-of-pairs-of-interchangeable-rectangles | Python3 solution - Faster than 100%, less than 100% | elainefaith0314 | 0 | 40 | number of pairs of interchangeable rectangles | 2,001 | 0.451 | Medium | 27,928 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/1458484/Python-Bruteforce | class Solution:
def maxProduct(self, s: str) -> int:
subs = []
n = len(s)
def dfs(curr, ind, inds):
if ind == n:
if curr == curr[::-1]:
subs.append((curr, inds))
return
dfs(curr+s[ind], ind+1, inds|{ind})
... | maximum-product-of-the-length-of-two-palindromic-subsequences | Python - Bruteforce | ajith6198 | 3 | 232 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,929 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/1461446/Python3-bitmask-dp | class Solution:
def maxProduct(self, s: str) -> int:
@cache
def lps(mask):
"""Return length of longest palindromic sequence."""
if not mask: return 0
if not mask & (mask-1): return 1
lo = int(log2(mask & ~(mask-1))) # least significan... | maximum-product-of-the-length-of-two-palindromic-subsequences | [Python3] bitmask dp | ye15 | 1 | 62 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,930 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/1458482/PYTHON-Simple-solution-backtracking | class Solution:
def maxProduct(self, s: str) -> int:
self.answer = 0
def dfs(i, word, word2):
if i >= len(s):
if word == word[::-1] and word2 == word2[::-1]:
self.answer = max(len(word) * len(word2), self.answer)
return
... | maximum-product-of-the-length-of-two-palindromic-subsequences | [PYTHON] - Simple solution, backtracking✅ | just_4ina | 1 | 282 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,931 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/2844906/python-3-Bitmask | class Solution:
def maxProduct(self, s: str) -> int:
def create_string(v):
res=[]
for i in range(n):
if 1<<i&v:
res.append(s[i])
if res==res[::-1]:
pal[v]=len(res)
pal=dict()
n=len(s)
for i in range(1,pow(2,n)):
create_string(i)
res=0
for x in pal:
for y in pal:
if not x... | maximum-product-of-the-length-of-two-palindromic-subsequences | [python 3] Bitmask | gabhay | 0 | 1 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,932 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/2313099/Python-3-Bitmask | class Solution:
def maxProduct(self, s: str) -> int:
def create_string(v):
res=[]
for i in range(n):
if 1<<i&v:
res.append(s[i])
if res==res[::-1]:
pal[v]=len(res)
pal=dict()
n=len(s)
for i in range(1,pow(2,n)):
create_string(i)
res=0
for x in pal:
for y in pal:
if not x... | maximum-product-of-the-length-of-two-palindromic-subsequences | [Python 3] Bitmask | gabhay | 0 | 28 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,933 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/1769323/Python3-Bitmask-%2B-DP | class Solution:
def maxProduct(self, s: str) -> int:
res, N = 0, len(s)
for i in range((1<<N)-1):
res = max(res, self.mpl(s,i) * self.mpl(s,(1<<N)-1-i))
return res
def mpl(self,s,i): # max palindrome subsequence length
t = ''
for j in range(len(s)):
... | maximum-product-of-the-length-of-two-palindromic-subsequences | [Python3] Bitmask + DP | Rainyforest | 0 | 104 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,934 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-subsequences/discuss/1467087/Python-dfs%2Bmemo-came-up-in-contest | class Solution:
def maxProduct(self, s: str) -> int:
if len(s) < 2:
return 0
if len(s) == 2:
return 1
self.ans = -1
visited = set()
def dfs(arr1, arr2, idx):
if tuple(sorted([arr1, arr2]) + [idx]) in visited:
return
... | maximum-product-of-the-length-of-two-palindromic-subsequences | Python dfs+memo came up in contest | ScoutBoi | 0 | 192 | maximum product of the length of two palindromic subsequences | 2,002 | 0.533 | Medium | 27,935 |
https://leetcode.com/problems/smallest-missing-genetic-value-in-each-subtree/discuss/1461767/Python3-dfs | class Solution:
def smallestMissingValueSubtree(self, parents: List[int], nums: List[int]) -> List[int]:
ans = [1] * len(parents)
if 1 in nums:
tree = {}
for i, x in enumerate(parents):
tree.setdefault(x, []).append(i)
k = nums.i... | smallest-missing-genetic-value-in-each-subtree | [Python3] dfs | ye15 | 0 | 89 | smallest missing genetic value in each subtree | 2,003 | 0.443 | Hard | 27,936 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1471015/Python-Clean-and-concise.-Dictionary-T.C-O(N) | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
seen = defaultdict(int)
counter = 0
for num in nums:
tmp, tmp2 = num - k, num + k
if tmp in seen:
counter += seen[tmp]
if tmp2 in seen:
counter += s... | count-number-of-pairs-with-absolute-difference-k | [Python] Clean & concise. Dictionary T.C O(N) | asbefu | 36 | 3,700 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,937 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1471015/Python-Clean-and-concise.-Dictionary-T.C-O(N) | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
seen = defaultdict(int)
counter = 0
for num in nums:
counter += seen[num-k] + seen[num+k]
seen[num] += 1
return counter | count-number-of-pairs-with-absolute-difference-k | [Python] Clean & concise. Dictionary T.C O(N) | asbefu | 36 | 3,700 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,938 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2101510/2-Approach-Python-Solution-O(n)-and-O(n2) | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
hash = {}
for i in nums:
if i in hash:
hash[i] +=1
else:
hash[i] = 1
for i in hash:
if i+k in h... | count-number-of-pairs-with-absolute-difference-k | 2 Approach Python Solution O(n) & O(n2) | Brillianttyagi | 7 | 298 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,939 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2101510/2-Approach-Python-Solution-O(n)-and-O(n2) | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
i = 0
j = len(nums)-1
nums.sort()
while i<len(nums):
if abs(nums[j]-nums[i])==k:
count+=1
j-=1
... | count-number-of-pairs-with-absolute-difference-k | 2 Approach Python Solution O(n) & O(n2) | Brillianttyagi | 7 | 298 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,940 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1470878/Python-3-Simple-solution-or-Explained! | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
for i, j in permutations(nums, 2):
if (i - j) == k: count += 1
return count | count-number-of-pairs-with-absolute-difference-k | [Python 3] Simple solution | Explained! | JK0604 | 6 | 880 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,941 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1958493/Python-easy-to-read-and-understand-or-hashmap | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
d = collections.defaultdict(int)
ans = 0
for num in nums:
d[num] = d.get(num, 0) + 1
for key in d:
if (key + k) in d:
ans += d[key] * d[key + k]
return ans | count-number-of-pairs-with-absolute-difference-k | Python easy to read and understand | hashmap | sanial2001 | 3 | 272 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,942 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1633099/Easy-O(n)-Python-Solution. | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
pairs = {}
for num in nums:
if num+k not in pairs:
pairs[num+k] = [0,0]
if num not in pairs:
pairs[num] = [0,0]
pairs[num][0] +=1
... | count-number-of-pairs-with-absolute-difference-k | Easy O(n) Python Solution. | manassehkola | 3 | 316 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,943 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1818715/Ez-to-Understand-oror-hashmap-and-brute-force | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
d={}
co=0
for i in nums:
if i in d:
d[i]+=1
else:
d[i]=1
for i in nums:
if k+i in d:
co=co+d[k+i]
return co | count-number-of-pairs-with-absolute-difference-k | Ez to Understand || hashmap and brute force | ashu_py22 | 2 | 101 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,944 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1818715/Ez-to-Understand-oror-hashmap-and-brute-force | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
co=0
for i in range(len(nums)):
for j in range(0, len(nums)):
if nums[i]+k==nums[j]:
co+=1
return co | count-number-of-pairs-with-absolute-difference-k | Ez to Understand || hashmap and brute force | ashu_py22 | 2 | 101 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,945 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1818715/Ez-to-Understand-oror-hashmap-and-brute-force | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
co=0
for i in range(len(nums)):
if nums[i]+k in nums:
co+=nums.count(nums[i]+k)
return co | count-number-of-pairs-with-absolute-difference-k | Ez to Understand || hashmap and brute force | ashu_py22 | 2 | 101 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,946 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1757985/PYTHON-ONE-LINER | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
return sum(nums.count(num+k) for num in nums if num+k in nums) | count-number-of-pairs-with-absolute-difference-k | PYTHON ONE-LINER | vijayvardhan6 | 2 | 170 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,947 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1574982/Using-counter-99.89-speed | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
cnt = Counter(nums)
return sum(cnt[key] * cnt[key + k]
for key in sorted(cnt.keys()) if key + k in cnt) | count-number-of-pairs-with-absolute-difference-k | Using counter, 99.89% speed | EvgenySH | 2 | 293 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,948 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2538343/very-easy-python-solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count=0
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if abs(nums[i]-nums[j])==k:
count+=1
return count | count-number-of-pairs-with-absolute-difference-k | very easy python solution | Sneh713 | 1 | 20 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,949 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2285518/Python-or-O(norn)-or-hashmapor-Similar-to-Two-sum | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
hashmap = {}
answer = 0
for i in range(len(nums)):
if nums[i]+k in hashmap.keys():
answer += hashmap[nums[i]+k]
if nums[i]-k in hashmap.keys():
answer += hashm... | count-number-of-pairs-with-absolute-difference-k | Python | O(n|n) | hashmap| Similar to Two-sum | user7457RV | 1 | 57 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,950 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1721809/Faster-than-99.6.-Defaultdict-O(n) | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
d = defaultdict(int)
for el in nums:
if el - k > 0:
d[el-k] += 1
s = 0
for el in nums:
s += d[el]
return s | count-number-of-pairs-with-absolute-difference-k | Faster than 99.6%. Defaultdict O(n) | mygurbanov | 1 | 235 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,951 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1470954/Python3-solution-hash-map-102ms-O(N2) | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
nums = sorted(nums)
ans = 0
s = set(nums)
for n in s:
ans += nums.count(n)*nums.count(n+k)
return ans | count-number-of-pairs-with-absolute-difference-k | Python3 solution - hash map, 102ms, O(N^2) | elainefaith0314 | 1 | 181 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,952 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2848044/python | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
cnt, ans = Counter(nums), 0
for key in cnt:
if key + k in cnt:
ans += cnt[key] * cnt[key + k]
return ans | count-number-of-pairs-with-absolute-difference-k | python | xy01 | 0 | 1 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,953 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2816372/One-Liner-Python-Solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
return len([True for i in nums for j in nums if i < j and abs(i - j) == k]) | count-number-of-pairs-with-absolute-difference-k | One-Liner Python Solution | PranavBhatt | 0 | 2 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,954 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2815260/Easy-way | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
res=0
for i in nums:
for j in nums:
if i-j==k:
res=res+1
return res | count-number-of-pairs-with-absolute-difference-k | Easy way | nishithakonuganti | 0 | 1 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,955 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2788652/Simple-Python-Solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
for i in nums:
for j in nums:
if i-j == k:
count = count + 1
return count | count-number-of-pairs-with-absolute-difference-k | Simple Python Solution | dnvavinash | 0 | 2 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,956 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2723599/Python3-Solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
ans = 0
for i, num in enumerate(nums):
ans += sum([True for n in nums[i:] if abs(num-n) == k])
return ans | count-number-of-pairs-with-absolute-difference-k | Python3 Solution | sipi09 | 0 | 8 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,957 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2718428/Python3-O(n)-easy-solution-without-sorting | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
arr = [0] * 101
for num in nums:
arr[num] += 1
ans = 0
for i in range(101):
if i + k > 100:
break
ans += arr[i] * arr[i + k]
return ans | count-number-of-pairs-with-absolute-difference-k | [Python3] O(n) easy solution without sorting | huangweijing | 0 | 10 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,958 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2640142/Python3-Counter | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
c = Counter(nums)
tot = 0
for low in c:
tot += c[low] * c[low + k]
return tot | count-number-of-pairs-with-absolute-difference-k | Python3 Counter | godshiva | 0 | 4 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,959 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2559563/simple-python-solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
pairs = 0
for x in range(len(nums)):
for y in range(x+1,len(nums)):
if abs(nums[x]-nums[y]) == k:
pairs += 1
return pairs | count-number-of-pairs-with-absolute-difference-k | simple python solution | maschwartz5006 | 0 | 40 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,960 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2545142/EASY-PYTHON3-SOLUTION | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
for i in range(len(nums)):
for j in range(i,len(nums)):
if abs(nums[i] - nums[j])==k:
count+=1
return count | count-number-of-pairs-with-absolute-difference-k | 🔥 EASY PYTHON3 SOLUTION 🔥 | rajukommula | 0 | 34 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,961 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2484944/Easy-hashmap-solution-beats-93 | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
dic = {}
count = 0
for i in range(len(nums)):
if nums[i] not in dic:
dic[nums[i]] = 1
else:
dic[nums[i]] += 1
for j in nums:
if j + k in dic... | count-number-of-pairs-with-absolute-difference-k | Easy hashmap solution beats 93% | aruj900 | 0 | 84 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,962 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2404852/2006.-Count-Number-of-Pairs-With-Absolute-Difference-K%3A-One-liner | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
return sum([1 for number_a in nums for number_b in nums if number_a-number_b==k]) | count-number-of-pairs-with-absolute-difference-k | 2006. Count Number of Pairs With Absolute Difference K: One liner | rogerfvieira | 0 | 17 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,963 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2384312/FASTER-Than-50-EASY-Python-Solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
c=0
for i in range(0,len(nums)-1):
for j in range(i+1,len(nums)):
if abs(nums[i]-nums[j])==k: c=c+1
return c | count-number-of-pairs-with-absolute-difference-k | FASTER Than 50% EASY Python Solution | keertika27 | 0 | 33 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,964 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2372950/Python-faster-than-98 | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
diffList=[0]*101
maxNum = 0
for i in nums:
diffList[i]+=1
if i > maxNum:
maxNum=i
result = 0
for i in range(maxNum-k+1):
result+=diffList[i]*diffLis... | count-number-of-pairs-with-absolute-difference-k | Python faster than 98% | Shrey-Modi | 0 | 60 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,965 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2326297/Python3-Solution-with-using-hashmap | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
d = collections.defaultdict(int)
res = 0
for num in nums:
res += d[num - k]
res += d[num + k]
d[num] += 1
return res | count-number-of-pairs-with-absolute-difference-k | [Python3] Solution with using hashmap | maosipov11 | 0 | 46 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,966 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2239513/Python3-brute-force | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
l=[]
for i in range(0, len(nums)):
for j in range(i+1, len(nums)):
if abs(nums[i]-nums[j]) == k:
l.append([nums[i],nums[j]])
return len(l) | count-number-of-pairs-with-absolute-difference-k | Python3 brute force | psnakhwa | 0 | 13 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,967 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2195256/222-ms-faster-than-62.05-of-Python3 | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
for i in range(len(nums)):
for j in range(i,len(nums)):
if abs(nums[i]-nums[j])==k:
count+=1
return count | count-number-of-pairs-with-absolute-difference-k | 222 ms, faster than 62.05% of Python3 | EbrahimMG | 0 | 35 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,968 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2182498/Python3-Easy-Peasy | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
n= len(nums)
count=0
for i in range(0,n):
for j in range(i+1,n):
if(abs(nums[i]-nums[j])==k):
count+=1
return count | count-number-of-pairs-with-absolute-difference-k | Python3 Easy Peasy | hvt1998 | 0 | 54 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,969 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2178830/Python-simple-solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
ans = 0
for i in range(len(nums)):
for j in range(i,len(nums)):
if i == j: continue
if abs(nums[i]-nums[j]) == k:
ans += 1
return ans | count-number-of-pairs-with-absolute-difference-k | Python simple solution | StikS32 | 0 | 51 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,970 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/2043388/Python3-easy-to-understand | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
count = 0
for i,_ in enumerate(nums):
for j,_ in enumerate(nums):
if nums[i] - nums[j] == k:
count += 1
return count | count-number-of-pairs-with-absolute-difference-k | [Python3] easy to understand | Shiyinq | 0 | 68 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,971 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1881238/Python3-filter-and-map-solution | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
res = 0
for i in range(len(nums)):
ft = filter(lambda j: j==k, map(lambda n: abs(n-nums[i]), nums[i:]))
res += len([x for x in ft])
else:
return res | count-number-of-pairs-with-absolute-difference-k | Python3 filter and map solution | khRay13 | 0 | 83 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,972 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1851426/2-Lines-Python-Solution-oror-50-Faster-oror-Memory-less-than-85 | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
ans=0
for i in range(len(nums)-1):
for j in range(i+1,len(nums)):
if abs(nums[j]-nums[i])==k: ans+=1
return ans | count-number-of-pairs-with-absolute-difference-k | 2-Lines Python Solution || 50% Faster || Memory less than 85% | Taha-C | 0 | 93 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,973 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1673682/Python3-oror-List-Comprehension | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
return len([(item1, nums[j]) for i, item1 in enumerate(nums) for j in range(i+1, len(nums)) if abs(item1 - nums[j]) == k]) | count-number-of-pairs-with-absolute-difference-k | Python3 || List Comprehension | user9295j | 0 | 120 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,974 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1671829/Python3-O(n)-IF-you-can't-look-forward-just-walk-and-look-behind | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
m, cnt = {nums[0] : 1}, 0
for i in range(1, len(nums)):
# Solution for the equation |x - y| = k
# either x - y = k or x - y = -k
t1, t2 = nums[i] - k, nums[i] + k
if... | count-number-of-pairs-with-absolute-difference-k | Python3 O(n), IF you can't look forward just walk and look behind | Hejita | 0 | 120 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,975 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1555575/Python3-faster-than-97.93-solutions | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
nums_dict: dict = {}
for num in nums:
if num not in nums_dict:
nums_dict[num] = 1
else:
nums_dict[num] += 1
count: int = 0
for key in ... | count-number-of-pairs-with-absolute-difference-k | Python3, faster than 97.93% solutions | sirenescx | 0 | 190 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,976 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1521114/Python-from-brute-force-to-linear-time | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
return sum(abs(nums[j] - n) == k for i, n in enumerate(nums) for j in range(i, len(nums))) | count-number-of-pairs-with-absolute-difference-k | Python, from brute force to linear time | mousun224 | 0 | 116 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,977 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1521114/Python-from-brute-force-to-linear-time | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
d = collections.defaultdict(list)
cnt = 0
for i, n in enumerate(nums):
cnt += (len(d[n + k]))
if n > k:
cnt += len(d[n - k])
d[n] += i,
return cnt | count-number-of-pairs-with-absolute-difference-k | Python, from brute force to linear time | mousun224 | 0 | 116 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,978 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1502559/The-most-simple-python-solution%3A-O(n)-time-O(n)-space-with-hash | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
n = len(nums)
h = defaultdict(int)
h[nums[0]] += 1
res = 0
for i in range(1, n):
res += h[nums[i]+k] + h[nums[i]-k]
h[nums[i]] += 1
return res | count-number-of-pairs-with-absolute-difference-k | The most simple python solution: O(n) time, O(n) space with hash | byuns9334 | 0 | 230 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,979 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1471583/Python-O(N)O(N)-solution-with-a-defaultdict-for-indices | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
n_idxs = defaultdict(list)
for j, n in enumerate(nums):
n_idxs[n].append(j)
return sum(sum(j > i for j in n_idxs[n - k]) + sum(j > i for j in n_idxs[n + k])
for i, n in enum... | count-number-of-pairs-with-absolute-difference-k | Python, O(N)/O(N) solution with a defaultdict for indices | blue_sky5 | 0 | 59 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,980 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1471581/Python-brute-force-one-liner | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
return sum(abs(nums[i] - nums[j]) == k for i in range(len(nums)) for j in range(i+1, len(nums))) | count-number-of-pairs-with-absolute-difference-k | Python, brute force one liner | blue_sky5 | 0 | 66 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,981 |
https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/discuss/1471579/Python3-freq-table | class Solution:
def countKDifference(self, nums: List[int], k: int) -> int:
ans = 0
freq = defaultdict(int)
for x in nums:
ans += freq[x - k] + freq[x + k]
freq[x] += 1
return ans | count-number-of-pairs-with-absolute-difference-k | [Python3] freq table | ye15 | 0 | 70 | count number of pairs with absolute difference k | 2,006 | 0.823 | Easy | 27,982 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/1470895/Python-Sorting.-Easy-to-understand-and-clean-T.C%3A-O(n-log-n)-S.C%3A-O(N) | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
"""
The idea is to:
1st sort the numbers
2nd Create a counter to save the frequency of each number
3nd iterate the array and for each number check if the double exists.
... | find-original-array-from-doubled-array | [Python] Sorting. Easy to understand and clean T.C: O(n log n) S.C: O(N) | asbefu | 13 | 981 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,983 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578083/python-easy-peasy-solution-using-queue-and-sorting | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
changed.sort()
stk,res=deque([]),[]
for i in changed:
if stk and stk[0]*2==i:
b=stk.popleft()
res.append(b)
else:
stk.append(i)
return... | find-original-array-from-doubled-array | python easy-peasy solution using queue and sorting | benon | 8 | 589 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,984 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577455/Python-Solution | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 == 1:
return []
data = Counter(changed)
result = []
for k in sorted(data):
if data[k] < 0:
return []
elif k == 0:
x, y... | find-original-array-from-doubled-array | Python Solution | hgalytoby | 3 | 241 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,985 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577455/Python-Solution | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 == 1:
return []
data = Counter(changed)
result = []
for k in sorted(data):
if data[k] < 0:
return []
value = k * 2
while d... | find-original-array-from-doubled-array | Python Solution | hgalytoby | 3 | 241 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,986 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577373/Python-or-Math-%2B-Sort-%2B-Counter | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 != 0:
return []
changed.sort(reverse = True)
unpaired = Counter(); original = []
for i, num in enumerate(changed):
if num != 0:
if unpaired[2 * n... | find-original-array-from-doubled-array | Python | Math + Sort + Counter | sr_vrd | 3 | 310 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,987 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2581541/Python3-or-Easy-solution-Explained-or-HashMap-or-97-Faster | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
ll = len(changed)
if ll % 2 != 0: return []
result, hashmap = [], {}
# Push numbers in hashmap {number: count}
for n in changed: hashmap[n] = 1 + hashmap.get(n, 0)
for num in sorted(hashmap.... | find-original-array-from-doubled-array | Python3 | Easy solution Explained | HashMap | 97% Faster | abdoohossamm | 2 | 53 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,988 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2580151/Python-or-Two-solutions-using-Stack-and-Dictionary-with-explanation-or-Faster-than-95 | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 == 1: return []
changed.sort()
res, stack = deque(), deque()
for num in changed:
if stack and num == stack[0] * 2:
res.append(stack.popleft(... | find-original-array-from-doubled-array | Python | Two solutions using Stack and Dictionary with explanation | Faster than 95% | ahmadheshamzaki | 2 | 47 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,989 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2580151/Python-or-Two-solutions-using-Stack-and-Dictionary-with-explanation-or-Faster-than-95 | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 == 1: return []
changed.sort()
count = Counter(changed)
res = deque()
for num in changed:
if count[num] == 0:
continue
... | find-original-array-from-doubled-array | Python | Two solutions using Stack and Dictionary with explanation | Faster than 95% | ahmadheshamzaki | 2 | 47 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,990 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2595725/Very-Simple-Python-(Self-explanatory)%3A | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed)%2!=0:return []
changed.sort()
c=Counter(changed)
ans=[]
if c[0]%2==0:
ans+=[0]*(c[0]//2)
for i in c:
if i==0 or c[i]==0:
... | find-original-array-from-doubled-array | Very Simple Python (Self explanatory): | goxy_coder | 1 | 73 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,991 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2579404/Python-Simple-Python-Solution-Using-Dictionary-or-Sorting | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2 != 0:
return []
changed = sorted(changed)
result = []
d = {}
for i in changed:
if i not in d:
d[i] = 1
else:
d[i] = d[i] + 1
for num in changed:
double = num * 2
if num in d and d... | find-original-array-from-doubled-array | [ Python ] ✅✅ Simple Python Solution Using Dictionary | Sorting 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 1 | 33 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,992 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578848/Simple-%22python%22-Solution | class Solution:
def findOriginalArray(self, changed):
c = Counter(changed)
# int = [0,0,0,0]
if c[0]%2:
return []
for x in sorted(c): # c = [1:1,2:1,3:1,4:1,6:1,8:1]
if c[x] > c[2*x]: # [6,3,4,1] = c [1:1,3:1,4:1,6:1]
return []
c[2*... | find-original-array-from-doubled-array | Simple "python" Solution | anandchauhan8791 | 1 | 111 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,993 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2578000/simple-python-using-map | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if(len(changed) % 2):
return []
changed = sorted(changed)
h = {}
for i in changed:
if(i not in h):
h[i] = 0
h[i] += 1
ans = []
for i in changed:
if(i == 0):
if(h[i]%2):
return []
else:
count... | find-original-array-from-doubled-array | simple python using map | jagdishpawar8105 | 1 | 17 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,994 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2577792/python3-or-dictionary-or-explained-or-easy-to-understand | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n%2==1: return [] # len of changed array = 2*x (x = len of original array)
d={}
for e in changed: d[e] = d.get(e, 0)+1 # using dictionary for optimization
arr = []... | find-original-array-from-doubled-array | python3 | dictionary | explained | easy to understand | H-R-S | 1 | 32 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,995 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2101593/EXPLAINED-oror-O(1)-SPACE-oror-O(N-LOGN)-TIME-oror-PYTHON | class Solution:
def findOriginalArray(self, a: List[int]) -> List[int]:
if len(a)%2:
return []
a.sort()
w,s,d=0,0,1
while d<len(a):
if a[d]%2==0:
while s<d-1 and a[s]<(a[d]//2):
... | find-original-array-from-doubled-array | ✔️EXPLAINED || ✔️O(1) SPACE || O(N LOGN) TIME || 🐍PYTHON🐍 | karan_8082 | 1 | 153 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,996 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/1501076/Using-Counter-to-sort-the-keys-95-speed | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
if len(changed) % 2:
return []
ans = []
cnt = Counter(changed)
if 0 in cnt:
if cnt[0] % 2:
return []
else:
ans.extend([0] * (cnt[0] // 2))... | find-original-array-from-doubled-array | Using Counter to sort the keys, 95% speed | EvgenySH | 1 | 238 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,997 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/1471585/Python3-freq-table | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
freq = Counter(changed)
ans = []
for x in sorted(freq):
if freq[x]:
if x and freq[x] <= freq[2*x]:
ans.extend([x] * freq[x])
freq[2*x] -= freq[... | find-original-array-from-doubled-array | [Python3] freq table | ye15 | 1 | 45 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,998 |
https://leetcode.com/problems/find-original-array-from-doubled-array/discuss/2824173/Python3-or-Sorting-%2B-Hashmap | class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n % 2 != 0:
return []
freq = Counter(changed)
ans = []
changed.sort()
for i in range(n):
if changed[i] % 2 != 0:
if freq[changed[... | find-original-array-from-doubled-array | [Python3] | Sorting + Hashmap | swapnilsingh421 | 0 | 1 | find original array from doubled array | 2,007 | 0.409 | Medium | 27,999 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.