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|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1383632/Python3-%3A-simple-93-faster
|
class Solution:
def countGoodSubstrings(self, s: str) -> int:
count=0
for i in range(0,len(s)-2):
if len(set(s[i:i+3])) == 3:
count+=1
return count
|
substrings-of-size-three-with-distinct-characters
|
Python3 : simple 93% faster
|
p_a
| 0
| 52
|
substrings of size three with distinct characters
| 1,876
| 0.703
|
Easy
| 26,600
|
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1339461/Python-O(n)-Sliding-Window-Technique-Hashmap
|
class Solution(object):
def countGoodSubstrings(self, s):
"""
:type s: str
:rtype: int
"""
if len(s) < 3:
return 0
count = 0
i, j = 0, 0
has = {}
while j < len(s):
if j-i == 3:
count += 1
i += 1
continue
if (s[j] not in has) or (s[j] in has and has[s[j]] < i):
has[s[j]] = j
else:
if j-i == 3:
count += 1
i = has[s[j]] + 1
has[s[j]] = j
j += 1
if j-i == 3:
count += 1
return count
|
substrings-of-size-three-with-distinct-characters
|
Python O(n) Sliding Window Technique Hashmap
|
Sibu0811
| 0
| 145
|
substrings of size three with distinct characters
| 1,876
| 0.703
|
Easy
| 26,601
|
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1327994/Python-Easy-Simple-O(n)
|
class Solution(object):
def countGoodSubstrings(self, s):
count=0
for each in range(2,len(s)+1):
if len(set(s[each-3:each])) == 3:
count+=1
return count
|
substrings-of-size-three-with-distinct-characters
|
Python Easy Simple O(n)
|
akashadhikari
| 0
| 50
|
substrings of size three with distinct characters
| 1,876
| 0.703
|
Easy
| 26,602
|
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1255635/Python3-or-Generalized-Solution-or-o(n)
|
class Solution:
def countGoodSubstrings(self, s: str) -> int:
seen = set()
req_len = 3
start = 0
ans = 0
for i in range(len(s)):
if i-start == req_len:
seen.discard(s[start])
start+=1
while s[i] in seen:
seen.discard(s[start])
start += 1
seen.add(s[i])
if len(seen) == req_len:
ans += 1
return ans
|
substrings-of-size-three-with-distinct-characters
|
Python3 | Generalized Solution | o(n)
|
Sanjaychandak95
| 0
| 30
|
substrings of size three with distinct characters
| 1,876
| 0.703
|
Easy
| 26,603
|
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1238893/PythonPython3-solution-in-6-lines-with-explanation
|
class Solution:
def countGoodSubstrings(self, s: str) -> int:
cnt = 0 #to count the distinct character substring of length 3
for i in range(len(s)-2):
subStr = s[i:i+3] #compute the substring
if len(set(subStr)) == 3: #if set of subStr length is equal to 3 then increment cnt by 1
cnt += 1
return cnt
|
substrings-of-size-three-with-distinct-characters
|
Python/Python3 solution in 6 lines with explanation
|
prasanthksp1009
| 0
| 41
|
substrings of size three with distinct characters
| 1,876
| 0.703
|
Easy
| 26,604
|
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1238864/Python%3A-Find-all-substrings-of-length-3-and-then-find-unique.
|
class Solution:
def countGoodSubstrings(self, s: str) -> int:
substring = []
for i in range(0,len(s)-2):
a = s[i:i+3]
substring.append(a)
ans = 0
for i in substring:
if len(set(i)) == 3:
ans += 1
return ans
|
substrings-of-size-three-with-distinct-characters
|
Python: Find all substrings of length 3 and then find unique.
|
harshhx
| 0
| 437
|
substrings of size three with distinct characters
| 1,876
| 0.703
|
Easy
| 26,605
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2087728/Python-Easy-To-Understand-Code-oror-Beginner-Friendly-oror-Brute-Force
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
pair_sum = []
nums.sort()
for i in range(len(nums)//2):
pair_sum.append(nums[i]+nums[len(nums)-i-1])
return max(pair_sum)
|
minimize-maximum-pair-sum-in-array
|
Python Easy To Understand Code || Beginner Friendly || Brute Force
|
Shivam_Raj_Sharma
| 5
| 207
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,606
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1725067/Python-or-Sort-and-Two-Pointers
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
i = 0
j = len(nums) - 1
res = 0
while i < j:
res = max(res, nums[i] + nums[j])
i += 1
j -= 1
return res
|
minimize-maximum-pair-sum-in-array
|
Python | Sort and Two Pointers
|
jgroszew
| 4
| 258
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,607
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1292471/Python-fast-and-pythonic
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
return max([value+nums[-index] for index, value in enumerate(nums[:len(nums)//2], 1)])
|
minimize-maximum-pair-sum-in-array
|
[Python] fast and pythonic
|
cruim
| 2
| 340
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,608
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2327399/Python-or-Easy-and-Fast-Solution-or-Two-pointer
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
ans = 0
nums.sort()
l, r = 0, len(nums)-1
while l<r:
ans = max(ans, nums[l]+nums[r])
l += 1
r -= 1
return ans
|
minimize-maximum-pair-sum-in-array
|
Python | Easy & Fast Solution | Two pointer
|
desalichka
| 1
| 46
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,609
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1653815/Python-simple-two-pointers-solution-after-sort
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
res = 0
l, r = 0, n-1
while l < r:
res = max(res, nums[l] + nums[r])
l += 1
r -= 1
return res
|
minimize-maximum-pair-sum-in-array
|
Python simple two-pointers solution after sort
|
byuns9334
| 1
| 72
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,610
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1359976/Sort-and-zip-98-speed
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
half_len_nums = len(nums) // 2
return max(a + b for a, b in zip(nums[:half_len_nums:],
nums[-1:half_len_nums - 1: -1]))
|
minimize-maximum-pair-sum-in-array
|
Sort and zip, 98% speed
|
EvgenySH
| 1
| 157
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,611
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1276736/Python3-simple-solution-using-sorting-and-single-while-loop-beats-90-users
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
i,j = 0, len(nums)-1
res = []
while i < j:
res.append(nums[i]+nums[j])
i += 1
j -= 1
return max(res)
|
minimize-maximum-pair-sum-in-array
|
Python3 simple solution using sorting and single while loop beats 90% users
|
EklavyaJoshi
| 1
| 68
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,612
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1245842/python3-Simple-solution-beats-99.43-of-python3-submissions.
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
return max([nums[x]+nums[-x-1] for x in range(len(nums)//2)])
|
minimize-maximum-pair-sum-in-array
|
[python3] Simple solution, beats 99.43 % of python3 submissions.
|
SushilG96
| 1
| 181
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,613
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1238919/Python-oror-sort
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
pairs = []
nums = sorted(nums)
n = len(nums)
for i in range(len(nums)//2):
a = [nums[i],nums[n-i-1]]
pairs.append(a)
sum_ = []
for i,j in pairs:
sum_.append(i+j)
return max(sum_)
|
minimize-maximum-pair-sum-in-array
|
Python || sort
|
harshhx
| 1
| 87
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,614
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1238865/PythonPython3-solution-in-5-lines-with-explanation
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort() # sort the numbers
lis = [] # to store the res and it takes O(n) space complexity
for i in range(len(nums)//2): # traverse the loop to length(nums)/2 times
lis.append(nums[i]+nums[~i]) #add the indexes 0+(-1),1+(-2)...goes on till n//2 times
return max(lis) # return the maximum element in the list
|
minimize-maximum-pair-sum-in-array
|
Python/Python3 solution in 5 lines with explanation
|
prasanthksp1009
| 1
| 95
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,615
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1238865/PythonPython3-solution-in-5-lines-with-explanation
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()# sort the numbers
maxi = 0 # to store the maximum value and it takes only O(1) space complexity
for i in range(len(nums)//2):# traverse the loop to length(nums)/2 times
maxi = max(maxi,nums[i]+nums[~i])#add the indexes 0+(-1),1+(-2)...goes on till n//2 times and compute the maxi value simultaneously
return maxi
|
minimize-maximum-pair-sum-in-array
|
Python/Python3 solution in 5 lines with explanation
|
prasanthksp1009
| 1
| 95
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,616
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2845420/Python-easy-solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
n=len(nums)
ot=[]
for i in range(n//2):
ot.append(nums[i]+nums[n-i-1])
return max(ot)
|
minimize-maximum-pair-sum-in-array
|
Python easy solution
|
patelhet050603
| 0
| 2
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,617
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2834978/Python3-2-liner-beats-97.2888.60
|
class Solution:
def minPairSum(self, nums):
nums.sort()
return max(nums[i]+nums[-1*i-1] for i in range(len(nums)))
|
minimize-maximum-pair-sum-in-array
|
[Python3] 2-liner, beats 97.28%/88.60%
|
U753L
| 0
| 3
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,618
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2821014/Simple-two-pointers-solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
i = 0
j = len(nums)-1
max_sum = 0
while i < j:
max_sum = max(max_sum, nums[i] + nums[j])
i += 1
j -= 1
return max_sum
|
minimize-maximum-pair-sum-in-array
|
Simple two pointers solution
|
khaled_achech
| 0
| 1
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,619
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2805309/Python-Solution-EASY-TO-UNDERSTAND
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
m=0
nums.sort()
n=len(nums)
for i in range(n//2):
m=max(m,nums[i]+nums[n-i-1])
return m
|
minimize-maximum-pair-sum-in-array
|
Python Solution - EASY TO UNDERSTAND
|
T1n1_B0x1
| 0
| 4
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,620
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2801513/Fast-and-memory-efficient-solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
h = []
heapq.heapify(nums)
for i in range(len(nums)):
h.append(heapq.heappop(nums))
right = len(h)-1
m = 0
for left in range(len(h)//2):
m = max(m, (h[left]+h[right]))
right -= 1
return m
|
minimize-maximum-pair-sum-in-array
|
Fast and memory efficient solution
|
AndreySmirnov
| 0
| 1
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,621
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2785585/Python3-solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
res = []
i, j = 0, len(nums) - 1
while i < j:
res.append(nums[i] + nums[j])
i += 1
j -= 1
return max(res)
|
minimize-maximum-pair-sum-in-array
|
Python3 solution
|
mediocre-coder
| 0
| 2
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,622
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2782495/Python3-Solution-with-explanation-beats-99-in-memory
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
max_sum=0
for i in range(len(nums)//2):
max_sum = max(max_sum, nums.pop(0)+nums.pop(len(nums)-1))
return max_sum
|
minimize-maximum-pair-sum-in-array
|
Python3 Solution with explanation, beats 99% in memory
|
sipi09
| 0
| 2
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,623
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2777553/Python-simple-solution-using-sort
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
maxi = 0
for i in range(n//2):
print(nums[n-1])
if nums[i]+nums[n-1-i]>maxi:
maxi = nums[i]+nums[n-1-i]
return maxi
|
minimize-maximum-pair-sum-in-array
|
Python simple solution using sort
|
Rajeev_varma008
| 0
| 2
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,624
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2719124/Python-Sorting-O(nlogn)
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
ans = []
for i in range(len(nums)//2):
ans.append(nums[i] + nums[~i])
return max(ans)
|
minimize-maximum-pair-sum-in-array
|
Python Sorting O(nlogn)
|
axce1
| 0
| 6
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,625
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2689639/Simple-Python-Solution-or-Sorting-or-Two-Pointers
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
maxi=0
n=len(nums)
for i in range(n//2):
maxi=max(maxi, nums[i]+nums[-i-1])
return maxi
|
minimize-maximum-pair-sum-in-array
|
Simple Python Solution | Sorting | Two Pointers
|
Siddharth_singh
| 0
| 7
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,626
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2643880/Python-or-Simple-Solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
maxVal = 0
j = len(nums)-1
for i in range(len(nums)//2):
maxVal = max(maxVal, nums[i]+nums[j])
j-=1
return maxVal
|
minimize-maximum-pair-sum-in-array
|
Python | Simple Solution
|
naveenraiit
| 0
| 4
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,627
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2410988/Simple-python-code-with-explanation
|
class Solution:
#using sorting and two pointers approach
def minPairSum(self, nums: List[int]) -> int:
#create a new list to store sum of pairs
sumofpairs = []
#sort the elements
nums.sort()
#let l pointer be at 0th index
l = 0
#let r pointer be at last index
r = len(nums)-1
#(l<r) is the Condition to break
while l < r:
#add sum of first element and last element to sumofpairs list
sumofpairs.append(nums[l] + nums[r])
#increase the l pointer by 1
l = l + 1
#decrease the r pointer by 1
r = r - 1
#return the maximum of sumof pairs-->list
return max(sumofpairs)
|
minimize-maximum-pair-sum-in-array
|
Simple python code with explanation
|
thomanani
| 0
| 14
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,628
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2242405/Python-Solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
return max([nums[i]+nums[~i] for i in range(len(nums)//2)])
|
minimize-maximum-pair-sum-in-array
|
Python Solution
|
SakshiMore22
| 0
| 39
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,629
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2175328/Python-or-Easy-solution-using-two-pointers-technique
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
# ///// TC: O(nlogn) and SC: O(n) ////
nums.sort()
res = []
l,r = 0, len(nums) - 1
while l < r:
pairSum = nums[l] + nums[r]
res.append(pairSum)
l += 1
r -= 1
return max(res)
|
minimize-maximum-pair-sum-in-array
|
Python | Easy solution using two pointers technique
|
__Asrar
| 0
| 33
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,630
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2174776/Python3-Very-simple-solution-with-sort
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
## RC ##
## APPROACH: MATH ##
nums.sort()
res = 0
n = len(nums)
for i in range(n//2):
res = max(nums[i] + nums[n-1-i], res)
return res
|
minimize-maximum-pair-sum-in-array
|
[Python3] Very simple solution with sort
|
101leetcode
| 0
| 21
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,631
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1991533/Python-Solution-or-Sorting-and-Two-Pointers-or-Clean-Code
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
N = len(nums)
maxPairSum = 0
left = 0
right = N-1
while left < right:
maxPairSum = max(maxPairSum,nums[left] + nums[right])
left += 1
right -= 1
return maxPairSum
|
minimize-maximum-pair-sum-in-array
|
Python Solution | Sorting and Two Pointers | Clean Code
|
Gautam_ProMax
| 0
| 41
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,632
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1706581/Understandable-code-for-beginners-like-me-in-python-!!
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
pairsum,maxSum=0,0
nums.sort()
nums_len=len(nums)
half_len=nums_len//2
for index in range(half_len):
pairsum=nums[index]+nums[nums_len-index-1]
if(pairsum>maxSum):
maxSum=pairsum
return maxSum
|
minimize-maximum-pair-sum-in-array
|
Understandable code for beginners like me in python !!
|
kabiland
| 0
| 64
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,633
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1705074/Simple-python3-solution-with-description-and-example
|
class Solution:
def minPairSum(self, nums: List[int]) -> int: #eg - [3,5,2,3]
nums = sorted(nums) # sort list in ascending order => [2,3,3,5]
i = 0 # two pointer i and j => nums[i] = 2 and nums[j] = 5
j = len(nums)-1
lis = []
while(i<j):
lis.append(nums[i]+nums[j]) # sum ith and jth element (2,5), (3,3) pairs
i+=1
j-=1
return max(lis) # return maximum number in list => lis= [2+5, 3+3], [7,6] => ans = 7
|
minimize-maximum-pair-sum-in-array
|
Simple python3 solution with description and example
|
KratikaRathore
| 0
| 61
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,634
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1561707/Py3-Simple-solution-using-zip
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
res = 0
n = len(nums)
if n%2 == 0:
nums.sort()
for x, y in zip(nums[:n//2], nums[n//2:][::-1]):
res = max(res, x+y)
return res
|
minimize-maximum-pair-sum-in-array
|
[Py3] Simple solution using zip
|
ssshukla26
| 0
| 66
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,635
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1521050/Python-Heap
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
max_heap, min_heap = [-i for i in nums], nums
heapify(min_heap)
heapify(max_heap)
return max([
-heappop(max_heap) + heappop(min_heap)
for i in range(len(nums) // 2)
])
|
minimize-maximum-pair-sum-in-array
|
[Python] Heap
|
dev-josh
| 0
| 62
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,636
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1494755/Python-3-oror-Faster-than-95-ororSimple-2-ptr
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
n=len(nums)
i,j = 0,n-1
nums.sort()
max_sum = float("-inf")
while(i<n//2):
max_sum = max(nums[i]+nums[j],max_sum)
i+=1
j-=1
return max_sum
|
minimize-maximum-pair-sum-in-array
|
Python 3 || Faster than 95% ||Simple 2 ptr
|
ana_2kacer
| 0
| 113
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,637
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1459604/Python3-Solution-with-using-sorting
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
_max = 0
for idx in range(len(nums) // 2):
_max = max(_max, nums[idx] + nums[len(nums) - 1 - idx])
return _max
|
minimize-maximum-pair-sum-in-array
|
[Python3] Solution with using sorting
|
maosipov11
| 0
| 131
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,638
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1387779/Python-oror-O(n)-oror-Count-Sort-oror-Easy-Solution-oror-beat-100-speed-and-memory
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
self.myCountSort(nums)
asc = nums[0:len(nums)//2]
desc = nums[len(nums)//2:][::-1]
max_val, temp = 0, 0
for asc_val,desc_val in zip(asc,desc):
temp = asc_val+desc_val
max_val = max(max_val, temp)
return max_val
def myCountSort(self, the_list):
max_val = max(the_list)
count = [0] * (max_val+1)
for item in the_list:
count[item] += 1
index = 0
for i in range(len(count)):
while count[i] > 0:
the_list[index] = i
index += 1
count[i] -= 1
return the_list
|
minimize-maximum-pair-sum-in-array
|
Python || O(n) || Count Sort || Easy Solution || beat 100% speed and memory
|
zafarman
| 0
| 173
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,639
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1268312/Python-faster-than-95-O(nlogn)
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums = sorted(nums)
ans = nums[0]
for i in range(len(nums)//2):
pair_sum = nums[i] + nums[len(nums) -i - 1]
if pair_sum > ans:
ans = pair_sum
return ans
|
minimize-maximum-pair-sum-in-array
|
Python faster than 95% O(nlogn)
|
user5573CJ
| 0
| 84
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,640
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1239052/Python3-greedy
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
return max(nums[i] + nums[~i] for i in range(len(nums)//2))
|
minimize-maximum-pair-sum-in-array
|
[Python3] greedy
|
ye15
| 0
| 35
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,641
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1239052/Python3-greedy
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
ans = 0
for i in range(len(nums)//2):
ans = max(ans, nums[i] + nums[~i])
return ans
|
minimize-maximum-pair-sum-in-array
|
[Python3] greedy
|
ye15
| 0
| 35
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,642
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/2309773/Python-oror-Time-O(n)-two-pointer-solution
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
res = 0
nums.sort()
i,j = 0 , len(nums)-1
while(i<j):
res = max(res , nums[i]+nums[j])
i+=1
j-=1
return res
|
minimize-maximum-pair-sum-in-array
|
Python || Time O(n) two-pointer solution
|
ronipaul9972
| -1
| 34
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,643
|
https://leetcode.com/problems/minimize-maximum-pair-sum-in-array/discuss/1657614/Python-oror-O(nlogn)-time-oror-O(1)-spaceoror-sorting-oror-2-pointers
|
class Solution:
def minPairSum(self, nums: List[int]) -> int:
nums.sort()
start, end = 0, len(nums) - 1
max_sum = 0
while start <= end:
curr_sum = nums[start] + nums[end]
max_sum = max(max_sum,curr_sum)
start += 1
end -= 1
return max_sum
#TC -> O(n*logn)
#SC -> O(1)
|
minimize-maximum-pair-sum-in-array
|
Python || O(nlogn) time || O(1) space|| sorting || 2 pointers
|
s_m_d_29
| -1
| 95
|
minimize maximum pair sum in array
| 1,877
| 0.803
|
Medium
| 26,644
|
https://leetcode.com/problems/get-biggest-three-rhombus-sums-in-a-grid/discuss/1239929/python-oror-100-faster-oror-well-explained-oror-Simple-approach
|
class Solution:
def getBiggestThree(self, grid: List[List[int]]) -> List[int]:
def calc(l,r,u,d):
sc=0
c1=c2=(l+r)//2
expand=True
for row in range(u,d+1):
if c1==c2:
sc+=grid[row][c1]
else:
sc+=grid[row][c1]+grid[row][c2]
if c1==l:
expand=False
if expand:
c1-=1
c2+=1
else:
c1+=1
c2-=1
return sc
m=len(grid)
n=len(grid[0])
heap=[]
for i in range(m):
for j in range(n):
l=r=j
d=i
while l>=0 and r<=n-1 and d<=m-1:
sc=calc(l,r,i,d)
l-=1
r+=1
d+=2
if len(heap)<3:
if sc not in heap:
heapq.heappush(heap,sc)
else:
if sc not in heap and sc>heap[0]:
heapq.heappop(heap)
heapq.heappush(heap,sc)
heap.sort(reverse=True)
return heap
|
get-biggest-three-rhombus-sums-in-a-grid
|
🐍 {python} || 100% faster || well-explained || Simple approach
|
abhi9Rai
| 6
| 721
|
get biggest three rhombus sums in a grid
| 1,878
| 0.464
|
Medium
| 26,645
|
https://leetcode.com/problems/get-biggest-three-rhombus-sums-in-a-grid/discuss/1442967/Python-3-or-DP-Matrix-Padding-Prefix-Sum-or-Explanation
|
class Solution:
def getBiggestThree(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0])
dp = [[[0, 0]] * (n+2) for _ in range(m+2)]
ans = []
for i in range(1, m+1):
for j in range(1, n+1): # [i, j] will be the bottom vertex
ans.append(grid[i-1][j-1])
dp[i][j] = [grid[i-1][j-1], grid[i-1][j-1]]
dp[i][j][0] += dp[i-1][j-1][0] # dp: major diagonal
dp[i][j][1] += dp[i-1][j+1][1] # dp: minor diagonal
for win in range(1, min(m, n)):
x1, y1 = i-win, j-win # left vertex
x2, y2 = i-win, j+win # right vertex
x3, y3 = i-win-win, j # top vertex
if not (all(1 <= x < m+1 for x in [x1, x2, x3]) and all(1 <= y < n+1 for y in [y1, y2, y3])):
break
b2l = dp[i][j][0] - dp[x1-1][y1-1][0] # bottom node to left node (node sum), major diagonal
b2r = dp[i][j][1] - dp[x2-1][y2+1][1] # bottom node to right node (node sum), minor diagonal
l2t = dp[x1][y1][1] - dp[x3-1][y3+1][1] # left node to top node (node sum), minor diagonal
r2t = dp[x2][y2][0] - dp[x3-1][y3-1][0] # right node to top node (node sum), major diagonal
vertices_sum = grid[i-1][j-1] + grid[x1-1][y1-1] + grid[x2-1][y2-1] + grid[x3-1][y3-1]
cur = b2l + b2r + l2t + r2t - vertices_sum # sum(edges) - sum(4 vertices)
ans.append(cur)
return sorted(set(ans), reverse=True)[:3] # unique + sort reverse + keep only first 3
|
get-biggest-three-rhombus-sums-in-a-grid
|
Python 3 | DP, Matrix Padding, Prefix Sum | Explanation
|
idontknoooo
| 2
| 659
|
get biggest three rhombus sums in a grid
| 1,878
| 0.464
|
Medium
| 26,646
|
https://leetcode.com/problems/get-biggest-three-rhombus-sums-in-a-grid/discuss/1360208/Brute-force-75-speed
|
class Solution:
def getBiggestThree(self, grid: List[List[int]]) -> List[int]:
first = second = third = 0
def update_sums(val):
nonlocal first, second, third
if val > first:
third = second
second = first
first = val
elif first > val > second:
third = second
second = val
elif second > val > third:
third = val
rows, cols = len(grid), len(grid[0])
max_size = max(rows // 2, cols // 2) + 1
for r in range(rows):
for c in range(cols):
update_sums(grid[r][c])
for n in range(1, max_size):
if c + 2 * n < cols and r - n >= 0 and r + n < rows:
sum_rhombus = (grid[r][c] + grid[r][c + 2 * n] +
grid[r - n][c + n] + grid[r + n][c + n])
end_c = c + 2 * n
for i in range(1, n):
sum_rhombus += (grid[r + i][c + i] +
grid[r - i][c + i] +
grid[r + i][end_c - i] +
grid[r - i][end_c - i])
update_sums(sum_rhombus)
else:
break
biggest = {first, second, third} - {0}
return sorted(biggest, reverse=True)
|
get-biggest-three-rhombus-sums-in-a-grid
|
Brute force, 75% speed
|
EvgenySH
| 1
| 337
|
get biggest three rhombus sums in a grid
| 1,878
| 0.464
|
Medium
| 26,647
|
https://leetcode.com/problems/get-biggest-three-rhombus-sums-in-a-grid/discuss/1242382/1878.-Get-Biggest-Three-Rhombus-Sums-in-a-Grid-oror-python-solution
|
class Solution(object):
def getBiggestThree(self, grid):
def calc(l,r,u,d):
ssum=0
expand=True
c1=c2=(l+r)//2
for row in range(u,d+1):
if c1==c2:
ssum+=grid[row][c1]
else :
ssum+=grid[row][c1]+grid[row][c2]
if c1==l:
expand=False
if expand:
c1-=1
c2+=1
else :
c1+=1
c2-=1
return ssum
m=len(grid)
n=len(grid[0])
pq=[]
for i in range(m):
for j in range(n):
l=r=j
d=i
while l>=0 and r<n and d<m:
ssum=calc(l,r,i,d)
l-=1
r+=1
d+=2
if len(pq)<3:
if ssum not in pq:
heapq.heappush(pq,ssum)
else :
if ssum not in pq and ssum>pq[0]:
heapq.heappop(pq)
heapq.heappush(pq,ssum)
pq.sort(reverse=True)
return pq
```
|
get-biggest-three-rhombus-sums-in-a-grid
|
1878. Get Biggest Three Rhombus Sums in a Grid || python solution
|
aayush_chhabra
| 1
| 511
|
get biggest three rhombus sums in a grid
| 1,878
| 0.464
|
Medium
| 26,648
|
https://leetcode.com/problems/get-biggest-three-rhombus-sums-in-a-grid/discuss/1463534/Python3-Passes-116117-cases-please-help
|
class Solution:
def getBiggestThree(self, grid: List[List[int]]) -> List[int]:
run = set()
n = len(grid)
m = len(grid[0])
# Loop through every grid item
for i in range(n):
for j in range(m):
# Calculate the max rhombus around the current grid item
res = self.count(grid, i, j, len(grid), len(grid[0]))
# Add to list of sums
run.add(res)
res = []
# Stupid way of getting biggest 3,
# a heap would work better but the problem
# isn't here
return sorted(run, reverse=True)[:3]
# This function calculates the width of the largest
# rhombus that can be made around the given grid row and col.
# Then it calculates each rhombus that can be made up until that max width,
# keeping track of what the largest sum is amongst all the created rhombuses,
# and returns the max
def count(self, grid, row, col, n, m):
# Calculate how wide the largest rhombus can be.
# it is the smaller number of how close are you to the
# nearest edge of row, or nearest edge of column
distance = min(n - row, m - col)
# set the max perimeter sum to the current number
global_max = grid[row][col]
# From 1 to the width of the biggest rhombus that can be made
for i in range(1, distance):
# check 4 corners, if not 4 corners, no rhombus can be made at current distance
if row - i >= 0 and row + i < n and col - i >= 0 and col + i < m:
top_point = row - i
bottom_point = row + i
# Add up the vertices (points)
local_max = grid[row-i][col] + grid[row+i][col] + grid[row][col-i] + grid[row][col+i]
# For 1 up to the width of the current rhombus,
# add up the perimeter
for j in range(1,i):
local_max += grid[top_point + j][col - j] + grid[top_point + j][col+j]
local_max += grid[bottom_point - j][col-j] + grid[bottom_point - j][col + j]
# Save the bigger perimeter sum
global_max = max(global_max, local_max)
else:
break
return global_max
|
get-biggest-three-rhombus-sums-in-a-grid
|
Python3 Passes 116/117 cases, please help
|
walshyb
| 0
| 290
|
get biggest three rhombus sums in a grid
| 1,878
| 0.464
|
Medium
| 26,649
|
https://leetcode.com/problems/get-biggest-three-rhombus-sums-in-a-grid/discuss/1239060/Python3-prefix-sums
|
class Solution:
def getBiggestThree(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0]) # dimensions
anti, diag = {}, {}
for i in range(m):
for j in range(n):
key = i+j
if key not in anti: anti[key] = [0]
anti[key].append(anti[key][-1] + grid[i][j])
key = i-j
if key not in diag: diag[key] = [0]
diag[key].append(diag[key][-1] + grid[i][j])
def fn(i, j, k):
"""Return sum of k diagonal elements starting from (i, j)"""
if i >= j: return diag[i-j][j+k] - diag[i-j][j]
return diag[i-j][i+k] - diag[i-j][i]
def gn(i, j, k):
"""Return sum of k anti-diagonal elements starting from (i, j)"""
if i+j < n: return anti[i+j][i+k] - anti[i+j][i]
return anti[i+j][n-1-j+k] - anti[i+j][n-1-j]
ans = set()
for i in range(m):
for j in range(n):
ans.add(grid[i][j])
for ii in range(i+2, m, 2):
r = (ii-i)//2
if j-r < 0 or j+r >= n: break
val = 0
val += fn(i, j, r+1)
val += gn(i, j, r+1)
val += fn((ii+i)//2, j-r, r+1)
val += gn((ii+i)//2, j+r, r+1)
val -= grid[i][j] + grid[(ii+i)//2][j-r] + grid[(ii+i)//2][j+r] + grid[ii][j]
ans.add(val)
return sorted(ans, reverse=True)[:3]
|
get-biggest-three-rhombus-sums-in-a-grid
|
[Python3] prefix sums
|
ye15
| 0
| 111
|
get biggest three rhombus sums in a grid
| 1,878
| 0.464
|
Medium
| 26,650
|
https://leetcode.com/problems/minimum-xor-sum-of-two-arrays/discuss/1238641/Bit-Mask
|
class Solution:
def minimumXORSum(self, a: List[int], b: List[int]) -> int:
@cache
def dp(mask: int) -> int:
i = bin(mask).count("1")
if i >= len(a):
return 0
return min((a[i] ^ b[j]) + dp(mask + (1 << j))
for j in range(len(b)) if mask & (1 << j) == 0)
return dp(0)
|
minimum-xor-sum-of-two-arrays
|
Bit Mask
|
votrubac
| 103
| 6,400
|
minimum xor sum of two arrays
| 1,879
| 0.448
|
Hard
| 26,651
|
https://leetcode.com/problems/minimum-xor-sum-of-two-arrays/discuss/1239072/Python3-bit-mask-dp
|
class Solution:
def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
@cache
def fn(mask, k):
"""Return min xor sum."""
if not mask: return 0
ans = inf
for i in range(n):
if mask & (1<<i):
ans = min(ans, (nums1[i]^nums2[k]) + fn(mask^(1<<i), k+1))
return ans
return fn((1<<n)-1, 0)
|
minimum-xor-sum-of-two-arrays
|
[Python3] bit-mask dp
|
ye15
| 1
| 60
|
minimum xor sum of two arrays
| 1,879
| 0.448
|
Hard
| 26,652
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1241968/Minus-49
|
class Solution:
def isSumEqual(self, first: str, second: str, target: str) -> bool:
def op(s: str): return "".join(chr(ord(ch) - 49) for ch in s)
return int(op(first)) + int(op(second)) == int(op(target))
|
check-if-word-equals-summation-of-two-words
|
Minus 49
|
votrubac
| 18
| 1,200
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,653
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1241973/Python3-or-Time%3A-O(n)-or-Space%3A-O(1)-or-Clear.-Explanation!
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
mapping = {
'a':'0', 'b':'1', 'c':'2', 'd':'3', 'e':'4', 'f':'5', 'g':'6', 'h':'7', 'i':'8', 'j':'9'
}
def decoding(s):
nonlocal mapping
return int(''.join(list(map((lambda i:mapping[i]),list(s)))))
return decoding(firstWord) + decoding(secondWord) == decoding(targetWord)
|
check-if-word-equals-summation-of-two-words
|
Python3 | Time: O(n) | Space: O(1) | Clear. Explanation!
|
antonyakovlev
| 9
| 683
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,654
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1239984/Python-oror-simple-using-ascii-conversion
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
a, b, c = [],[],[]
a_,b_,c_="","",""
base = ord("a")
def getList(l,s):
for i in s:
l.append(ord(i)-base)
return l
def getString(x,l):
for i in l:
x += str(i)
return x
a = getList(a,firstWord)
b = getList(b,secondWord)
c = getList(c,targetWord)
a_ = getString(a_,a)
b_ = getString(b_,b)
c_ = getString(c_,c)
return(( int(a_)+int(b_)) == int(c_))
|
check-if-word-equals-summation-of-two-words
|
Python || simple using ascii conversion
|
harshhx
| 4
| 492
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,655
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1357032/Easy-Python-Solutions-(3-Approaches-Speed-98-Memory-100)
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
alphabets = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
target_val = 0
temp = ""
for i in targetWord:
temp += str(alphabets.index(i))
target_val = int(temp)
temp = ""
for i in firstWord:
temp += str(alphabets.index(i))
target_val -= int(temp)
temp = ""
for i in secondWord:
temp += str(alphabets.index(i))
target_val -= int(temp)
if target_val:
return False
else:
return True
|
check-if-word-equals-summation-of-two-words
|
Easy Python Solutions (3 Approaches - Speed 98%, Memory 100%)
|
the_sky_high
| 3
| 108
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,656
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1357032/Easy-Python-Solutions-(3-Approaches-Speed-98-Memory-100)
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
alphabets = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
target_val = 0
temp = ""
for i in targetWord:
temp += str(alphabets.index(i))
target_val = int(temp)
temp = ""
for i in firstWord:
temp += str(alphabets.index(i))
target_val -= int(temp)
temp = ""
for i in secondWord:
temp += str(alphabets.index(i))
target_val -= int(temp)
if target_val:
return False
else:
return True
|
check-if-word-equals-summation-of-two-words
|
Easy Python Solutions (3 Approaches - Speed 98%, Memory 100%)
|
the_sky_high
| 3
| 108
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,657
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1357032/Easy-Python-Solutions-(3-Approaches-Speed-98-Memory-100)
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
alphabets = {'a': '0', 'b': '1', 'c': '2', 'd': '3', 'e': '4', 'f': '5', 'g': '6', 'h': '7', 'i': '8', 'j': '9'}
target_val = 0
temp = ""
for i in targetWord:
temp += alphabets[i]
target_val = int(temp)
temp = ""
for i in firstWord:
temp += alphabets[i]
target_val -= int(temp)
temp = ""
for i in secondWord:
temp += alphabets[i]
target_val -= int(temp)
if target_val:
return False
else:
return True
|
check-if-word-equals-summation-of-two-words
|
Easy Python Solutions (3 Approaches - Speed 98%, Memory 100%)
|
the_sky_high
| 3
| 108
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,658
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2186388/Check-if-Word-Equals-Summation-of-Two-Words
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
wordList = [firstWord, secondWord, targetWord]
sumList = []
for word in wordList:
sumList.append(self.calcSum(word))
return sumList[0]+sumList[1]==sumList[2]
def calcSum(self, word):
s = ""
for letter in word:
s+=str(ord(letter)-97)
return int(s)
|
check-if-word-equals-summation-of-two-words
|
Check if Word Equals Summation of Two Words
|
pratik_patra
| 1
| 31
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,659
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1791063/Simple-python3-solution-or-82-faster-or-Easy-to-understand
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
s = "abcdefghij"
fir = str()
sec = str()
tar = str()
for i in firstWord:
fir += str(s.index(i))
for i in secondWord:
sec += str(s.index(i))
for i in targetWord:
tar += str(s.index(i))
return int(fir)+int(sec) == int(tar)
|
check-if-word-equals-summation-of-two-words
|
Simple python3 solution | 82% faster | Easy to understand
|
Coding_Tan3
| 1
| 36
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,660
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1698961/inner-function-based
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
def get_number(s):
num = 0
for c in s:
num = num * 10 + ord(c) - ord('a')
return num
return get_number(firstWord) + get_number(secondWord) == get_number(targetWord)
|
check-if-word-equals-summation-of-two-words
|
inner function based
|
snagsbybalin
| 1
| 17
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,661
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1266027/Python3-O(mn)-time-O(1)-space
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
dictionary = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
firstWordConvert, secondWordConvert, targetWordConvert = '', '', ''
for letter in firstWord:
firstWordConvert += str(dictionary.index(letter))
for letter in secondWord:
secondWordConvert += str(dictionary.index(letter))
for letter in targetWord:
targetWordConvert += str(dictionary.index(letter))
return True if int(firstWordConvert) + int(secondWordConvert) == int(targetWordConvert) else False
|
check-if-word-equals-summation-of-two-words
|
Python3 O(mn) time O(1) space
|
peterhwang
| 1
| 157
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,662
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2825043/Python-Simple-Solution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
sum1 = ""
for i in range(0, len(firstWord)):
if firstWord[i] == 'a':
if sum1 != "0":
sum1 += str(ord(firstWord[i]) - 97)
else:
sum1 += str(ord(firstWord[i]) - 97)
sum2 = ""
for i in range(0, len(secondWord)):
if secondWord[i] == 'a':
if sum2 != "0":
sum2 += str(ord(secondWord[i]) - 97)
else:
sum2 += str(ord(secondWord[i]) - 97)
sum3 = ""
for i in range(0, len(targetWord)):
if targetWord[i] == 'a':
if sum3 != "0":
sum3 += str(ord(targetWord[i]) - 97)
else:
sum3 += str(ord(targetWord[i]) - 97)
return int(sum1) + int(sum2) == int(sum3)
|
check-if-word-equals-summation-of-two-words
|
Python Simple Solution
|
PranavBhatt
| 0
| 1
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,663
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2812302/Python-or-simple-solution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
def encode(s):
return int("".join([str(ord(char) - 97) for char in s]))
return encode(firstWord) + encode(secondWord) == encode(targetWord)
|
check-if-word-equals-summation-of-two-words
|
Python | simple solution
|
kawamataryo
| 0
| 1
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,664
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2773715/Python-or-LeetCode-or-1880.-Check-if-Word-Equals-Summation-of-Two-Words
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
map1 = {'a':0, 'b':1, 'c':2, 'd':3, 'e':4, 'f':5, 'g':6, 'h':7, 'i':8, 'j':9}
s1 = ""
s2 = ""
s3 = ""
for i in firstWord:
s1 += str(map1[i])
for j in secondWord:
s2 += str(map1[j])
for k in targetWord:
s3 += str(map1[k])
return int(s1) + int(s2) == int(s3)
|
check-if-word-equals-summation-of-two-words
|
Python | LeetCode | 1880. Check if Word Equals Summation of Two Words
|
UzbekDasturchisiman
| 0
| 1
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,665
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2773715/Python-or-LeetCode-or-1880.-Check-if-Word-Equals-Summation-of-Two-Words
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
firstNum, secondNum, targetNum = "", "", ""
for char in firstWord:
firstNum += str(ord(char) - ord('a'))
for char in secondWord:
secondNum += str(ord(char) - ord('a'))
for char in targetWord:
targetNum += str(ord(char) - ord('a'))
return int(firstNum) + int(secondNum) == int(targetNum)
|
check-if-word-equals-summation-of-two-words
|
Python | LeetCode | 1880. Check if Word Equals Summation of Two Words
|
UzbekDasturchisiman
| 0
| 1
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,666
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2773715/Python-or-LeetCode-or-1880.-Check-if-Word-Equals-Summation-of-Two-Words
|
class Solution:
def f(self, word: str) -> int:
wordNum = ""
for i in word:
wordNum += str(ord(i) - ord('a'))
return int(wordNum)
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return self.f(firstWord) + self.f(secondWord) == self.f(targetWord)
|
check-if-word-equals-summation-of-two-words
|
Python | LeetCode | 1880. Check if Word Equals Summation of Two Words
|
UzbekDasturchisiman
| 0
| 1
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,667
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2710939/Python-oror-Simple-Approach-using-dictionary-oror-Beats-88
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
v={'a':0,'b':1,'c':2,'d':3,'e':4,'f':5,'g':6,'h':7,'i':8,'j':9}
l=[]
l1=[]
l2=[]
for i in firstWord:
l.append(v.get(i))
for i in secondWord:
l1.append(v.get(i))
for i in targetWord:
l2.append(v.get(i))
res=int("".join(map(str,l)))
res1=int("".join(map(str,l1)))
fin=int("".join(map(str,l2)))
if(res+res1==fin):
return True
else:
return False
|
check-if-word-equals-summation-of-two-words
|
Python || Simple Approach using dictionary || Beats 88%
|
shivammenda2002
| 0
| 4
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,668
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2609661/Solution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
fn=fs=tw = ""
for i in range(len(firstWord)):
fn += str(ord(firstWord[i])-97)
for j in range(len(secondWord)):
fs += str(ord(secondWord[j])-97)
for k in range(len(targetWord)):
tw += str(ord(targetWord[k])-97)
return ((int(fn)+int(fs)) == int(tw))
|
check-if-word-equals-summation-of-two-words
|
Solution
|
fiqbal997
| 0
| 8
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,669
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2520949/Easy-Python-Solution-for-Beginners
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
#Creating a map to convert alphabets to numbers
dicAlpha = {
'a':'0','b':'1','c':'2','d':'3','e':'4','f':'5','g':'6','h':'7','i':'8','j':'9'
}
newFirstWord = ""
newSecondWord = ""
newTargetWord = ""
# Converting alphabets to numbers , example this loop will convert "abc" -> "021"
for i in range(len(firstWord)):
newFirstWord+=dicAlpha[firstWord[i]]
for i in range(len(secondWord)):
newSecondWord+=dicAlpha[secondWord[i]]
for i in range(len(targetWord)):
newTargetWord+=dicAlpha[targetWord[i]]
if int(newFirstWord) + int(newSecondWord) == int(newTargetWord):
return True
else:
return False
|
check-if-word-equals-summation-of-two-words
|
Easy Python Solution for Beginners
|
Ron99
| 0
| 23
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,670
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2324275/Python-Simple
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
num1 = ''
num2 = ''
for i in firstWord:
num1+=str(ord(i)-ord('a'))
for i in secondWord:
num2+=str(ord(i)-ord('a'))
target=''
for i in targetWord:
target+=str(ord(i)-ord('a'))
return int(target) == (int(num1)+int(num2))
|
check-if-word-equals-summation-of-two-words
|
Python Simple
|
sunakshi132
| 0
| 17
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,671
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2298827/PYTHON-oror-FAASTER-THAN-88-oror-SPACE-LESS-THAN-93oror-BASIC
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
def num(a):
n=0
for f in a:
n=n*10 +(ord(f)-97)
return n
if num(targetWord)==num(firstWord)+num(secondWord):
return(True)
else:
return (False)
|
check-if-word-equals-summation-of-two-words
|
PYTHON || FAASTER THAN 88% || SPACE LESS THAN 93%|| BASIC
|
vatsalg2002
| 0
| 12
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,672
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2215263/MINUS-97
|
'''class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
ans=False
first=''
second=''
target=''
for i in firstWord:
first=first+str(ord(i)-97)
for i in secondWord:
second=second+str(ord(i)-97)
for i in targetWord:
target=target+str(ord(i)-97)
add=int(first)+int(second)
if add==int(target):
ans=True
return ans
pritn(isSumEqual(firstWord,secondWord,targetWord))'''
|
check-if-word-equals-summation-of-two-words
|
MINUS 97
|
keertika27
| 0
| 10
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,673
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2152519/Python-or-Very-Easy-elegant-and-intuitive-approach-using-cashing
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
table = {'a':0,'b':1,'c':2,'d':3,'e':4,'f':5,'g':6,'h':7,'i':8,'j':9}
num1 = 0
for i in firstWord:
num1 = num1 * 10 + table[i]
# print(num1)
num2 = 0
for i in secondWord:
num2 = num2 * 10 + table[i]
# print(num2)
target = 0
for i in targetWord:
target = target * 10 + table[i]
if num1 + num2 == target:
return True
return False
|
check-if-word-equals-summation-of-two-words
|
Python | Very Easy, elegant and intuitive approach using cashing
|
__Asrar
| 0
| 30
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,674
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/2045091/Python-simple-solution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
def convert_word(word):
from string import ascii_lowercase as al
converted = ''
for i in word:
converted += str(al.index(i))
return int(converted)
return convert_word(firstWord) + convert_word(secondWord) == convert_word(targetWord)
|
check-if-word-equals-summation-of-two-words
|
Python simple solution
|
StikS32
| 0
| 24
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,675
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1942280/Python-Clean-and-Simple-%2B-One-Liner
|
class Solution:
def isSumEqual(self, w1, w2, target):
def val(w): return int("".join([str(ord(c)-ord("a")) for c in w]))
return val(w1)+val(w2) == val(target)
|
check-if-word-equals-summation-of-two-words
|
Python - Clean and Simple + One-Liner
|
domthedeveloper
| 0
| 41
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,676
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1942280/Python-Clean-and-Simple-%2B-One-Liner
|
class Solution:
def isSumEqual(self, w1, w2, target):
return (lambda val : val(w1)+val(w2) == val(target))(lambda w : int("".join([str(ord(c)-ord("a")) for c in w])))
|
check-if-word-equals-summation-of-two-words
|
Python - Clean and Simple + One-Liner
|
domthedeveloper
| 0
| 41
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,677
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1887290/Python-one-liner-with-memory-usage-less-than-82
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return int(''.join(str(ord(x)-97) for x in firstWord)) + int(''.join(str(ord(x)-97) for x in secondWord)) == int(''.join(str(ord(x)-97) for x in targetWord))
|
check-if-word-equals-summation-of-two-words
|
Python one liner with memory usage less than 82%
|
alishak1999
| 0
| 39
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,678
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1868904/Python-dollarolution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
s, count = '', 0
for i in firstWord:
s += str(ord(i)-97)
count += int(s)
s = ''
for i in secondWord:
s += str(ord(i)-97)
count += int(s)
s = ''
for i in targetWord:
s += str(ord(i)-97)
return count == int(s)
|
check-if-word-equals-summation-of-two-words
|
Python $olution
|
AakRay
| 0
| 33
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,679
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1849669/python3orsimpleorshortorO(N)or
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
int_str = lambda s: int(''.join([str(ord(letter) - ord('a')) for letter in s]))
return int_str(firstWord) + int_str(secondWord) == int_str(targetWord)
|
check-if-word-equals-summation-of-two-words
|
python3|simple|short|O(N)|🐍🐍
|
YaBhiThikHai
| 0
| 29
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,680
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1846959/Easiest-Python3-Solution-oror-100-Faster-oror-Easy-to-understand-Python-code
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
x=0
y=0
z=0
valueAlpha=0
alpha=""
for i in firstWord:
alpha=i
valueAlpha=ord(alpha)-97
x=x*10+valueAlpha
for i in secondWord:
alpha=i
valueAlpha=ord(alpha)-97
y=y*10+valueAlpha
for i in targetWord:
alpha=i
valueAlpha=ord(alpha)-97
z=z*10+valueAlpha
return x+y==z
|
check-if-word-equals-summation-of-two-words
|
Easiest Python3 Solution || 100% Faster || Easy to understand Python code
|
RatnaPriya
| 0
| 36
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,681
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1718566/python3-or-one-line-solution-or-faster-than-71.43
|
class Solution:
def generate_num(self, words) -> int:
return int("".join([str(ord(l) % 97) for l in words]))
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return (self.generate_num(firstWord) + self.generate_num(secondWord)) == self.generate_num(targetWord)
|
check-if-word-equals-summation-of-two-words
|
python3 | one line solution | faster than 71.43%
|
khalidhassan3011
| 0
| 43
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,682
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1534691/String-translate-method-90-speed
|
class Solution:
d = {97 + i: 48 + i for i in range(10)}
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return (int(firstWord.translate(Solution.d)) +
int(secondWord.translate(Solution.d)) ==
int(targetWord.translate(Solution.d)))
|
check-if-word-equals-summation-of-two-words
|
String translate method, 90% speed
|
EvgenySH
| 0
| 65
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,683
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1503964/3-line-solution-wo-explicit-type-casting-in-Python
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
a = ord('a')
to_num = lambda s: sum((ord(c) - a) * (10 ** i) for i, c in enumerate(reversed(s)))
return sum(map(to_num, (firstWord, secondWord))) == to_num(targetWord)
|
check-if-word-equals-summation-of-two-words
|
3-line solution w/o explicit type-casting in Python
|
mousun224
| 0
| 53
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,684
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1496330/Python-Unicode
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
first, second, target_tot = 0, 0, 0
for fw in firstWord:
first = first * 10 + (ord(fw) - ord('a'))
for sw in secondWord:
second = second * 10 + (ord(sw) - ord('a'))
for tw in targetWord:
target_tot = target_tot * 10 + (ord(tw) - ord('a'))
return first + second == target_tot
|
check-if-word-equals-summation-of-two-words
|
Python Unicode
|
jlee9077
| 0
| 60
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,685
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1406726/Python3-Faster-then-84-of-the-Solutions
|
class Solution:
def convert(self,s):
x = ''
for i in s:
x = x + str(ord(i)-97)
return int(x)
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return self.convert(firstWord)+self.convert(secondWord)==self.convert(targetWord)
|
check-if-word-equals-summation-of-two-words
|
Python3 - Faster then 84% of the Solutions
|
harshitgupta323
| 0
| 78
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,686
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1302642/Python3-Simple-solution-beats-97-of-the-submissions
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
charMap = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4 , 'f': 5, 'g': 6, 'h': 7, 'i':8, 'j': 9}
firstWordMap = ''.join([ str(charMap[x]) for x in firstWord ])
secondWordMap = ''.join( [ str(charMap[x]) for x in secondWord ] )
targetWordMap = ''.join( [ str(charMap[x]) for x in targetWord ] )
return bool(int(firstWordMap) + int(secondWordMap) == int(targetWordMap))
|
check-if-word-equals-summation-of-two-words
|
[Python3] Simple solution, beats 97% of the submissions
|
GauravKK08
| 0
| 38
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,687
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1281996/Python-Simple
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return self.word_to_int(firstWord) + self.word_to_int(secondWord) == self.word_to_int(targetWord)
def word_to_int(self, s):
return int(''.join(str(ord(char) - 97)
for char in s))
|
check-if-word-equals-summation-of-two-words
|
Python Simple
|
peatear-anthony
| 0
| 72
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,688
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1275996/Python-3-one-liner!-Use-ord-function(28-ms-faster-than-91.33)
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
return (int("".join([str(ord(i)-97) for i in firstWord]))+int("".join([str(ord(i)-97) for i in secondWord]))) == int("".join([str(ord(i)-97) for i in targetWord]))
|
check-if-word-equals-summation-of-two-words
|
Python 3 one liner! Use ord function(28 ms, faster than 91.33%)
|
lin11116459
| 0
| 45
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,689
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1275996/Python-3-one-liner!-Use-ord-function(28-ms-faster-than-91.33)
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
a=int("".join([str(ord(i)-97) for i in firstWord]))
b=int("".join([str(ord(i)-97) for i in secondWord]))
c=int("".join([str(ord(i)-97) for i in targetWord]))
return (a+b) == c
|
check-if-word-equals-summation-of-two-words
|
Python 3 one liner! Use ord function(28 ms, faster than 91.33%)
|
lin11116459
| 0
| 45
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,690
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1266966/Easy-and-Efficient-Python-Solution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
li = [chr(i) for i in range(97, 107)]
def calculate(word):
ans = ''
for i in word:
ans += str(li.index(i))
return int(ans)
sum1 = calculate(firstWord)
sum2 = calculate(secondWord)
sum3 = calculate(targetWord)
return sum1+sum2 == sum3
|
check-if-word-equals-summation-of-two-words
|
Easy & Efficient Python Solution
|
sanu2sid
| 0
| 81
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,691
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1265351/or-Python-or-Solution-using-String-Methods
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
words = [firstWord, secondWord, targetWord]
for i in range(len(words)):
convert = words[i].maketrans('abcdefghij', '0123456789')
words[i] = words[i].translate(convert)
return int(words[0])+int(words[1]) == int(words[2])
|
check-if-word-equals-summation-of-two-words
|
| Python | Solution using String Methods
|
bruadarach
| 0
| 46
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,692
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1260062/Python-Solutions-and-Explanations-Noobie-version-Equivalent-Lambda-Version-and-extra-thoughts.
|
class Solution(object):
def getWordValue(self, word):
""" Return a new string such that each character is represented by it's numeric counter described in the problem statement"""
newWord=''
for letter in word:
newWord+=(str(ord(letter)-97))
return newWord
def isSumEqual(self, firstWord, secondWord, targetWord):
"""
:type firstWord: str
:type secondWord: str
:type targetWord: str
:rtype: bool
"""
# 1. Get the sum of the firstWord & secondWord
sumValue = int(self.getWordValue(firstWord))+int(self.getWordValue(secondWord))
# 2. Get the value of the targetWord
targetValue = int(self.getWordValue(targetWord))
# 3. return comparison boolean
return sumValue == targetValue
|
check-if-word-equals-summation-of-two-words
|
Python Solutions & Explanations - Noobie version, Equivalent Lambda Version & extra thoughts.
|
antonmdv
| 0
| 48
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,693
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1260062/Python-Solutions-and-Explanations-Noobie-version-Equivalent-Lambda-Version-and-extra-thoughts.
|
class Solution(object):
def isSumEqual(self, firstWord, secondWord, targetWord):
"""
:type firstWord: str
:type secondWord: str
:type targetWord: str
:rtype: bool
"""
numeric_value = lambda s: int(''.join([str(ord(letter)-97) for letter in s]))
return numeric_value(firstWord) + numeric_value(secondWord) == numeric_value(targetWord)
|
check-if-word-equals-summation-of-two-words
|
Python Solutions & Explanations - Noobie version, Equivalent Lambda Version & extra thoughts.
|
antonmdv
| 0
| 48
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,694
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1259787/python-3-%3A-super-easy-solution-%3A-)
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
f = "".join([str(ord(i)-97) for i in firstWord ]) # ord("a") = 97
s = "".join([str(ord(i)-97) for i in secondWord ])
t = "".join([str(ord(i)-97) for i in targetWord ])
return int(f) + int(s) == int(t)
|
check-if-word-equals-summation-of-two-words
|
python 3 : super easy solution : )
|
rohitkhairnar
| 0
| 118
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,695
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1255097/python-3-using-ord()
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
def val(s):
res=""
for i in range(len(s)):
res+=str(ord(s[i])-ord('a'))
return int(res)
return (val(targetWord)==val(firstWord)+val(secondWord))
|
check-if-word-equals-summation-of-two-words
|
python 3 using ord()
|
janhaviborde23
| 0
| 86
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,696
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1242418/Python3-easy-solution
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
s = "abcdefghij"
x1 = ''
x2 = ''
x3 = ''
for i in firstWord:
x1 += str(s.index(i))
x1 = int(x1)
for i in secondWord:
x2 += str(s.index(i))
x2 = int(x2)
for i in targetWord:
x3 += str(s.index(i))
x3 = int(x3)
if x1+x2==x3:a
return True
else:
return False
|
check-if-word-equals-summation-of-two-words
|
Python3 easy solution
|
Sanyamx1x
| 0
| 35
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,697
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1242016/Python-simple-or-Hashmap
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
memo = {'a':'0', 'b':'1', 'c':'2', 'd':'3','e':'4', 'f':'5', 'g':'6', 'h':'7', 'i':'8','j':'9'}
def manipulate(word, value):
for i in word:
value += memo[i]
return int(value)
return manipulate(firstWord, "") + manipulate(secondWord,"") == manipulate(targetWord,"")
|
check-if-word-equals-summation-of-two-words
|
Python simple | Hashmap
|
dee7
| 0
| 50
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,698
|
https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/discuss/1241607/PythonPython3-solution-with-explanation
|
class Solution:
def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
alpha = 'abcdefghijklmnopqrstuvwxyz'
firstNum,secondNum,targetNum = '','',''
for i in firstWord:
firstNum += str(alpha.index(i))
for i in secondWord:
secondNum += str(alpha.index(i))
for i in targetWord:
targetNum += str(alpha.index(i))
#print(firstNum,secondNum,targetNum)
return int(firstNum)+int(secondNum) == int(targetNum)
|
check-if-word-equals-summation-of-two-words
|
Python/Python3 solution with explanation
|
prasanthksp1009
| 0
| 74
|
check if word equals summation of two words
| 1,880
| 0.738
|
Easy
| 26,699
|
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.