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https://leetcode.com/problems/arithmetic-subarrays/discuss/1406723/Python3-solution-clean-code
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: def arithmetic(arr): # function to check if the sequence is arithmetic for i in range(len(arr)): if i <= len(arr)-2 and i >= 1: if arr[i+1] -...
arithmetic-subarrays
Python3 solution clean code
FlorinnC1
0
52
arithmetic subarrays
1,630
0.8
Medium
23,700
https://leetcode.com/problems/arithmetic-subarrays/discuss/1370794/python3-solution-using-formula
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: visited = [] for start, end in zip(l, r): seq = nums[start:end+1] # using formula for sum of arithmetic sequence form = (max(seq) + min(seq)) * len(seq)...
arithmetic-subarrays
python3 solution using formula
G0udini
0
53
arithmetic subarrays
1,630
0.8
Medium
23,701
https://leetcode.com/problems/arithmetic-subarrays/discuss/1174539/python-3-solution
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: temp=[] for i in range(len(l)): p=nums[l[i]:r[i]+1] p.sort() flag=True for i in range(2,len(p)): if p[i-1]-p[i-2]!=p[i]-p[i-1...
arithmetic-subarrays
python 3 solution
janhaviborde23
0
93
arithmetic subarrays
1,630
0.8
Medium
23,702
https://leetcode.com/problems/arithmetic-subarrays/discuss/1110072/Python-or-Simple-Clean-Solution
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: def check(arr): diff=arr[1]-arr[0] for i in range(len(arr)-1): if arr[i+1]-arr[i]!=diff: return False return True res...
arithmetic-subarrays
Python | Simple Clean Solution
rajatrai1206
0
70
arithmetic subarrays
1,630
0.8
Medium
23,703
https://leetcode.com/problems/arithmetic-subarrays/discuss/948654/Intuitive-approach
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: def is_a_arithmetic_subarray(a, b): sorted_nums = sorted(nums[a:b+1]) p = sorted_nums[1] - sorted_nums[0] for i in range(2, len(sorted_nums)): if...
arithmetic-subarrays
Intuitive approach
puremonkey2001
0
41
arithmetic subarrays
1,630
0.8
Medium
23,704
https://leetcode.com/problems/arithmetic-subarrays/discuss/909086/Python3-5-line-sorting
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: ans = [] for ll, rr in zip(l, r): seq = sorted(nums[ll:rr+1]) ans.append(len(set(seq[i] - seq[i-1] for i in range(1, len(seq)))) == 1) return ans
arithmetic-subarrays
[Python3] 5-line sorting
ye15
0
150
arithmetic subarrays
1,630
0.8
Medium
23,705
https://leetcode.com/problems/arithmetic-subarrays/discuss/909086/Python3-5-line-sorting
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: def fn(seq): """Return True if seq is an arithmetic subarray.""" mn, mx = min(seq), max(seq) if mn == mx: return True # edge case step = (...
arithmetic-subarrays
[Python3] 5-line sorting
ye15
0
150
arithmetic subarrays
1,630
0.8
Medium
23,706
https://leetcode.com/problems/arithmetic-subarrays/discuss/1393675/Python3-Easy-sort-and-set-solutions-beat-90%2B
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: ''' Time: O(L*nlogn * n) ''' def check(arr): n = len(arr) if n < 2: return False d = arr[1] - arr[0] for i in...
arithmetic-subarrays
[Python3] Easy sort and set solutions beat 90%+
nightybear
-1
52
arithmetic subarrays
1,630
0.8
Medium
23,707
https://leetcode.com/problems/arithmetic-subarrays/discuss/1393675/Python3-Easy-sort-and-set-solutions-beat-90%2B
class Solution: def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: ''' Time: O(L*n) ''' def check(arr): min_val, max_val, set_val = min(arr), max(arr), set(arr) if len(arr) != len(set_val): r...
arithmetic-subarrays
[Python3] Easy sort and set solutions beat 90%+
nightybear
-1
52
arithmetic subarrays
1,630
0.8
Medium
23,708
https://leetcode.com/problems/arithmetic-subarrays/discuss/1284752/Python-Easy-Solution-Faster-Than-99.02
class Solution: def checker(self,nums): i = 2 n = len(nums) diff = nums[1] - nums[0] while i < n: if nums[i] - nums[i-1] != diff: return False i += 1 return True def checkArithmeticSubarrays(self, nums: List[int], left: Li...
arithmetic-subarrays
Python Easy Solution Faster Than 99.02%
paramvs8
-1
55
arithmetic subarrays
1,630
0.8
Medium
23,709
https://leetcode.com/problems/path-with-minimum-effort/discuss/909094/Python3-bfs-and-Dijkstra
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: m, n = len(heights), len(heights[0]) queue = {(0, 0): 0} # (0, 0) maximum height so far seen = {(0, 0): 0} # (i, j) -> heights ans = inf while queue: newq = {} # ne...
path-with-minimum-effort
[Python3] bfs & Dijkstra
ye15
2
219
path with minimum effort
1,631
0.554
Medium
23,710
https://leetcode.com/problems/path-with-minimum-effort/discuss/909094/Python3-bfs-and-Dijkstra
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: m, n = len(heights), len(heights[0]) # dimensions seen = {(0, 0): 0} pq = [(0, 0, 0)] while pq: h, i, j = heappop(pq) if i == m-1 and j == n-1: return h for ii, jj in (...
path-with-minimum-effort
[Python3] bfs & Dijkstra
ye15
2
219
path with minimum effort
1,631
0.554
Medium
23,711
https://leetcode.com/problems/path-with-minimum-effort/discuss/909094/Python3-bfs-and-Dijkstra
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: m, n = len(heights), len(heights[0]) def fn(x): """Return True if given effort is enough.""" stack = [(0, 0)] # i|j|h starting from top-left seen = set() while sta...
path-with-minimum-effort
[Python3] bfs & Dijkstra
ye15
2
219
path with minimum effort
1,631
0.554
Medium
23,712
https://leetcode.com/problems/path-with-minimum-effort/discuss/1989967/Python-3-easy-Solution-oror-Faster-than-98.5-oror-Min-Heap-oror-O(mn.log(mn))
class Solution: def minimumEffortPath(self, grid: List[List[int]]) -> int: # Base case if len(grid)==1 and len(grid[0])==1: return 0 row , col = len(grid) , len(grid[0]) visited = set() # set : store indices (i,j) which we have visited already before reaching de...
path-with-minimum-effort
Python 3 easy Solution || Faster than 98.5% || Min-Heap || O(mn.log(mn))
Laxman_Singh_Saini
1
63
path with minimum effort
1,631
0.554
Medium
23,713
https://leetcode.com/problems/path-with-minimum-effort/discuss/1693676/Python-two-methods%3A-union-find-and-DFS
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: m,n = len(heights),len(heights[0]) parent = list(range(m*n)) rank = [1 for _ in range(m*n)] def find(node): if parent[node]!=node: parent[node] = find(parent[node]) r...
path-with-minimum-effort
Python two methods: union find & DFS
1579901970cg
1
253
path with minimum effort
1,631
0.554
Medium
23,714
https://leetcode.com/problems/path-with-minimum-effort/discuss/1693676/Python-two-methods%3A-union-find-and-DFS
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: lo,hi = 0,10**6-1 m,n = len(heights),len(heights[0]) def dfs(i,j,pre,target): if (i,j) in visited: return False if i<0 or j<0 or i>m-1 or j>n-1 or abs(heights[i][j]-pre)>targ...
path-with-minimum-effort
Python two methods: union find & DFS
1579901970cg
1
253
path with minimum effort
1,631
0.554
Medium
23,715
https://leetcode.com/problems/path-with-minimum-effort/discuss/2808461/Python-3-heap-(Priority-Queue)-simple-solution-with-comments
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: visited = set() h = [] drt = [[-1, 0], [0, -1], [1, 0], [0, 1]] max_diff = 0 rows_n = len(heights) cols_n = len(heights[0]) heappush(h,...
path-with-minimum-effort
Python 3 - heap (Priority Queue) - simple solution - with comments
noob_in_prog
0
4
path with minimum effort
1,631
0.554
Medium
23,716
https://leetcode.com/problems/path-with-minimum-effort/discuss/2708595/Djikstra-simple-BFS
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: n,m = len(heights),len(heights[0]) directions = [(0,1),(1,0),(-1,0),(0,-1)] effort = [[float("inf")]*m for _ in range(n)] effort[0][0] = 0 minHeap = [] heappush(minHeap,(0,0,0)) ...
path-with-minimum-effort
Djikstra simple BFS
shriyansnaik
0
8
path with minimum effort
1,631
0.554
Medium
23,717
https://leetcode.com/problems/path-with-minimum-effort/discuss/2653509/Dijkstra-and-binary-search
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: nrows, ncols = len(heights), len(heights[0]) dp = [] dp2 = [] for _ in range(nrows): dp.append([float("inf")] * ncols) dp2.append([float("inf")] * ncols) dp[0][0] = 0...
path-with-minimum-effort
Dijkstra and binary search
sticky_bits
0
6
path with minimum effort
1,631
0.554
Medium
23,718
https://leetcode.com/problems/path-with-minimum-effort/discuss/2653509/Dijkstra-and-binary-search
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: nrows, ncols = len(heights), len(heights[0]) def calcEffort(r, c, r2, c2): return abs(heights[r][c]-heights[r2][c2]) left, right = 0, 10000000 visited = set() def dfs(eff, r...
path-with-minimum-effort
Dijkstra and binary search
sticky_bits
0
6
path with minimum effort
1,631
0.554
Medium
23,719
https://leetcode.com/problems/path-with-minimum-effort/discuss/2614714/Python3-or-Why-Am-I-getting-TLE-With-DFS-and-Binary-Search
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: m, n = len(heights), len(heights[0]) #To minimize effort, we can consider full range of possible effort values along any path! #Minimum would be 0 for route where all entries have same height and max would be maxim...
path-with-minimum-effort
Python3 | Why Am I getting TLE With DFS and Binary Search?
JOON1234
0
13
path with minimum effort
1,631
0.554
Medium
23,720
https://leetcode.com/problems/path-with-minimum-effort/discuss/2321793/Python3-or-BinarySearch-%2B-BFS
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: l,h=0,1e6+1 while l<h: mid=(l+h)//2 if self.isPossible(heights,mid): h=mid else: l=mid+1 return int(h) def isPossible(self,heights,limit): ...
path-with-minimum-effort
[Python3] | BinarySearch + BFS
swapnilsingh421
0
50
path with minimum effort
1,631
0.554
Medium
23,721
https://leetcode.com/problems/path-with-minimum-effort/discuss/1991399/ororPYTHON-SOL-oror-WELL-COMMENTED-oror-WELL-EXPLAINED-oror-DIJKSTRA'S-SOL-oror-EASY-TO-READ-oror
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: # implementing dijkstra's algorithm rows,cols = len(heights),len(heights[0]) # first mark every place unvisited vis = [[False for col in range(cols)] for row in range(rows)] # make minimum effort to...
path-with-minimum-effort
||PYTHON SOL || WELL COMMENTED || WELL EXPLAINED || DIJKSTRA'S SOL || EASY TO READ ||
reaper_27
0
60
path with minimum effort
1,631
0.554
Medium
23,722
https://leetcode.com/problems/path-with-minimum-effort/discuss/1989830/Python3-Solution-with-using-dfs-and-binary-search
class Solution: def can_reach_dest(self, x, y, mid, lrow, lcol, heights,visited): if x == lrow - 1 and y == lcol - 1: # target cell return True visited[x][y] = True # dfs part for dx, dy in [[0,1], [1,0], [0,-1], [-1, 0]]: _x = x + dx ...
path-with-minimum-effort
[Python3] Solution with using dfs and binary search
maosipov11
0
30
path with minimum effort
1,631
0.554
Medium
23,723
https://leetcode.com/problems/path-with-minimum-effort/discuss/1988810/Python
class Solution: def minimumEffortPath(self, hs: List[List[int]]) -> int: rs, cs = len(hs), len(hs[0]) # format: (max abs difference, row, col) heap = [(0, 0, 0)] # dict stores the maximum abs difference seen up to (row, col) - default +inf seen = defaultdict(lambda: ...
path-with-minimum-effort
Python
captainspongebob1
0
20
path with minimum effort
1,631
0.554
Medium
23,724
https://leetcode.com/problems/path-with-minimum-effort/discuss/1987784/Python3-solution
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: # return True if it's possible to go from (0,0) to (m-1, n-1) # using only edges of k or less cost. def is_there_a_path(k: int) -> bool: def dfs(x, y): if (x, y) == (m-1, n-1): ...
path-with-minimum-effort
Python3 solution
dalechoi
0
69
path with minimum effort
1,631
0.554
Medium
23,725
https://leetcode.com/problems/path-with-minimum-effort/discuss/1567897/Python3-Short-Dijkstra's
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: h = [(0, 0, 0, 0)] visited = set() while h: weight, x, y, max_diff = heappop(h) # weight is the transition weight, x y are the coordinates, and max_diff is the maximum diff on the path from 0 0 to the ...
path-with-minimum-effort
Python3 Short Dijkstra's
danwuSBU
0
147
path with minimum effort
1,631
0.554
Medium
23,726
https://leetcode.com/problems/path-with-minimum-effort/discuss/1096787/Python-AC
class Solution: def minimumEffortPath(self, heights: List[List[int]]) -> int: M, N = map(len, (heights, heights[0])) heap = [(0, 0, 0)] seen = set() result = 0 while heap: # Pop effort, i, j = heapq.heappop(heap) ...
path-with-minimum-effort
Python AC
dev-josh
0
192
path with minimum effort
1,631
0.554
Medium
23,727
https://leetcode.com/problems/rank-transform-of-a-matrix/discuss/913790/Python3-UF
class Solution: def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]: m, n = len(matrix), len(matrix[0]) # dimension # mapping from value to index mp = {} for i in range(m): for j in range(n): mp.setdefault(matrix[i][j], []).append...
rank-transform-of-a-matrix
[Python3] UF
ye15
5
465
rank transform of a matrix
1,632
0.41
Hard
23,728
https://leetcode.com/problems/rank-transform-of-a-matrix/discuss/2476234/Python-Union-Find-%2B-Graph-%2B-Topological-Sort-%2B-BFS-Queue-heavily-commented
class Solution: def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]: m = len(matrix) n = len(matrix[0]) def find_root(x: int, y: int): if parent[x][y] == (x, y): return (x, y) else: r = find_root(parent[x]...
rank-transform-of-a-matrix
[Python] Union Find + Graph + Topological Sort + BFS Queue, heavily commented
bbshark
0
77
rank transform of a matrix
1,632
0.41
Hard
23,729
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/963292/Python-1-liner
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: return sorted(sorted(nums,reverse=1),key=nums.count)
sort-array-by-increasing-frequency
Python 1-liner
lokeshsenthilkumar
22
1,500
sort array by increasing frequency
1,636
0.687
Easy
23,730
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1446521/Using-Counter-95-speed
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: lst = sorted([(key, freq) for key, freq in Counter(nums).items()], key=lambda tpl: (tpl[1], -tpl[0])) ans = [] for key, freq in lst: ans += [key] * freq return ans
sort-array-by-increasing-frequency
Using Counter, 95% speed
EvgenySH
3
478
sort array by increasing frequency
1,636
0.687
Easy
23,731
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1870522/Python-Clean-and-Concise!-Multiple-Solutions!-One-Liners!
class Solution: def frequencySort(self, nums): c = Counter(nums) nums.sort(reverse=True) nums.sort(key=lambda x: c[x]) return nums
sort-array-by-increasing-frequency
Python - Clean and Concise! Multiple Solutions! One-Liners!
domthedeveloper
2
388
sort array by increasing frequency
1,636
0.687
Easy
23,732
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1870522/Python-Clean-and-Concise!-Multiple-Solutions!-One-Liners!
class Solution: def frequencySort(self, nums): return (lambda c : sorted(sorted(nums, reverse=True), key=lambda x: c[x]))(Counter(nums))
sort-array-by-increasing-frequency
Python - Clean and Concise! Multiple Solutions! One-Liners!
domthedeveloper
2
388
sort array by increasing frequency
1,636
0.687
Easy
23,733
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1870522/Python-Clean-and-Concise!-Multiple-Solutions!-One-Liners!
class Solution: def frequencySort(self, nums): c = Counter(nums) nums.sort(key=lambda x: (c[x], -x)) return nums
sort-array-by-increasing-frequency
Python - Clean and Concise! Multiple Solutions! One-Liners!
domthedeveloper
2
388
sort array by increasing frequency
1,636
0.687
Easy
23,734
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1870522/Python-Clean-and-Concise!-Multiple-Solutions!-One-Liners!
class Solution: def frequencySort(self, nums): return (lambda c : sorted(nums, key=lambda x: (c[x], -x)))(Counter(nums))
sort-array-by-increasing-frequency
Python - Clean and Concise! Multiple Solutions! One-Liners!
domthedeveloper
2
388
sort array by increasing frequency
1,636
0.687
Easy
23,735
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1851859/simple-python-dictionary
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: d = defaultdict(int) for i in nums: d[i] += 1 output = [] for i in sorted(d.items(), key=lambda kv: (kv[1],-kv[0])): output.extend([i[0]]*i[1]) return output
sort-array-by-increasing-frequency
simple python dictionary
gasohel336
2
281
sort array by increasing frequency
1,636
0.687
Easy
23,736
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1411793/Python3-Faster-than-94.95-of-the-Solutions
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: count = {} for num in nums: if num not in count: count[num] = 1 else: count[num] += 1 l = [(k,v) for k,v in count.items()] l.sort(key = lambda x:x[0], revers...
sort-array-by-increasing-frequency
Python3 - Faster than 94.95% of the Solutions
harshitgupta323
2
162
sort array by increasing frequency
1,636
0.687
Easy
23,737
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1999997/easy-commented-python-code
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: d = {} count = 0 output = [] nums.sort() # create a dictonary key = frequency and value = list of all numbers with that freq. for i in set(nums): # print(d) if nums.count(i)...
sort-array-by-increasing-frequency
easy commented python code
dakash682
1
279
sort array by increasing frequency
1,636
0.687
Easy
23,738
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1605085/Python3-Sorting-with-Counter
class Solution: def frequencySort(self,nums): res=[] for item,count in Counter(sorted(nums)).most_common()[::-1]: res+=[item]*count return res
sort-array-by-increasing-frequency
Python3 Sorting with Counter
description
1
336
sort array by increasing frequency
1,636
0.687
Easy
23,739
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1582660/Python-Easy-to-Understand-Counter-Beats-95-Runtime
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: counter = Counter(nums) sorted_count = [[key] * count for key, count in sorted(counter.items(), key=lambda kv: (kv[1], -kv[0]))] return itertools.chain(*sorted_count)
sort-array-by-increasing-frequency
[Python] Easy to Understand - Counter - Beats 95% Runtime
matthewaj
1
599
sort array by increasing frequency
1,636
0.687
Easy
23,740
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2811224/Easy-To-Understand-or-Beginner-Level-Code-or-Python
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: ans = list() r = Counter(nums) r = Counter(nums).most_common() def sorting(x): return x[0] def sorting1(y): return y[1] r.sort(key=sorting, reverse=True) r.sort(k...
sort-array-by-increasing-frequency
Easy To Understand | Beginner Level Code | Python
beingdillig
0
4
sort array by increasing frequency
1,636
0.687
Easy
23,741
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2775514/PYTHON-1-LINER
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: return sorted(sorted(nums,reverse=True),key=lambda x:nums.count(x))
sort-array-by-increasing-frequency
PYTHON 1-LINER 🤞
ganesh_5I4
0
7
sort array by increasing frequency
1,636
0.687
Easy
23,742
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2774034/Python-1-line-solution.-But-can-we-use-python-built-in-functions
class Solution(object): def frequencySort(self, nums): """ :type nums: List[int] :rtype: List[int] """ return sorted(nums, key=lambda x:(nums.count(x), -x))
sort-array-by-increasing-frequency
Python 1 line solution. But can we use python built-in functions???
csgogogo9527
0
6
sort array by increasing frequency
1,636
0.687
Easy
23,743
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2767841/Python-easy-code-solution-for-beginners.
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: nums.sort() value = [] ans = [] num = nums[::-1] c = Counter(num) for i in c: value.append(c[i]) value.sort() for i in value: for key, value in c.items(): ...
sort-array-by-increasing-frequency
Python easy code solution for beginners.
anshu71
0
19
sort array by increasing frequency
1,636
0.687
Easy
23,744
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2746649/O(nlogn)-time-or-O(n)-space
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: d = {} for i in nums: d[i]=d.get(i, 0)+1 nums.sort(key = lambda x:(d[x], -x)) return nums
sort-array-by-increasing-frequency
O(nlogn) time | O(n) space
wakadoodle
0
11
sort array by increasing frequency
1,636
0.687
Easy
23,745
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2674704/Easy-Python-Solution-Using-Dictionary
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: dic={} for i in nums: if i not in dic: dic[i]=1 else: dic[i]+=1 arr=nums arr=sorted(arr,key=lambda x:(dic[x],-x)) return arr
sort-array-by-increasing-frequency
Easy Python Solution Using Dictionary
ankitr8055
0
9
sort array by increasing frequency
1,636
0.687
Easy
23,746
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2454721/python3-using-dictionary-with-set-faster-than-84
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: ans=[] from collections import Counter counts=Counter(nums) for v in sorted(set(counts.values())): temp_k=[] for k in counts.keys(): if counts[k]==v: ...
sort-array-by-increasing-frequency
[python3] using dictionary with set, faster than 84%
hhlinwork
0
74
sort array by increasing frequency
1,636
0.687
Easy
23,747
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/2379861/Sort-Array-by-Increasing-Frequency
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: x = {} for i in nums: if i in x : x[i]+=1 else: x[i]=1 def swap(i,j): nums[i],nums[j]= nums[j],nums[i] for i in range(len(nums)): for...
sort-array-by-increasing-frequency
Sort Array by Increasing Frequency
dhananjayaduttmishra
0
79
sort array by increasing frequency
1,636
0.687
Easy
23,748
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1862050/Python-(Simple-Approach-and-Beginner-Friendly)
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: res = {} result = [] for i in sorted(set(nums)): res[i] = nums.count(i) print(res) for i in (reversed(sorted(res, reverse = True, key = res.get))): for _ in range(nums.count(i)): ...
sort-array-by-increasing-frequency
Python (Simple Approach and Beginner-Friendly)
vishvavariya
0
261
sort array by increasing frequency
1,636
0.687
Easy
23,749
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1818670/2-Lines-Python-Solution-oror-70-Faster-oror-Memory-less-than-98
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: return list(chain(*[[idx]*val for idx,val in sorted(Counter(nums).items(),key=lambda x:(x[1],-x[0]))]))
sort-array-by-increasing-frequency
2-Lines Python Solution || 70% Faster || Memory less than 98%
Taha-C
0
191
sort array by increasing frequency
1,636
0.687
Easy
23,750
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1591936/Python-3-Counter-solution
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: c = collections.Counter(nums) return sorted(nums, key=lambda num: (c[num], -num))
sort-array-by-increasing-frequency
Python 3 Counter solution
dereky4
0
664
sort array by increasing frequency
1,636
0.687
Easy
23,751
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1571984/python-ordereddict-custom-sort
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: dic={} for i in nums: dic[i]=dic.get(i,0)+1 ord_dic = OrderedDict(sorted(dic.items(), key=lambda t: t[0],reverse=True)) _ord = OrderedDict(sorted(ord_dic.items(),key=lambda t: t[1])) idx=0 ...
sort-array-by-increasing-frequency
python ordereddict custom sort
rashmirashmitha32
0
241
sort array by increasing frequency
1,636
0.687
Easy
23,752
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1555503/Python3-Solution-with-using-sorting
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: c = collections.Counter(nums) return sorted(nums, key=lambda val: (c[val], -val))
sort-array-by-increasing-frequency
[Python3] Solution with using sorting
maosipov11
0
191
sort array by increasing frequency
1,636
0.687
Easy
23,753
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1542633/Python-solution-48-ms-90-better-runtime-and-95-better-memory-usage
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: num_count = defaultdict(int) for num in nums: num_count[num] += 1 count_nums = defaultdict(set) for num, count in num_count.items(): count_nums[count].add(num) output = [] f...
sort-array-by-increasing-frequency
Python solution 48 ms 90% better runtime and 95% better memory usage
akshaykumar19002
0
336
sort array by increasing frequency
1,636
0.687
Easy
23,754
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1537740/Python-Easy-Solution
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: nums.sort() c = collections.Counter(nums).most_common(len(nums)) res = [] while c: l = c[-1][1] n = c[-1][0] for i in range(l): res.ap...
sort-array-by-increasing-frequency
Python Easy Solution
aaffriya
0
352
sort array by increasing frequency
1,636
0.687
Easy
23,755
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1053368/Faster-then-85-of-Solution-and-less-than-58.42-memory-usage
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: #nums = [-1,1,-6,4,5,-6,1,4,1] dict2 = self.dict1(nums) keys = list(self.dict1(nums).keys()) revers = {} for i in range(0,len(keys)): if dict2[keys[i]] not in revers : revers[di...
sort-array-by-increasing-frequency
Faster then 85% of Solution and less than 58.42% memory usage
xevb
0
122
sort array by increasing frequency
1,636
0.687
Easy
23,756
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1051038/Python3-simple-solution-using-dictionary
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: nums = sorted(nums,reverse = True) d = {} x = [] for i in nums: d[i] = d.get(i,0) + 1 d = {i:j for i,j in sorted(d.items(),key=lambda x: x[1])} for i,j in d.items(): x += [i...
sort-array-by-increasing-frequency
Python3 simple solution using dictionary
EklavyaJoshi
0
134
sort array by increasing frequency
1,636
0.687
Easy
23,757
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1036375/python3-using-OrderedDict
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: d = collections.OrderedDict() nums.sort(reverse=True) for el in nums: if el in d: d[el] += 1 else: d[el] = 1 rez = [] i = 0 dist = len(d) ...
sort-array-by-increasing-frequency
python3 using OrderedDict
dmo2412
0
194
sort array by increasing frequency
1,636
0.687
Easy
23,758
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/1026766/Python3-sorting-O(NlogN)
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: freq = {} for x in nums: freq[x] = 1 + freq.get(x, 0) return sorted(nums, key=lambda x: (freq[x], -x))
sort-array-by-increasing-frequency
[Python3] sorting O(NlogN)
ye15
0
120
sort array by increasing frequency
1,636
0.687
Easy
23,759
https://leetcode.com/problems/sort-array-by-increasing-frequency/discuss/965286/Faster-than-98-Python3
class Solution: def frequencySort(self, nums: List[int]) -> List[int]: counter = [(x[0], len(list(x[1]))) for x in groupby(sorted(nums))] sorted_new_array = sorted(counter, key=lambda x: (x[1], -x[0])) nums = [] for i in sorted_new_array: nums += ([i[0]] * i[1]) r...
sort-array-by-increasing-frequency
Faster than 98% Python3
WiseLin
-1
599
sort array by increasing frequency
1,636
0.687
Easy
23,760
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2806260/Python-or-Easy-Peasy-Code-or-O(n)
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: l = [] for i in points: l.append(i[0]) a = 0 l.sort() for i in range(len(l)-1): if l[i+1] - l[i] > a: a = l[i+1] - l[i] return a
widest-vertical-area-between-two-points-containing-no-points
Python | Easy Peasy Code | O(n)
bhuvneshwar906
0
2
widest vertical area between two points containing no points
1,637
0.842
Medium
23,761
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2789278/Python-1-line-code
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: return max([t - s for s, t in zip(sorted([r[0] for r in points]), sorted([r[0] for r in points])[1:])])
widest-vertical-area-between-two-points-containing-no-points
Python 1 line code
kumar_anand05
0
4
widest vertical area between two points containing no points
1,637
0.842
Medium
23,762
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2778374/Python-easy-O(nlogn)-time-solution
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: points.sort(key = lambda x:x[0]) res = 0 for i in range(len(points)-1): res = max(res, points[i+1][0] - points[i][0]) return res
widest-vertical-area-between-two-points-containing-no-points
Python easy O(nlogn) time solution
byuns9334
0
2
widest vertical area between two points containing no points
1,637
0.842
Medium
23,763
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2698327/Python3-Simple-Solution
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: res = 0 points.sort(key = lambda x: x[0]) prev = points[0][0] for x, y in points[1:]: res = max(res, x - prev) prev = x return res
widest-vertical-area-between-two-points-containing-no-points
Python3 Simple Solution
mediocre-coder
0
3
widest vertical area between two points containing no points
1,637
0.842
Medium
23,764
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2673727/Python3-few-line-solution-memory-beats%3A-91.64
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: xs = sorted({x[0] for x in points}) if len(xs) == 1: return 0 else: return max(xs[i+1]-xs[i] for i in range(0,len(xs)-1))
widest-vertical-area-between-two-points-containing-no-points
Python3 few line solution - memory beats: 91.64%
sipi09
0
5
widest vertical area between two points containing no points
1,637
0.842
Medium
23,765
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2614520/Python-solution-with-sorting-oror-N(logN)-Time-complexity
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: # Sorting points on x co-ordinate points.sort(key = lambda x: x[0]) max_dist = 0 # Get the max value by comparing consecutive values for i in range(1, len(points)): max_dist = max(ma...
widest-vertical-area-between-two-points-containing-no-points
Python solution with sorting || N(logN) Time complexity
vanshika_2507
0
9
widest vertical area between two points containing no points
1,637
0.842
Medium
23,766
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2586554/Python3-or-Step-by-Step-or-O(1)-Space-in-2-lines
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: points.sort(key=lambda a: a[0]) return max((v2[0] - v1[0] for v1, v2 in zip(points, islice(points,1,None))))
widest-vertical-area-between-two-points-containing-no-points
Python3 | Step-by-Step | O(1) Space in 2 lines
dimitars
0
12
widest vertical area between two points containing no points
1,637
0.842
Medium
23,767
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2586554/Python3-or-Step-by-Step-or-O(1)-Space-in-2-lines
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: # Note: # This problem is solved by finding the largest # difference between each consecutive point # in the X axis. # 1. The list is sorted using the first element # of each...
widest-vertical-area-between-two-points-containing-no-points
Python3 | Step-by-Step | O(1) Space in 2 lines
dimitars
0
12
widest vertical area between two points containing no points
1,637
0.842
Medium
23,768
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2407274/Python3-Solution-with-using-sorting
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: points.sort(key=lambda x: x[0]) res = 0 for i in range(len(points) - 1): res = max(res, points[i + 1][0] - points[i][0]) return res
widest-vertical-area-between-two-points-containing-no-points
[Python3] Solution with using sorting
maosipov11
0
12
widest vertical area between two points containing no points
1,637
0.842
Medium
23,769
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2385059/Python3-or-Sort-Only-Using-X-Coords
class Solution: #T.C = O(n + n + n) -> O(n) #S.C = O(n + n) -> O(n) def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: #Approach: Disregard y coordinates! #Record x coordinates of every point in points! #Sort the x coordinates! For every pair of adjacent x coordinate v...
widest-vertical-area-between-two-points-containing-no-points
Python3 | Sort Only Using X Coords
JOON1234
0
4
widest vertical area between two points containing no points
1,637
0.842
Medium
23,770
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/2120855/python-sort-using-lambda
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: points.sort(key = lambda x : x[0]) ans = [] for i in range(len(points)-1): val = abs(points[i][0] - points[i+1][0]) ans.append(val) return max(ans)...
widest-vertical-area-between-two-points-containing-no-points
python sort using lambda
somendrashekhar2199
0
16
widest vertical area between two points containing no points
1,637
0.842
Medium
23,771
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/1573651/python-solution-or-faster-than-94
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: L,max = [],0 for i in range(len(points)): L.append(points[i][0]) L = list(set(L)) L.sort() for i in range(len(L)-1): if L[i+1] - L...
widest-vertical-area-between-two-points-containing-no-points
python solution | faster than 94%
anandanshul001
0
71
widest vertical area between two points containing no points
1,637
0.842
Medium
23,772
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/1573651/python-solution-or-faster-than-94
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: points.sort() idx = max_width = 0 while idx < (len(points)-1): jdx = idx + 1 if points[jdx][0] - points[idx][0] > max_width : ...
widest-vertical-area-between-two-points-containing-no-points
python solution | faster than 94%
anandanshul001
0
71
widest vertical area between two points containing no points
1,637
0.842
Medium
23,773
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/1101479/Python3-1-dim-problem
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: vals = sorted(x for x, _ in points) return max(vals[i] - vals[i-1] for i in range(1, len(vals)))
widest-vertical-area-between-two-points-containing-no-points
[Python3] 1-dim problem
ye15
0
71
widest vertical area between two points containing no points
1,637
0.842
Medium
23,774
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/1101479/Python3-1-dim-problem
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: points.sort(key=lambda x: x[0]) ans = 0 for i in range(1, len(points)): ans = max(ans, points[i][0] - points[i-1][0]) return ans
widest-vertical-area-between-two-points-containing-no-points
[Python3] 1-dim problem
ye15
0
71
widest vertical area between two points containing no points
1,637
0.842
Medium
23,775
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/1008850/python3-easy-example
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: sort_point = sorted(points, key=lambda x: x[0]) return max([abs(sort_point[i][0] - sort_point[i+1][0]) for i in range(len(sort_point) - 1)])
widest-vertical-area-between-two-points-containing-no-points
python3 easy example
MartenMink
0
58
widest vertical area between two points containing no points
1,637
0.842
Medium
23,776
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/984967/Python3-%3A-faster-than-98.54-of-Python3-online-submissions
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: y = sorted([i for i,j in points]) m = 0 temp = 0 for i in range(1,len(y)): temp = y[i] - y[i-1] m = max(temp,m) return m
widest-vertical-area-between-two-points-containing-no-points
Python3 : faster than 98.54% of Python3 online submissions
abhijeetmallick29
0
93
widest vertical area between two points containing no points
1,637
0.842
Medium
23,777
https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/discuss/919487/explanation-The-simplest-Python-without-Y
class Solution: def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int: last = None mx = None for i in sorted(points): i = i[0] if last is not None and (mx is None or i - last > mx): mx = i - last last = i return mx
widest-vertical-area-between-two-points-containing-no-points
[explanation] The simplest Python without Y
c0ntender
0
72
widest vertical area between two points containing no points
1,637
0.842
Medium
23,778
https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/1101671/Python3-top-down-and-bottom-up-dp
class Solution: def countSubstrings(self, s: str, t: str) -> int: m, n = len(s), len(t) @cache def fn(i, j, k): """Return number of substrings ending at s[i] and t[j] with k=0/1 difference.""" if i < 0 or j < 0: return 0 if s[i] == t[j]: return...
count-substrings-that-differ-by-one-character
[Python3] top-down & bottom-up dp
ye15
5
303
count substrings that differ by one character
1,638
0.714
Medium
23,779
https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/1101671/Python3-top-down-and-bottom-up-dp
class Solution: def countSubstrings(self, s: str, t: str) -> int: m, n = len(s), len(t) dp0 = [[0]*(n+1) for _ in range(m+1)] # 0-mismatch dp1 = [[0]*(n+1) for _ in range(m+1)] # 1-mismatch ans = 0 for i in range(m): for j in range(n): if ...
count-substrings-that-differ-by-one-character
[Python3] top-down & bottom-up dp
ye15
5
303
count substrings that differ by one character
1,638
0.714
Medium
23,780
https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/1186959/Python3-simple-solution
class Solution: def countSubstrings(self, s: str, t: str) -> int: count = 0 for i in range(len(s)): for j in range(len(t)): x,y = i,j d = 0 while x < len(s) and y < len(t): if s[x] != t[y]: d += 1...
count-substrings-that-differ-by-one-character
Python3 simple solution
EklavyaJoshi
1
235
count substrings that differ by one character
1,638
0.714
Medium
23,781
https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/2839579/Python-(Simple-DP)
class Solution: def countSubstrings(self, s, t): n, m = len(s), len(t) match = [[0 for _ in range(m+1)] for _ in range(n+1)] matchone = [[0 for _ in range(m+1)] for _ in range(n+1)] for i in range(1,n+1): for j in range(1,m+1): if s[i-1] == t[j-1]: ...
count-substrings-that-differ-by-one-character
Python (Simple DP)
rnotappl
0
2
count substrings that differ by one character
1,638
0.714
Medium
23,782
https://leetcode.com/problems/count-substrings-that-differ-by-one-character/discuss/1498318/Python3-or-O(MN*max(MN))
class Solution: def countSubstrings(self, s: str, t: str) -> int: ans=0 for i in range(len(s)): for j in range(len(t)): x=i y=j d=0 while x<len(s) and y<len(t): if s[x]!=t[y]: d+=1...
count-substrings-that-differ-by-one-character
[Python3] | O(MN*max(MN))
swapnilsingh421
0
179
count substrings that differ by one character
1,638
0.714
Medium
23,783
https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/discuss/1101522/Python3-top-down-dp
class Solution: def numWays(self, words: List[str], target: str) -> int: freq = [defaultdict(int) for _ in range(len(words[0]))] for word in words: for i, c in enumerate(word): freq[i][c] += 1 @cache def fn(i, k): """Return number o...
number-of-ways-to-form-a-target-string-given-a-dictionary
[Python3] top-down dp
ye15
2
169
number of ways to form a target string given a dictionary
1,639
0.429
Hard
23,784
https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/discuss/2836353/Python3-Recursion-to-memoization-to-optimization-with-complete-explaination
class Solution: def numWays(self, words: List[str], target: str) -> int: MOD = 10**9+7 n,m = len(target),len(words[0]) def dfs(i,j): if j == n: return 1 if i == m: return 0 if n-j > m-i: return 0 res = 0 for word in words:...
number-of-ways-to-form-a-target-string-given-a-dictionary
[Python3] Recursion to memoization to optimization with complete explaination
shriyansnaik
0
7
number of ways to form a target string given a dictionary
1,639
0.429
Hard
23,785
https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/discuss/2836353/Python3-Recursion-to-memoization-to-optimization-with-complete-explaination
class Solution: def numWays(self, words: List[str], target: str) -> int: MOD = 10**9+7 n,m = len(target),len(words[0]) dp = {} def dfs(i,j): if j == n: return 1 if i == m: return 0 if n-j > m-i: return 0 if (i,j) in dp: return dp[(i,j...
number-of-ways-to-form-a-target-string-given-a-dictionary
[Python3] Recursion to memoization to optimization with complete explaination
shriyansnaik
0
7
number of ways to form a target string given a dictionary
1,639
0.429
Hard
23,786
https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/discuss/2836353/Python3-Recursion-to-memoization-to-optimization-with-complete-explaination
class Solution: def numWays(self, words: List[str], target: str) -> int: MOD = 10**9+7 n,m = len(target),len(words[0]) frequency = defaultdict(int) for word in words: for i,ch in enumerate(word): frequency[(ch,i)]+=1 dp = {} def d...
number-of-ways-to-form-a-target-string-given-a-dictionary
[Python3] Recursion to memoization to optimization with complete explaination
shriyansnaik
0
7
number of ways to form a target string given a dictionary
1,639
0.429
Hard
23,787
https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/discuss/2353648/python-3-or-top-down-1-D-dp-or-O(mn)O(n)
class Solution: def numWays(self, words: List[str], target: str) -> int: m, n = len(target), len(words[0]) wordChars = [] for j in range(n): wordChars.append(collections.Counter(word[j] for word in words)) prev = [1] * (n + 1) for i in range(m - 1, -1, -1): ...
number-of-ways-to-form-a-target-string-given-a-dictionary
python 3 | top down 1-D dp | O(mn)/O(n)
dereky4
0
416
number of ways to form a target string given a dictionary
1,639
0.429
Hard
23,788
https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/discuss/925002/bottom-up-top-down-dynamic-programming.-explained
class Solution: def numWays(self, words: List[str], target: str) -> int: @lru_cache(None) # dfs(i, j) is number of ways to construct taget[:i+1] using chars with index at most j in the word in the words dictionary. def dfs(i, j): if i < 0: return 1 if j < 0...
number-of-ways-to-form-a-target-string-given-a-dictionary
bottom up/ top down dynamic programming. explained
ytb_algorithm
0
201
number of ways to form a target string given a dictionary
1,639
0.429
Hard
23,789
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/918382/Python3-2-line-O(N)
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: mp = {x[0]: x for x in pieces} i = 0 while i < len(arr): if (x := arr[i]) not in mp or mp[x] != arr[i:i+len(mp[x])]: return False i += len(mp[x]) return True
check-array-formation-through-concatenation
[Python3] 2-line O(N)
ye15
9
705
check array formation through concatenation
1,640
0.561
Easy
23,790
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/918382/Python3-2-line-O(N)
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: mp = {x[0]: x for x in pieces} return sum((mp.get(x, []) for x in arr), []) == arr
check-array-formation-through-concatenation
[Python3] 2-line O(N)
ye15
9
705
check array formation through concatenation
1,640
0.561
Easy
23,791
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/996914/Python-Simple-O(n)-solution-based-on-the-hints
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: for i, piece in enumerate(pieces): for j,num in enumerate(piece): try: pieces[i][j] = arr.index(num) except: return False pieces.sort() retur...
check-array-formation-through-concatenation
[Python] Simple O(n) solution based on the hints
KevinZzz666
4
342
check array formation through concatenation
1,640
0.561
Easy
23,792
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/1237002/Python3-Intuitive-solution-by-hash-map
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: dic = {} for piece in pieces: dic[piece[0]] = piece idx = 0 while idx < len(arr): key = arr[idx] if key in dic: if arr[idx: idx + len(dic[key])...
check-array-formation-through-concatenation
Python3 Intuitive solution by hash map
georgeqz
2
138
check array formation through concatenation
1,640
0.561
Easy
23,793
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/1095488/Python3-simple-solution-by-two-approaches
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: s = ''.join(map(str, arr)) for i in pieces: if ''.join(map(str, i)) not in s or not i[0] in arr: return False return True
check-array-formation-through-concatenation
Python3 simple solution by two approaches
EklavyaJoshi
2
75
check array formation through concatenation
1,640
0.561
Easy
23,794
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/1095488/Python3-simple-solution-by-two-approaches
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: for i in pieces: if i[0] not in arr: return False pieces.sort(key=lambda a:arr.index(a[0])) res = [] for i in pieces: res += i return True if res =...
check-array-formation-through-concatenation
Python3 simple solution by two approaches
EklavyaJoshi
2
75
check array formation through concatenation
1,640
0.561
Easy
23,795
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/2523031/C%2B%2BPython-Solution
class Solution: def canFormArray(self, arr, pieces): d = {x[0]: x for x in pieces} return list(chain(*[d.get(num, []) for num in arr])) == arr
check-array-formation-through-concatenation
C++/Python Solution
arpit3043
1
68
check array formation through concatenation
1,640
0.561
Easy
23,796
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/1515782/Python-the-most-optimal-solution%3A-O(n)-time-O(n)-space-with-hashmap
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: res = [] n = len(arr) indexes = {} for i in range(len(pieces)): indexes[pieces[i][0]] = i idx = 0 while idx < n: if arr[idx] not in indexes: ...
check-array-formation-through-concatenation
Python the most optimal solution: O(n) time, O(n) space with hashmap
byuns9334
1
112
check array formation through concatenation
1,640
0.561
Easy
23,797
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/1135159/Python-Faster-than-99
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: p = {val[0]: val for val in pieces} i = 0 while i < len(arr): if arr[i] in p: for val in p[arr[i]]: if i == len(arr) or arr[i] != val: ...
check-array-formation-through-concatenation
[Python] Faster than 99%
FooMan
1
168
check array formation through concatenation
1,640
0.561
Easy
23,798
https://leetcode.com/problems/check-array-formation-through-concatenation/discuss/1019541/python3-solution-without-converting-into-string
class Solution: def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool: j,i=0,0 l=[] while i<len(arr) and j<len(pieces): if arr[i] in pieces[j]: for k in range(len(pieces[j])): l.append(pieces[j][k]) ...
check-array-formation-through-concatenation
python3 solution without converting into string
_SID_
1
98
check array formation through concatenation
1,640
0.561
Easy
23,799