Datasets:
kaysss
/
leetcode-problem-solutions

Dataset card Data Studio Files Community
question_slug
stringlengths
3
77
title
stringlengths
1
183
slug
stringlengths
12
45
summary
stringlengths
1
160
⌀
author
stringlengths
2
30
certification
stringclasses
2 values
created_at
stringdate
2013-10-25 17:32:12
2025-04-12 09:38:24
updated_at
stringdate
2013-10-25 17:32:12
2025-04-12 09:38:24
hit_count
int64
0
10.6M
has_video
bool
2 classes
content
stringlengths
4
576k
upvotes
int64
0
11.5k
downvotes
int64
0
358
tags
stringlengths
2
193
comments
int64
0
2.56k
determine-color-of-a-chessboard-square
JavaScript Simple Solution 99.1% faster
javascript-simple-solution-991-faster-by-t2z1
\n/**\n * @param {string} coords\n * @return {boolean}\n */\nconst squareIsWhite = function (coords) {\n const letters = [\'a\', \'b\', \'c\', \'d\', \'e\',
js_pro
NORMAL
2022-10-07T09:05:37.744952+00:00
2024-11-24T07:15:39.979684+00:00
595
false
```\n/**\n * @param {string} coords\n * @return {boolean}\n */\nconst squareIsWhite = function (coords) {\n const letters = [\'a\', \'b\', \'c\', \'d\', \'e\', \'f\', \'g\', \'h\'];\n const [l, n] = coords.split(\'\');\n\n return (letters.indexOf(l) % 2 == 0 && n % 2 == 0) \n || (letters.indexOf(l) % 2 == 1 && n % 2 == 1);\n};\n\n```
2
0
['JavaScript']
1
determine-color-of-a-chessboard-square
Java | Math | Faster Than 100% Java Submissions
java-math-faster-than-100-java-submissio-givh
\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n Character str = coordinates.charAt(0);\n int num = Character.getNumer
Divyansh__26
NORMAL
2022-09-14T15:30:29.822855+00:00
2022-09-14T15:30:29.822897+00:00
173
false
```\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n Character str = coordinates.charAt(0);\n int num = Character.getNumericValue(coordinates.charAt(1));\n int flag=0;\n if(str==\'a\' || str==\'c\' || str==\'e\' || str==\'g\'){\n if(num%2==0)\n flag=1;\n }\n else{\n if(num%2!=0)\n flag=1;\n }\n if(flag==0)\n return false;\n return true;\n }\n}\n```\nKindly upvote if you like the code.
2
0
['Math', 'String', 'Java']
0
determine-color-of-a-chessboard-square
Simple if-else faster than 83%
simple-if-else-faster-than-83-by-aruj900-zl7k
\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n black = "aceg"\n white = "bdfh"\n if coordinates[0] in black an
aruj900
NORMAL
2022-08-29T10:13:12.823343+00:00
2022-08-29T10:13:12.823390+00:00
373
false
```\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n black = "aceg"\n white = "bdfh"\n if coordinates[0] in black and int(coordinates[1]) % 2 == 1:\n return False\n elif coordinates[0] in white and int(coordinates[1]) % 2 == 0:\n return False\n else:\n return True\n```
2
0
['Python', 'Python3']
0
determine-color-of-a-chessboard-square
[Python] One Line Solution || Faster than 94%
python-one-line-solution-faster-than-94-2tz5h
Python Code -\n\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n alphabets=\'abcdefgh\'\n return True if (alphabets.inde
rksreeraj
NORMAL
2022-08-16T11:49:43.901082+00:00
2022-08-16T11:49:43.901128+00:00
104
false
# Python Code -\n```\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n alphabets=\'abcdefgh\'\n return True if (alphabets.index(coordinates[0]) + int(coordinates[1])) % 2 == 0 else False\n```
2
0
['Python']
0
determine-color-of-a-chessboard-square
[PYTHON 3] EASY | SELF EXPLANITORY
python-3-easy-self-explanitory-by-omkarx-l53h
```\nclass Solution:\n def squareIsWhite(self, c: str) -> bool:\n e,o = ["b","d","f","h"], ["a","c","e","g"]\n if int(c[-1]) % 2 == 0:\n
omkarxpatel
NORMAL
2022-07-13T14:41:03.033188+00:00
2022-07-13T14:41:03.033225+00:00
610
false
```\nclass Solution:\n def squareIsWhite(self, c: str) -> bool:\n e,o = ["b","d","f","h"], ["a","c","e","g"]\n if int(c[-1]) % 2 == 0:\n if c[0] in e: return False\n else: return True\n else:\n if c[0] in e: return True\n else: return False\n
2
0
['Python', 'Python3']
0
determine-color-of-a-chessboard-square
✔ C++ || one-liner easy O(1)
c-one-liner-easy-o1-by-_aditya_hegde-snlw
Intuition:-\n\n-> You can easily get the pattern in the chessboard.\n-> sum( x, y ) of any co-ordinate is even -> Its black square (return 0)\n-> sum( x, y ) of
_Aditya_Hegde_
NORMAL
2022-06-18T09:38:36.421469+00:00
2022-06-18T09:38:36.421508+00:00
59
false
**Intuition:-**\n\n-> You can easily get the **pattern** in the chessboard.\n-> sum( x, y ) of any co-ordinate is **even** -> Its **black square** (return 0)\n-> sum( x, y ) of any co-ordinate is **odd** -> its **while square** (return 1)\n-> also I have used **bitwise AND operator** to check odd or even. (&)\n-> I have used **ternary operator** also here.\n\n**Complexity:-**\n\nTime=O(1) \nSpace=O(1) \n\nHoping you understood the solution :)\nPlease **upvote** if you found it usefull !\nThank you..\n\n```\nclass Solution {\npublic:\n bool squareIsWhite(string s) {\n return ((s[0]+s[1])&1)?1:0;\n }\n};\n```
2
0
['C']
0
determine-color-of-a-chessboard-square
Easy python solution
easy-python-solution-by-rupeshmohanty-eu8q
\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n odds = [\'a\',\'c\',\'e\',\'g\']\n evens = [\'b\',\'d\',\'f\',\'h\']\n
rupeshmohanty
NORMAL
2022-06-07T14:40:06.585294+00:00
2022-06-07T14:40:06.585343+00:00
99
false
```\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n odds = [\'a\',\'c\',\'e\',\'g\']\n evens = [\'b\',\'d\',\'f\',\'h\']\n \n if coordinates[0] in odds:\n if int(coordinates[1]) % 2 != 0:\n return False\n else:\n return True\n else:\n if int(coordinates[1]) % 2 != 0:\n return True\n else:\n return False\n```
2
0
['Python']
0
determine-color-of-a-chessboard-square
Java One-liner with explanation
java-one-liner-with-explanation-by-alexg-e66r
\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n return (coordinates.charAt(0) + coordinates.charAt(1)) % 2 != 0;\n }\n}\n
alexgornovoi
NORMAL
2022-03-19T16:56:02.384628+00:00
2022-03-19T16:58:50.524049+00:00
102
false
```\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n return (coordinates.charAt(0) + coordinates.charAt(1)) % 2 != 0;\n }\n}\n```\nThe ASCII decimal value for the letter "a" is odd and the ASCII decimal value for "b" is even and "c" is odd ... etc. As for the numbers the ASCII value for an odd number is odd and the ASCII number for an even number is even. (As can be seen by the table below)\n\n<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/1/1b/ASCII-Table-wide.svg/2560px-ASCII-Table-wide.svg.png" alt="closed_paren" title="Closed Parenthesis" width="400"/>\n\n\nHere we take the ASCII value of the char at postion 0 and the ASCII value of the char at postion 1 and check if their sum is odd if it is then we know that we are standing on a white square so we return true otherwise we return false.
2
0
['Java']
2
determine-color-of-a-chessboard-square
Python C one line
python-c-one-line-by-mrchuck-a5fe
py\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n return sum(map(ord,coordinates)) % 2\n\nc\nbool squareIsWhite(char * coordi
mrchuck
NORMAL
2022-02-26T13:25:30.600766+00:00
2022-02-26T13:26:54.491315+00:00
166
false
```py\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n return sum(map(ord,coordinates)) % 2\n```\n```c\nbool squareIsWhite(char * coordinates){\n return (*coordinates + *++coordinates) % 2;\n}\n```
2
0
['C', 'Python']
0
determine-color-of-a-chessboard-square
Easy Solution 100% fast
easy-solution-100-fast-by-deepank4-udiq
\nclass Solution {\npublic:\n bool squareIsWhite(string c) {\n\t//if sum of cordinates is even than color is black \n int x = c[0]-\'a\'+1;\n i
deepank4
NORMAL
2022-01-19T18:09:40.299367+00:00
2022-01-19T18:09:40.299411+00:00
177
false
```\nclass Solution {\npublic:\n bool squareIsWhite(string c) {\n\t//if sum of cordinates is even than color is black \n int x = c[0]-\'a\'+1;\n int y = c[1]-\'1\'+1;\n cout<<y << "\\n ";\n if((x+y)%2==0)\n return false;\n return true;\n }\n};\n```
2
0
['C', 'C++']
0
determine-color-of-a-chessboard-square
Beginner friendly JavaScript solution
beginner-friendly-javascript-solution-by-k7s1
Time Complexity : O(n)\n\n/**\n * @param {string} coordinates\n * @return {boolean}\n */\nvar squareIsWhite = function(coordinates) {\n let oddchar = "aceg",
HimanshuBhoir
NORMAL
2022-01-19T08:53:41.226948+00:00
2022-01-19T08:53:41.226981+00:00
127
false
**Time Complexity : O(n)**\n```\n/**\n * @param {string} coordinates\n * @return {boolean}\n */\nvar squareIsWhite = function(coordinates) {\n let oddchar = "aceg", evenchar = "bdfh", ch = coordinates.charAt(0), n = parseInt(coordinates.charAt(1)+"");\n if((n % 2 != 0 && oddchar.includes(ch)) || (n % 2 == 0 && evenchar.includes(ch))){\n return false;\n }\n return true;\n};\n```
2
0
['JavaScript']
0
determine-color-of-a-chessboard-square
Short || Clean || One Liner || Java
short-clean-one-liner-java-by-himanshubh-30ks
\njava []\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n return (coordinates.charAt(0)-\'a\'+ coordinates.charAt(1)-\'0\') %
HimanshuBhoir
NORMAL
2022-01-19T08:52:51.226840+00:00
2023-01-05T11:04:10.646269+00:00
74
false
\n```java []\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n return (coordinates.charAt(0)-\'a\'+ coordinates.charAt(1)-\'0\') % 2 == 0;\n }\n}\n```
2
0
['Java']
0
determine-color-of-a-chessboard-square
c++ 100 % faster
c-100-faster-by-mannu_story07-wixs
class Solution {\npublic:\n bool squareIsWhite(string coordinates) {\n int x = coordinates[0]-\'a\';\n int y = coordinates[1] - 1;\n if(
Mannu_story07
NORMAL
2021-10-02T05:49:31.265547+00:00
2021-10-02T05:49:31.265605+00:00
61
false
class Solution {\npublic:\n bool squareIsWhite(string coordinates) {\n int x = coordinates[0]-\'a\';\n int y = coordinates[1] - 1;\n if((x % 2 == 0 && y % 2 ==0 || x % 2 != 0 && y % 2 != 0)){\n return false;\n }\n else {\n return true;\n }\n }\n* 1. * };[](http://)
2
0
[]
0
determine-color-of-a-chessboard-square
[Java] 3 line simple %100 solution
java-3-line-simple-100-solution-by-zeldo-4uw7
if you like it pls upvote\n\n\n\nJava\n\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n int left = coordinates.charAt(0)-\'a\
zeldox
NORMAL
2021-08-29T22:48:11.378555+00:00
2021-08-29T22:48:11.378585+00:00
157
false
if you like it pls upvote\n\n\n\nJava\n```\nclass Solution {\n public boolean squareIsWhite(String coordinates) {\n int left = coordinates.charAt(0)-\'a\';\n int right = coordinates.charAt(1)-\'1\'; \n return (left %2 == 0 && right % 2 == 1) || (left %2 != 0 && right % 2 != 1);\n }\n}\n```\n
2
0
['Java']
0
determine-color-of-a-chessboard-square
C++ 100% fast
c-100-fast-by-harshattree-ep2o
\n bool squareIsWhite(string coordinates) {\n int a=coordinates[0]+coordinates[1];\n if(a%2==1)\n return true;\n return false;\n
HarshAttree
NORMAL
2021-07-22T05:16:50.723722+00:00
2021-07-22T05:16:50.723766+00:00
51
false
```\n bool squareIsWhite(string coordinates) {\n int a=coordinates[0]+coordinates[1];\n if(a%2==1)\n return true;\n return false;\n }\n```
2
0
[]
0
determine-color-of-a-chessboard-square
100% faster one line [Java] Solution with explanation
100-faster-one-line-java-solution-with-e-wlld
Imagine the x cordinate are also 1,2,3...so on. \n\n\nIdea is : Now, when we see this board we can find the logic that.\n- If x and y both are even Or odd then
HemantPatel
NORMAL
2021-07-02T04:31:36.505750+00:00
2021-07-02T04:31:36.505798+00:00
148
false
Imagine the x cordinate are also 1,2,3...so on. <br>\n<img src="https://assets.leetcode.com/users/images/36b5a71a-2639-4344-9edc-f88982b9096a_1625199149.2753234.png" alt="chess board image" width="300" height="auto">\n\n**Idea is :** Now, when we see this board we can find the logic that.\n- If x and y both are even Or odd then square is black.\n- Otherwise square is white.\n\nSo Pseudo code is :\n```\nif( x is even and y is even ) return false ( black square )\nelse if ( x is odd and y is odd ) return false ( black square )\nelse return true;\n```\n\nBut we can short it by this Logic :\nIf x and y both are even Or odd then (x + y) is even\nother wise (x + y) is odd\n```\nif((x + y) is odd) return true\nelse return false;\n```\n\nMy Solution is :\n\n\n\n\n```\n public boolean squareIsWhite(String coordinates) {\n return ((coordinates.charAt(0) - \'a\') + (coordinates.charAt(1) - \'1\')) % 2 != 0 ;\n }\n```
2
0
['Java']
0
determine-color-of-a-chessboard-square
C++ || One Liner || Easy understanding || With Comments
c-one-liner-easy-understanding-with-comm-iwpi
//Idea behind this to find a unique square number.\n//If that square number is even it will be white , else black.\n\nclass Solution {\npublic:\n bool square
ajaayush
NORMAL
2021-06-22T07:13:51.944594+00:00
2021-06-22T07:13:51.944637+00:00
152
false
//Idea behind this to find a unique square number.\n//If that square number is even it will be white , else black.\n```\nclass Solution {\npublic:\n bool squareIsWhite(string coordinates) {\n int i= (int)(coordinates[0]-\'a\'+1) + (int)(coordinates[1]-\'0\');\n if(i%2==0)\n return false;\n return true;\n }\n};\n```
2
0
['Math', 'C', 'C++']
0
determine-color-of-a-chessboard-square
PYTHON 1 LINER || WITH EXPLANATION
python-1-liner-with-explanation-by-mahad-wxm0
\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n \'\'\'\n check if file/x-coordinate and rank/y-coordinate are both eve
mahadrehan
NORMAL
2021-06-22T05:33:13.010244+00:00
2021-06-22T05:43:05.748527+00:00
80
false
```\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n \'\'\'\n check if file/x-coordinate and rank/y-coordinate are both even or are both odd\n must be odd rank with even file, or odd file with even rank for it to be a white square\n \'\'\'\n return int(coordinates[1])%2 != ord(coordinates[0])%2\n```\nHere the y-coordinate/rank is converted to an integer to check if its an odd or even rank. Using the python function ```ord()``` you get the ascii value of the x-coordinate/file to check if the file is an odd or even file (e.g C1, C is the 3rd file after A and B). It then returns True if the check evaluates to False since only 1 coordinate is even, meaning that the coordinate is a white square.
2
0
[]
0
determine-color-of-a-chessboard-square
JAVA ONE LINER
java-one-liner-by-tanmaymishrapersonal-xu6a
\nclass Solution {\n public boolean squareIsWhite(String coordinates) \n {\n return ((coordinates.charAt(0)-\'a\'+coordinates.charAt(1)-\'1\')%2!=0)
tanmaymishrapersonal
NORMAL
2021-06-12T18:25:15.159085+00:00
2021-06-12T18:25:15.159117+00:00
66
false
```\nclass Solution {\n public boolean squareIsWhite(String coordinates) \n {\n return ((coordinates.charAt(0)-\'a\'+coordinates.charAt(1)-\'1\')%2!=0);\n }\n}\n\n \n```
2
0
[]
2
determine-color-of-a-chessboard-square
Java || 1 Liner || 0ms || beats 100% || T.C - O(1) S.C - O(1)
java-1-liner-0ms-beats-100-tc-o1-sc-o1-b-rm6v
\n\tpublic boolean squareIsWhite(String coordinates) {\n \n char ch = coordinates.charAt(0);\n int num = coordinates.charAt(1);\n \n
LegendaryCoder
NORMAL
2021-05-18T17:08:00.715892+00:00
2021-05-18T17:08:00.715935+00:00
63
false
\n\tpublic boolean squareIsWhite(String coordinates) {\n \n char ch = coordinates.charAt(0);\n int num = coordinates.charAt(1);\n \n int val = (ch - \'a\') + (num - \'0\');\n return (val % 2 == 0) ? true : false;\n }
2
1
[]
0
determine-color-of-a-chessboard-square
Easy C# Solution
easy-c-solution-by-venendroid-x3vx
\n public bool SquareIsWhite(string coordinates)\n {\n //From the chess board we can see BLACK box are in ODD integer places for these columns \n
venendroid
NORMAL
2021-04-22T06:15:24.130890+00:00
2021-04-22T06:15:24.130929+00:00
59
false
```\n public bool SquareIsWhite(string coordinates)\n {\n //From the chess board we can see BLACK box are in ODD integer places for these columns \n var oddBlackList = new HashSet<char>(){\'a\', \'c\', \'e\', \'g\'};\n \n //Whole idea is:\n //From the give coordinates, if first char falls in oddBlackList \n //then we know white will be always at even block\n return oddBlackList.Contains(coordinates[0])? (coordinates[1] - \'0\')%2 == 0: (coordinates[1] - \'0\')%2 != 0;\n }\n```
2
0
[]
0
determine-color-of-a-chessboard-square
python one line sol (faster than 93%) (less mem than 100%)
python-one-line-sol-faster-than-93-less-6h4vm
\nclass Solution(object):\n def squareIsWhite(self, coordinates):\n return ((ord(coordinates[0])) + int(coordinates[1])) % 2\n
elayan
NORMAL
2021-04-15T11:00:10.235463+00:00
2021-04-15T11:00:10.235505+00:00
48
false
```\nclass Solution(object):\n def squareIsWhite(self, coordinates):\n return ((ord(coordinates[0])) + int(coordinates[1])) % 2\n```
2
1
[]
2
determine-color-of-a-chessboard-square
[C++] [Time-O(1)]
c-time-o1-by-turbotorquetiger-qmjg
Time Complexity: O(1)\nSpace Complexity: O(1)\n\n\nclass Solution \n{\npublic:\n bool squareIsWhite(string cood)\n {\n int f = cood[0]-\'a\';\n
ihavehiddenmyid
NORMAL
2021-04-10T15:23:10.874530+00:00
2021-04-10T15:23:10.874555+00:00
117
false
**Time Complexity: O(1)\nSpace Complexity: O(1)**\n\n```\nclass Solution \n{\npublic:\n bool squareIsWhite(string cood)\n {\n int f = cood[0]-\'a\';\n int s = cood[1]-\'0\';\n \n // f = a, c, e, g\n if ( f % 2 == 0 ) {\n if ( s % 2 == 0 ) {\n return true;\n }\n }\n // f = b, d, f, h\n else if ( f % 2 != 0 ) {\n if ( s % 2 != 0 ) {\n return true;\n }\n }\n return false;\n }\n};\n```\n\n**If you liked it, please upvote it**\n
2
0
['C']
0
determine-color-of-a-chessboard-square
Javascript, 100% faster, simple
javascript-100-faster-simple-by-mstaso1-vte3
\nvar squareIsWhite = function(coordinates) {\n const columns = [\'a\', \'b\', \'c\', \'d\', \'e\', \'f\', \'g\', \'h\']\n \n let i = 0;\n \n whi
mstaso1
NORMAL
2021-04-07T18:44:29.238181+00:00
2021-04-07T18:44:29.238213+00:00
114
false
```\nvar squareIsWhite = function(coordinates) {\n const columns = [\'a\', \'b\', \'c\', \'d\', \'e\', \'f\', \'g\', \'h\']\n \n let i = 0;\n \n while(coordinates[0] !== columns[i]) {\n i++;\n }\n \n let num = i + parseInt(coordinates[1]);\n \n \n return num % 2 === 0 \n \n};\n```
2
0
[]
1
determine-color-of-a-chessboard-square
[Python] 1-lines
python-1-lines-by-maiyude-7ccm
\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n return True if ((ord(coordinates[0]))+int(coordinates[1])) % 2 else False\n
maiyude
NORMAL
2021-04-03T16:46:43.207322+00:00
2021-04-03T16:46:43.207365+00:00
208
false
```\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n return True if ((ord(coordinates[0]))+int(coordinates[1])) % 2 else False\n```
2
0
['Python', 'Python3']
1
determine-color-of-a-chessboard-square
Simple Python 3 solution
simple-python-3-solution-by-swap24-jnc9
\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n x, y = ord(coordinates[0]) - 65, int(coordinates[1])\n return y % 2 ==
swap24
NORMAL
2021-04-03T16:28:22.022359+00:00
2021-04-03T16:28:22.022401+00:00
83
false
```\nclass Solution:\n def squareIsWhite(self, coordinates: str) -> bool:\n x, y = ord(coordinates[0]) - 65, int(coordinates[1])\n return y % 2 == 1 if x % 2 == 1 else y % 2 == 0\n```
2
0
[]
0
determine-color-of-a-chessboard-square
C# one liner | 0ms, beats 100%
c-one-liner-0ms-beats-100-by-bkga9sasfa-xoas
Complexity Time complexity: O(1) Space complexity: O(1) Code
bKGa9sAsfA
NORMAL
2025-03-07T10:03:28.191008+00:00
2025-03-07T10:03:28.191008+00:00
23
false
# Complexity - Time complexity: $$O(1)$$ - Space complexity: $$O(1)$$ # Code ```csharp [] public class Solution { public bool SquareIsWhite(string coordinates) { return ((coordinates[0] + coordinates[1]) & 1) == 1; } } ```
1
0
['C#']
0
determine-color-of-a-chessboard-square
Beats 100% , simple approach...
beats-100-simple-approach-by-arun_george-3hvo
IntuitionThe problem involves determining the color of a square on a chessboard given its coordinates. Chessboards have alternating colors, and the color of a s
Arun_George_
NORMAL
2025-03-05T16:55:30.003942+00:00
2025-03-06T16:12:03.709795+00:00
91
false
# Intuition The problem involves determining the color of a square on a chessboard given its coordinates. Chessboards have alternating colors, and the color of a square can be determined based on the combination of its letter (column) and number (row). ![image.png](https://assets.leetcode.com/users/images/df82ae80-9205-4172-8c7d-2bcfb210d161_1741277516.239935.png) # Approach 1. **Chessboard Pattern**: On a chessboard, the color of a square alternates between white and black. For example, 'a1' is black, 'a2' is white, 'a3' is black, and so on. 2. **Coordinate Analysis**: The letter part of the coordinate (a-h) and the number part (1-8) can be used to determine the color: - If the letter is in 'a', 'c', 'e', 'g' (odd columns), and the number is odd, the square is black. - If the letter is in 'a', 'c', 'e', 'g' (odd columns), and the number is even, the square is white. - If the letter is in 'b', 'd', 'f', 'h' (even columns), and the number is odd, the square is white. - If the letter is in 'b', 'd', 'f', 'h' (even columns), and the number is even, the square is black. 3. **Implementation**: We can use conditional statements to check the letter and number parts of the coordinate and return the appropriate color. # Complexity - **Time Complexity**: O(1) because the operations are constant time checks. - **Space Complexity**: O(1) as no additional space is used apart from the input. # Code ```python class Solution: def squareIsWhite(self, coordinates: str) -> bool: # Check if the letter is in 'a', 'c', 'e', 'g' if coordinates[0] in 'aceg': # If the number is odd, the square is black (False) if coordinates[1] in '1357': return False else: return True else: # If the letter is in 'b', 'd', 'f', 'h' and the number is odd, the square is white (True) if coordinates[1] in '1357': return True else: return False
1
0
['Python3']
0
delete-columns-to-make-sorted-iii
[Java/C++/Python] Maximum Increasing Subsequence
javacpython-maximum-increasing-subsequen-wio6
Intuition\n\nTake n cols as n elements, so we have an array of n elements.\n=> The final array has every row in lexicographic order.\n=> The elements in final s
lee215
NORMAL
2018-12-16T04:05:25.923137+00:00
2021-01-07T07:05:24.078617+00:00
8,771
false
# **Intuition**\n\nTake `n` cols as `n` elements, so we have an array of `n` elements.\n=> The final array has every row in lexicographic order.\n=> The elements in final state is in increasing order.\n=> The final elements is a sub sequence of initial array.\n=> Transform the problem to find the maximum increasing sub sequence.\n<br>\n\n# **Explanation**\nNow let\'s talking about how to find maximum increasing subsequence.\nHere we apply a O(N^2) dp solution.\n\n\n`dp[i]` means the longest subsequence ends with `i`-th element.\nFor all `i < j`, if `A[][i] < A[][j]`, then `dp[j] = max(dp[j], dp[i] + 1)`\n<br>\n\n# **Complexity**:\n`O(N^2)` to find increasing subsequence\n`O(M)` for comparing elements.\nSo Overall `O(MN^2)` time.\nSpace `O(N)` for dp.\n<br>\n\n**C++:**\n```cpp\n int minDeletionSize(vector<string>& A) {\n int m = A.size(), n = A[0].length(), res = n - 1, k;\n vector<int>dp(n, 1);\n for (int j = 0; j < n; ++j) {\n for (int i = 0; i < j; ++i) {\n for (k = 0; k < m; ++k) {\n if (A[k][i] > A[k][j]) break;\n }\n if (k == m && dp[i] + 1 > dp[j])\n dp[j] = dp[i] + 1;\n }\n res = min(res, n - dp[j]);\n }\n return res;\n }\n```\n\n**Java:**\n```java\n public int minDeletionSize(String[] A) {\n int m = A.length, n = A[0].length(), res = n - 1, k;\n int[] dp = new int[n];\n Arrays.fill(dp, 1);\n for (int j = 0; j < n; ++j) {\n for (int i = 0; i < j; ++i) {\n for (k = 0; k < m; ++k) {\n if (A[k].charAt(i) > A[k].charAt(j)) break;\n }\n if (k == m && dp[i] + 1 > dp[j])\n dp[j] = dp[i] + 1;\n }\n res = Math.min(res, n - dp[j]);\n }\n return res;\n }\n```\n**Python:**\n```py\n def minDeletionSize(self, A):\n n = len(A[0])\n dp = [1] * n\n for j in xrange(1, n):\n for i in xrange(j):\n if all(a[i] <= a[j] for a in A):\n dp[j] = max(dp[j], dp[i] + 1)\n return n - max(dp)\n```\n
151
3
[]
23
delete-columns-to-make-sorted-iii
C++ 7 lines, DP O(n * n * m)
c-7-lines-dp-on-n-m-by-votrubac-pe0i
A twist of the 300. Longest Increasing Subsequence problem.\n\nFor each position i,we track the maximum increasing subsequence. To do that, we analyze all j < i
votrubac
NORMAL
2018-12-19T04:54:05.665157+00:00
2018-12-19T04:54:05.665223+00:00
3,064
false
A twist of the [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/description/) problem.\n\nFor each position ```i```,we track the maximum increasing subsequence. To do that, we analyze all ```j < i```, and if ```A[j] < A[i]``` for *all strings* , then ```dp[i] = dp[j] + 1```.The runtime complexity is O(n * n * m), where n is the number of characters, and m is the number of strings.\n\nUnlike #300, we cannot use the binary search approach here because there is no stable comparasion (e.g. ```["ba", "ab"]``` are both "less" than each other).\n```\nint minDeletionSize(vector<string>& A) {\n vector<int> dp(A[0].size(), 1);\n for (auto i = 0; i < A[0].size(); ++i) {\n for (auto j = 0; j < i; ++j)\n for (auto k = 0; k <= A.size(); ++k) {\n if (k == A.size()) dp[i] = max(dp[i], dp[j] + 1);\n else if (A[k][j] > A[k][i]) break;\n }\n }\n return A[0].size() - *max_element(begin(dp), end(dp));\n}\n```
45
2
[]
7
delete-columns-to-make-sorted-iii
C++. Variation of LIS. Easy to understand.
c-variation-of-lis-easy-to-understand-by-yv99
\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& A) {\n int m = A.size();\n int n = A[0].size();\n vector<int> dp(A[0].
user5787i
NORMAL
2019-06-16T13:28:28.008161+00:00
2019-06-16T13:28:28.008199+00:00
1,367
false
```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& A) {\n int m = A.size();\n int n = A[0].size();\n vector<int> dp(A[0].size(), 1);\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < i; j++) {\n if(cmp(A, i, j)) {\n dp[i] = max(dp[i], dp[j] + 1);\n }\n }\n }\n return n - *max_element(dp.begin(), dp.end());\n }\n \n bool cmp(vector<string> &arr, int i, int j) {\n for(int k = 0; k < arr.size(); k++) {\n if(arr[k][j] > arr[k][i]) return 0;\n }\n return 1;\n }\n};\n```
13
0
[]
2
delete-columns-to-make-sorted-iii
TypeScript | Memory beats 100%
typescript-memory-beats-100-by-r9n-2wg9
Intuition\nKeep rows in order by deleting the minimum number of columns.\n\n# Approach\nDynamic Programming (DP): Track the longest sequence of valid columns.\n
r9n
NORMAL
2024-09-10T21:47:13.610396+00:00
2024-09-10T21:47:13.610422+00:00
37
false
# Intuition\nKeep rows in order by deleting the minimum number of columns.\n\n# Approach\nDynamic Programming (DP): Track the longest sequence of valid columns.\n\nCompare Columns: Check if one column can follow another in all rows.\n\nCompute Deletions: The result is the total columns minus the length of the longest valid sequence.\n\n# Complexity\n- Time complexity:\nO(C2(square) \xD7 R), checking pairs of columns for each row.\n\n- Space complexity:\nO(C), for the DP array.\n\n# Code\n```typescript []\nconst minDeletionSize = (strs: string[]): number => {\n const numRows = strs.length;\n const numCols = strs[0].length;\n const dp: number[] = new Array(numCols).fill(1);\n let minDeletions = numCols;\n\n for (let i = 0; i < numCols; ++i) {\n for (let j = 0; j < i; ++j) {\n let valid = true;\n for (let k = 0; k < numRows; ++k) {\n if (strs[k][j] > strs[k][i]) {\n valid = false;\n break;\n }\n }\n if (valid) {\n dp[i] = Math.max(dp[i], dp[j] + 1);\n }\n }\n minDeletions = Math.min(minDeletions, numCols - dp[i]);\n }\n\n return minDeletions;\n};\n\n```
9
0
['TypeScript']
0
delete-columns-to-make-sorted-iii
Python 3 || 6 lines, dp || T/S: 75% / 99%
python-3-6-lines-dp-ts-75-99-by-spauldin-v61m
\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n\n n = len(strs[0])\n\n isValid = lambda x: all(s[x] <= s[j] for s in s
Spaulding_
NORMAL
2023-08-17T20:28:48.267668+00:00
2024-05-31T18:55:55.578372+00:00
257
false
```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n\n n = len(strs[0])\n\n isValid = lambda x: all(s[x] <= s[j] for s in strs)\n\n dp = [1] * n\n\n for j in range(1, n):\n\n dp[j] = max((dp[x] for x in \n filter(isValid,range(j))), default = 0) + 1\n \n return n - max(dp)\n```\n[https://leetcode.com/problems/delete-columns-to-make-sorted-iii/submissions/1024237785/](https://leetcode.com/problems/delete-columns-to-make-sorted-iii/submissions/1024237785/)\n\nI could be wrong, but I think that time complexity is *O*(*MN*) and space complexity is *O*(*N*), in which *N* ~ `len(strs)` and *M* ~ `len(strs[0])`.
8
0
['Python3']
0
delete-columns-to-make-sorted-iii
Java simple LIS for all Strings. Faster than 98.76%
java-simple-lis-for-all-strings-faster-t-icp4
```\n\tpublic int minDeletionSize(String[] A) {\n int max= 1;\n int[] Lis= new int[A[0].length()];\n for(int i=0; i<A[0].length(); i++){\n
sunnydhotre
NORMAL
2021-02-13T01:07:07.638573+00:00
2021-02-13T01:07:07.638617+00:00
605
false
```\n\tpublic int minDeletionSize(String[] A) {\n int max= 1;\n int[] Lis= new int[A[0].length()];\n for(int i=0; i<A[0].length(); i++){\n Lis[i]= 1;\n for(int j=0; j<i; j++){\n if(checkLexicographic(A,i,j)){\n Lis[i]= Math.max(Lis[i],Lis[j]+1);\n }\n }\n max=Math.max(max,Lis[i]);\n }\n return A[0].length() - max;\n }\n public boolean checkLexicographic(String[] A,int i,int j){\n for(String str: A){\n if(str.charAt(i)<str.charAt(j)) return false;\n }\n return true;\n }
6
0
[]
0
delete-columns-to-make-sorted-iii
c++ solution based on graph theory and bfs
c-solution-based-on-graph-theory-and-bfs-52am
Regard this problem as the longest connection path problem for a graph, refer to the codes and comments below.\n\nclass Solution {\npublic:\n int minDeletion
jiayinggao
NORMAL
2022-05-27T17:50:05.632910+00:00
2022-05-27T17:50:05.632957+00:00
547
false
Regard this problem as the longest connection path problem for a graph, refer to the codes and comments below.\n``` \nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int single_size = strs.front().size();\n // printf("step1: establish the edges of the graph.\\n");\n vector<vector<int>> edges(single_size);\n for (int i = 0; i < single_size; i++){\n // as for index i, which index it can connect to\n for (int j = i + 1; j < single_size; j++){\n // take a look at the connection from index=i to index=j\n bool connect = true;\n for (int k = 0; k < strs.size(); k++){// scan each of the string in strs\n char former = strs[k][i], latter = strs[k][j];\n if (former > latter){\n connect = false;\n break;\n }\n }\n if (connect){// yes, the connection from index=i to index=j is satisfied, save it in edges\n edges[i].push_back(j);\n }\n }\n }\n // printf("step2: bfs search to find the longest connection path for the graph\\n");\n // printf("this part can also be solved by Dijkstra algorithm.\\n");\n vector<int> marked(single_size, 1);\n queue<pair<int, int>> units;\n for (int node = 0; node < single_size; node++){\n units.push({node, 1});\n }\n while(!units.empty()){\n int round_size = units.size();\n for (int i = 0; i < round_size; i++){\n auto unit = units.front(); units.pop();\n int node = unit.first, update_move = unit.second + 1;\n for (int j = 0; j < edges[node].size(); j++){\n int update_node = edges[node][j];\n if (update_move > marked[update_node]){\n marked[update_node] = update_move;\n units.push({update_node, update_move});\n }\n }\n }\n }\n // printf("step3: according to the vector<int> marked, select the longest path.\\n");\n int longest_move = 1;\n for (int node = 0; node < single_size; node++){\n longest_move = max(longest_move, marked[node]);\n }\n return single_size - longest_move;\n }\n};\n```
5
0
[]
0
delete-columns-to-make-sorted-iii
C++ Solution | Longest Increasing Subsequence
c-solution-longest-increasing-subsequenc-17hf
Make a vector of strings (in this case, v) in same column, and find maximum increasing subsequence among those strings. \nCheck function checks whether a string
harshnadar23
NORMAL
2021-09-03T05:29:01.442756+00:00
2021-09-03T05:29:35.143252+00:00
472
false
Make a vector of strings (in this case, v) in same column, and find maximum increasing subsequence among those strings. \nCheck function checks whether a string a is less than b according to the definition given. (every character should be lexographically less than or equal to corresponding one in b).\nI would suggest you to solve https://leetcode.com/problems/longest-increasing-subsequence/ first.\n```\nclass Solution {\npublic:\n bool check(string a, string b ){\n int n=a.size();\n for(int i=0;i<n;i++){\n if(a[i]<=b[i]) continue;\n else return false;\n }\n return true;\n }\n int minDeletionSize(vector<string>& strs) {\n vector<string> v;\n int n = strs.size();\n int m=strs[0].size();\n for(int i=0;i<m;i++){\n string s;\n for(int j=0;j<n;j++){\n s+=strs[j][i];\n }\n v.push_back(s);\n }\n vector<int> dp(m+1,1);\n \n for(int i=0;i<m;i++){\n for(int j=0;j<i;j++){\n if(check(v[j], v[i]) && dp[j]+1>dp[i]){\n dp[i]=dp[j]+1;\n }\n }\n }\n for(int i=0;i<m;i++) cout<<dp[i]<<endl;\n return m-(*max_element(dp.begin(), dp.end()));\n }\n};\n```
5
0
['Dynamic Programming']
0
delete-columns-to-make-sorted-iii
[Python3] top-down dp
python3-top-down-dp-by-ye15-qps3
\n\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n m, n = len(strs), len(strs[0]) # dimensions\n \n @cache \n
ye15
NORMAL
2021-06-08T03:40:02.347707+00:00
2021-06-08T03:40:26.624368+00:00
333
false
\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n m, n = len(strs), len(strs[0]) # dimensions\n \n @cache \n def fn(k, prev):\n """Return min deleted columns to make sorted."""\n if k == n: return 0 \n ans = 1 + fn(k+1, prev) # delete kth column\n if prev == -1 or all(strs[i][prev] <= strs[i][k] for i in range(m)): \n ans = min(ans, fn(k+1, k)) # retain kth column\n return ans \n \n return fn(0, -1)\n```
5
0
['Python3']
0
delete-columns-to-make-sorted-iii
Java Solution With Explanation
java-solution-with-explanation-by-naveen-1kfu
Idea\nPrereq https://leetcode.com/problems/longest-increasing-subsequence/discuss/342274/Intuitive-Java-Solution-With-Explanation\n\nNow, when filling the dp ar
naveen_kothamasu
NORMAL
2019-07-24T04:00:22.142200+00:00
2019-07-24T04:00:22.142232+00:00
642
false
**Idea**\n**Prereq** https://leetcode.com/problems/longest-increasing-subsequence/discuss/342274/Intuitive-Java-Solution-With-Explanation\n\nNow, when filling the `dp` array instead of looking at one string for comparing `i` and `j` indices, we need to look at all strings at those indices and confirm char at `i` can be prepended to char at `j` in all rows.\n\n```\npublic int minDeletionSize(String[] a) {\n int n = a[0].length(), res = n-1;\n int[] dp = new int[n];\n for(int k=0; k < a.length; k++){\n int max = 0;\n for(int i=n-1; i >= 0; i--){\n dp[i] = 1;\n for(int j=i+1; j < n; j++)\n if(compare(a, i, j) == -1)\n dp[i] = Math.max(dp[i], 1+dp[j]);\n max = Math.max(max, dp[i]);\n }\n res = Math.min(res, n-max);\n }\n return res;\n }\n private int compare(String[] a, int i, int j){\n for(int k=0; k < a.length; k++){\n if(a[k].charAt(i) > a[k].charAt(j))\n return 1;\n }\n return -1;\n }\n```
4
0
[]
0
delete-columns-to-make-sorted-iii
Solution
solution-by-deleted_user-d7gn
C++ []\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n \n int n = strs.size();\n int m = strs[0].length();\n\
deleted_user
NORMAL
2023-05-16T12:39:56.258657+00:00
2023-05-16T13:34:13.691015+00:00
377
false
```C++ []\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n \n int n = strs.size();\n int m = strs[0].length();\n\n vector<int> dp(m, 1);\n int res = 1;\n for(int i = 0; i < m; i++) {\n for(int j = 0; j < i; j++) {\n for(int k = 0; k <= n; k++) {\n if(k == n) {\n dp[i] = max(dp[i], dp[j] + 1);\n res = max(res, dp[i]);\n } else if (strs[k][j] > strs[k][i]) {\n break;\n }\n }\n }\n }\n return m - res;\n }\n};\n```\n\n```Python3 []\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n n = len(strs[0])\n dp = [1] * n\n for i in range(1, n):\n for j in range(i):\n valid = True\n for a in strs:\n if a[j] > a[i]: \n valid = False\n break\n if valid:\n dp[i] = max(dp[i], dp[j] + 1)\n return n - max(dp)\n```\n\n```Java []\nclass Solution {\n public int minDeletionSize(String[] strs) {\n \n int length = strs[0].length();\n int[] dp = new int[length];\n \n int max = 0;\n for (int i = 0; i < length; i++){\n dp[i] = 1;\n for (int j = 0; j < i; j++){\n if (checkLager(j,i, strs)){\n dp[i] = Math.max(dp[i], dp[j] + 1);\n }\n }\n max = Math.max(max, dp[i]);\n }\n return length - max;\n }\n public boolean checkLager(int j, int i, String[] strs){\n for (String s : strs){\n if (s.charAt(j) > s.charAt(i)){\n return false;\n }\n }\n return true;\n }\n}\n```\n
3
0
['C++', 'Java', 'Python3']
0
delete-columns-to-make-sorted-iii
Beats 98% || C++ || DP || Memoization
beats-98-c-dp-memoization-by-akash92-o71h
Intuition\n Describe your first thoughts on how to solve this problem. \nIf we can find the longest non decreasing common subsequence in all the strings, then w
akash92
NORMAL
2024-09-07T04:35:42.866345+00:00
2024-09-07T04:35:42.866388+00:00
107
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf we can find the longest non decreasing common subsequence in all the strings, then we can subtract its length from the total length of the string to get the minimum deletions.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe solution uses dynamic programming (DP) to solve the problem of finding the minimum number of columns to delete such that each row in the matrix remains lexicographically sorted. The function f explores two options at each column: either skip the column or include it if the strings remain lexicographically sorted. We use a DP table to store results for subproblems.\n\n# Complexity\n- Time complexity: $$O(n^2 * m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n^2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\n int m, n;\n int dp[101][102];\nprivate:\n int f(int curr, int prev, vector<string>& strs){\n if(curr == n) return 0;\n if(dp[curr][prev+1] != -1) return dp[curr][prev+1];\n int ans = 0;\n if(prev == -1){\n ans = max(f(curr+1,prev,strs), 1+f(curr+1,curr,strs));\n }\n else{\n bool flag = true;\n for(int i=0; i<m; i++){\n if(strs[i][prev] > strs[i][curr]){\n flag = false;\n break;\n }\n }\n if(flag){\n ans = max(f(curr+1,prev,strs), 1+f(curr+1,curr,strs));\n }\n else{\n ans = f(curr+1,prev,strs);\n }\n }\n\n return dp[curr][prev+1] = ans;\n }\npublic:\n int minDeletionSize(vector<string>& strs) {\n m = strs.size();\n n = strs[0].size();\n memset(dp, -1, sizeof(dp));\n return n-f(0,-1,strs);\n }\n};\n```
2
0
['Array', 'String', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++']
0
delete-columns-to-make-sorted-iii
[C++] Simple Dynamic Programming || Memoization
c-simple-dynamic-programming-memoization-oiqg
Approach: \n We need to ensure that after all the deletions, all the strings remain lexicographically sorted (increasing order i.e prev_char<=current_char). \n
yagni99
NORMAL
2024-05-02T06:21:47.720549+00:00
2024-05-02T06:32:09.298911+00:00
118
false
**Approach**: \n* We need to ensure that after all the deletions, all the strings remain lexicographically sorted (increasing order i.e ```prev_char<=current_char```). \n* We maintain current index and a previous index as our dp states. At each current index we choose whether to delete that index or not (deleting will delete all the chars at that index in all strings). \n* Now while deleting there\'s no condition to check, but for not deleting a character (keeping it as it is) we must ensure that the ```previous_char <= current_char``` for all the strings to ensure increasing order. \n\n```\n\tint rows,cols;\n vector<vector<int>>dp;\n \n int solve(vector<string>&v,int currentIndex,int prevIndex){\n \n //Base condition\n if(currentIndex==cols){\n return 0;\n }\n \n if(dp[currentIndex][prevIndex]!=-1)return dp[currentIndex][prevIndex];\n \n int ans=INT_MAX;\n if(prevIndex==cols){\n ans=min(solve(v,currentIndex+1,currentIndex),1+solve(v,currentIndex+1,prevIndex));\n }\n else{\n\t //If I choose to keep, it should be greater than equal to prev char for all the strings. \n int canKeep=1; \n for(int i=0;i<rows;i++){\n if(v[i][currentIndex]<v[i][prevIndex]){\n canKeep=0;\n break;\n }\n }\n \n\t\t\t//keep\n if(canKeep)ans=min(ans,solve(v,currentIndex+1,currentIndex));\n\t\t\t\n\t\t\t//delete\n ans=min(ans,1+solve(v,currentIndex+1,prevIndex));\n }\n \n return dp[currentIndex][prevIndex]=ans;\n }\n \n int minDeletionSize(vector<string>& strs) {\n rows=strs.size();\n cols=strs[0].size();\n dp.resize(cols+1,vector<int>(cols+1,-1));\n // (prevIndex = cols) shows that no prevIndex is chosen currently (better than using -1 to reduce extra lines of code)\n return solve(strs,0,cols);\n }\n```\n\n**Time Complexity**: O(n^3) = O(rows * cols * cols)\n**Space Complexity**: O(n^2) = O(cols) (recursion depth) + O(rows * cols) (for memoization)
2
0
['Dynamic Programming', 'Recursion', 'Memoization']
0
delete-columns-to-make-sorted-iii
C++| Recursion + Memo
c-recursion-memo-by-kumarabhi98-ofen
In simple words, a String in lexicographically increasing if last character is smaller than or equal to the current character in a string.\n1. Using this idea,
kumarabhi98
NORMAL
2022-03-25T13:58:10.906498+00:00
2022-03-25T14:00:22.369931+00:00
342
false
In simple words, a String in lexicographically increasing if last character is smaller than or equal to the current character in a string.\n1. Using this idea, try to delete every index keeping the record of `last` (closest previous index which is not deleted). \n1. There is also a possibility that we do not need to delete the current index if all the string is lexicographically increasing w.r.t `last` and `in`(current index).\n\nTake the min of the two result for every `in` and `last`. DP[i][j] represent minimum no. of deletion from `i`th index to end of string if the previous non-deleted index is `j`\n```\nclass Solution {\npublic:\n int dp[101][101];\n int dfs(vector<string>& strs,int n,int in,int last){ // in = current index, last = closest previous non-deleted index\n if(in>=n) return 0;\n if(dp[in][last]) return dp[in][last];\n \n bool st = 1; \n for(int i = 0; i<strs.size();++i){\n if(strs[i][in]<strs[i][last]) {st=0; break;} // check if strings are exicographically increasing\n\t\t\t //w.r.t to current and last index\n }\n \n int re = INT_MAX;\n int l = last;\n if(in==last) l=in+1;\n re = 1+dfs(strs,n,in+1,l); // delete the current index\n if(st) re = min(re, dfs(strs,n,in+1,in)); // if the strings are not lexicographically increasing,\n \t\t//than only choice we have is to delete the in\'th index\n \n return dp[in][last]=re;\n }\n int minDeletionSize(vector<string>& strs) {\n return dfs(strs,strs[0].size(),0,0);\n }\n};\n```
2
1
['Memoization', 'C']
0
delete-columns-to-make-sorted-iii
[C++] O(n^2) Alternate intuition (not LIS) Bottom-up DP. 16ms
c-on2-alternate-intuition-not-lis-bottom-sd2m
Define dp[i+1] as min deletions in strs[0...i] necessary to make every string in strs[0...i]\'s columns in lexicographical order where we are not deleting the
jossheim
NORMAL
2021-07-16T03:55:13.084624+00:00
2021-07-16T04:33:22.233589+00:00
252
false
Define `dp[i+1]` as min deletions in `strs[0...i]` necessary to make every string in `strs[0...i]`\'s columns in lexicographical order where we are **not deleting** the `i+1th` column (`strs[i]`).\n\n`m`: Number of strings in `strs`.\n\nRelation:\n```dp[i] = min {dp[k] + (i-k-1)} for all k in [0, i-1] such that strs[j][i-1] comes after strs[j][k-1] in lexicographic order for each j```\nWe can always remove all the columns `strs[0...i-2]` to have only the `i-1th` column so `dp[i]` will start with upper bound `i-1`.\n\nBase cases:\n`dp[0] = dp[1]= 0`.\n\nThe final answer is given by `dp[m+1]` where we assume that we have added the character `z` at the end of each string in `strs` since our dp condition says that we **must** pick up the `ith` column for `dp[i+1]` but we still want the freedom to delete the `mth` column(`strs[m-1]`). \n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int m = strs[0].size(), n= strs.size();\n vector<int> dp = vector<int>(m+2);\n for (int i = 2;i<=m+1;i++) {\n dp[i] = i-1;\n \n for (int k = 1;k <i;k++) { \n bool sat = true;\n if (i <= m) { // assuming we pushed \'z\' at the end of every string in strs, so below check is not needed for i == m+1\n for (auto & s: strs) { // checking if column i-1 is lexicographically larger or equal than column k-1.\n if (s[i-1] < s[k-1]) {\n sat = false;\n break;\n }\n } \n }\n \n if (sat) {\n dp[i] = min(dp[i], dp[k] + i-k-1);\n }\n }\n }\n return dp.back();\n }\n};\n```
2
0
[]
0
delete-columns-to-make-sorted-iii
Python3 - Longest Increasing Subsequence
python3-longest-increasing-subsequence-b-b1ar
For each column iterate all columns before it which forms a increasing sequence of characters which will give you the longest such subsequence. Then you can del
havingfun
NORMAL
2020-06-15T11:25:19.884179+00:00
2020-06-15T11:25:19.884215+00:00
277
false
For each column iterate all columns before it which forms a increasing sequence of characters which will give you the longest such subsequence. Then you can delete all other columns to get one of the longest subsequence.\n\nIntution -\nThink about how will you solve this problem if you have a single string ->\n1st Idea -> Sort the string. See the maximum length sorted substring in single string.\nThis can be easily applied to multiple strings.\n\n```\nclass Solution:\n def minDeletionSize(self, A: List[str]) -> int:\n dp = [1 for i in range(len(A[0]))]\n for i in range(1, len(A[0])):\n for j in range(i-1, -1, -1):\n poss = True\n for k in range(len(A)):\n if A[k][i] < A[k][j]:\n poss = False\n break\n if poss:\n dp[i] = max(dp[i], 1 + dp[j])\n return len(A[0]) - max(dp)\n```
2
0
[]
0
delete-columns-to-make-sorted-iii
easy python solution with memoization
easy-python-solution-with-memoization-by-uw69
for A = ["babca","bbazb"]\nfirst we conver it to\nb,b line 1\na,b line 2\nb,a line 3\nc,z line 4\na,b line 5\n\nwhether we should left line 1 or not is determin
bupt_wc
NORMAL
2018-12-16T04:11:44.940037+00:00
2018-12-16T04:11:44.940135+00:00
737
false
for A = ["babca","bbazb"]\nfirst we conver it to\n`b,b` line 1\n`a,b` line 2\n`b,a` line 3\n`c,z` line 4\n`a,b` line 5\n\nwhether we should left `line 1` or not is determined by which will get smaller `D.length`.\nif we do not delete this line, then the next line\'s element should all be larger than this line, otherwise we must delete next line\n\nwe define function `solve(A, index, start)` means the `D.length` if we use `start` as the first element in `A[index:]`.\n`index` means the line which we are considering\n`start` is a list, represent the last line we choose left.\n\nso if `any(A[index][i] < start[i])`, we delete this line, and analyse solve(A, index+1, start)\n\notherwise, if `all(A[index][i] >= start[i])`, we choose the min D.length in case `left A[index]` and case `delete A[index]`, e.g. `min(solve(A,index+1,tuple(A[index])), 1+solve(A,index+1, start))`\n\ncause there are many double count, so I use `tuple format` to represent `start`\n```python\nclass Solution(object):\n def minDeletionSize(self, A):\n cache = {}\n def solve(A, index, start):\n if index >= len(A): return 0\n if (index, start) in cache: return cache[index,start]\n \n if any(A[index][j] < start[j] for j in range(len(A[index]))):\n cache[index,start] = 1 + solve(A, index+1, start)\n return cache[index,start]\n \n cache[index,start] = min(solve(A,index+1,tuple(A[index])), 1+solve(A,index+1, start))\n return cache[index,start]\n \n return solve(list(zip(*A)), 0, tuple([\'a\']*len(A)))\n```
2
0
[]
1
delete-columns-to-make-sorted-iii
DP | Memoization | Pick/Notpick approach | C++
dp-memoization-picknotpick-approach-c-by-ezw9
\nclass Solution {\npublic:\n int dp[105][105];\n int solve(int ind, int n, int prev, vector<string>& strs) {\n if(ind == n) return 0;\n if(
siddh5599
NORMAL
2023-09-08T10:07:40.977690+00:00
2023-09-08T10:07:40.977720+00:00
318
false
```\nclass Solution {\npublic:\n int dp[105][105];\n int solve(int ind, int n, int prev, vector<string>& strs) {\n if(ind == n) return 0;\n if(prev >= 0 && dp[ind][prev] != -1) return dp[ind][prev];\n int minval = 150;\n int flag = 0;\n for(int i = 0;i < strs.size();i++) {\n if(prev == -1) break;\n if(prev >= 0 && strs[i][prev] > strs[i][ind]) {\n flag = 1;\n break;\n }\n }\n //pick\n if(flag == 0) minval = solve(ind+1, n, ind, strs);\n \n //not-pick\n minval = min(minval, 1+solve(ind+1, n, prev, strs));\n if(prev >= 0) return dp[ind][prev] = minval;\n return minval;\n }\n \n int minDeletionSize(vector<string>& strs) {\n memset(dp, -1, sizeof dp);\n int n = strs[0].size();\n return solve(0, n, -1, strs);\n }\n};\n```
1
0
['Dynamic Programming', 'Memoization']
1
delete-columns-to-make-sorted-iii
java easiest solution
java-easiest-solution-by-codewithlegend-vr3k
Code\n\nclass Solution {\npublic int minDeletionSize(String[] strs) {\n int n = strs.length;\n int m = strs[0].length();\n int[] dp = new int[m];\n
CodeWithLegend
NORMAL
2023-05-11T14:55:40.746191+00:00
2023-05-11T14:55:40.746276+00:00
105
false
# Code\n```\nclass Solution {\npublic int minDeletionSize(String[] strs) {\n int n = strs.length;\n int m = strs[0].length();\n int[] dp = new int[m];\n \n int overallMax = 0;\n for(int i=0;i<m;i++){\n dp[i] = 1;\n for(int j=0;j<i;j++){\n \n if(isValid(strs,i,j)){\n dp[i] = Math.max(dp[i],dp[j]+1);\n }\n \n }\n overallMax = Math.max(dp[i],overallMax);\n }\n \n return m-overallMax;\n}\n\nprivate boolean isValid(String[] strs ,int i,int j){\n for(int k=0;k<strs.length;k++){\n if(strs[k].charAt(j)>strs[k].charAt(i))\n return false;\n }\n return true;\n}\n}\n```
1
0
['Java']
0
delete-columns-to-make-sorted-iii
EASY UNDERSTAND 70% FASTER C++ SOLUTION
easy-understand-70-faster-c-solution-by-dwz0p
\nclass Solution {\npublic:\n int check(vector<string> &v,int i,int j){\n for(int k = 0; k < v.size(); k++){\n if(v[k][j]>v[k][i])\n
abhay_12345
NORMAL
2022-12-01T09:07:50.080864+00:00
2022-12-01T09:07:50.080905+00:00
427
false
```\nclass Solution {\npublic:\n int check(vector<string> &v,int i,int j){\n for(int k = 0; k < v.size(); k++){\n if(v[k][j]>v[k][i])\n return 0;\n }\n return 1;\n }\n int minDeletionSize(vector<string>& strs) {\n vector<int> dp(strs[0].length(),1);\n int ans = 0;\n for(int i = 0; i < strs[0].length(); i++){\n for(int j = 0; j < i; j++){\n if(check(strs,i,j)){\n dp[i] = max(dp[j]+1,dp[i]);\n }\n }\n ans = max(ans,dp[i]);\n }\n return strs[0].length()-ans;\n }\n};\n```
1
0
['Dynamic Programming', 'Depth-First Search', 'C', 'C++']
1
delete-columns-to-make-sorted-iii
[Python3] O(mn^2) DP
python3-omn2-dp-by-cava-uygs
Intuition\n Describe your first thoughts on how to solve this problem. \nThis problem is similar to longest non-decrease subseqence, except we need to consider
cava
NORMAL
2022-11-30T09:34:39.962255+00:00
2022-11-30T09:34:39.962299+00:00
183
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis problem is similar to longest non-decrease subseqence, except we need to consider all strings at the same time.\n\n\n# Code\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n m, n = len(strs), len(strs[0])\n dp = [1] * n\n for i in range(1, n):\n for j in range(i):\n if all(strs[k][i] >= strs[k][j] for k in range(m)):\n dp[i] = max(dp[i], dp[j] + 1)\n return n - max(dp)\n```
1
0
['Python3']
0
delete-columns-to-make-sorted-iii
JavaScript Solution
javascript-solution-by-crisdev2705-4m8m
Complexity\n- Memory: 42.8 MB\n\n- Runtime: 103 ms\n\n- Beats: 100%\n\n# Code\n\n/**\n * @param {string[]} strs\n * @return {number}\n */\nvar minDeletionSize =
crisdev2705
NORMAL
2022-11-18T04:46:40.934270+00:00
2022-11-18T04:48:42.819892+00:00
167
false
# Complexity\n- Memory: 42.8 MB\n\n- Runtime: 103 ms\n\n- Beats: 100%\n\n# Code\n```\n/**\n * @param {string[]} strs\n * @return {number}\n */\nvar minDeletionSize = function(strs) {\n if (strs.length === 0) return 0\n let dp = new Array(strs[0].length).fill(1)\n for (let i = 1; i < strs[0].length; i++) {\n for (let j = 0; j < i; j++) {\n let flag = true\n for (let k = 0; k < strs.length; k++) {\n if (strs[k][j] > strs[k][i]) {\n flag = false\n break\n }\n }\n if (flag) {\n dp[i] = Math.max(dp[i], dp[j] + 1)\n }\n }\n }\n return strs[0].length - Math.max(...dp)\n};\n```
1
0
['JavaScript']
0
delete-columns-to-make-sorted-iii
[C++] | MEMOIZATION | DYNAMIC PROGRAAMING
c-memoization-dynamic-prograaming-by-jat-q8bv
\nclass Solution {\npublic:\n int dp[101][101];\n int helper(vector<string>&strs, int i, int j){\n if(i==strs[0].length()){\n return 0;\
jatinbansal1179
NORMAL
2022-10-01T08:18:23.304192+00:00
2022-10-01T08:18:23.304234+00:00
318
false
```\nclass Solution {\npublic:\n int dp[101][101];\n int helper(vector<string>&strs, int i, int j){\n if(i==strs[0].length()){\n return 0;\n }\n if(dp[i][j]!=-1){\n return dp[i][j];\n }\n int ans = INT_MIN;\n bool b = false;\n \n for(int k = 0; k < strs.size();k++){\n if(strs[k][i]<strs[k][j]){\n b = true;\n break;\n }\n }\n if(b == false){\n \n ans = 1 + helper(strs,i+1,j);\n \n ans = min(ans,helper(strs,i+1,i));\n }\n else{\n ans = 1 + helper(strs,i+1,j);\n }\n return dp[i][j] = ans;\n \n \n }\n \n \n int minDeletionSize(vector<string>& strs) {\n // vector<char> last(strs.size(),\'A\');\n // if(strs[0].length()==1) return 0;\n memset(dp,-1,sizeof(dp));\n for(int i = 0; i < strs.size();i++){\n strs[i] = \'A\' + strs[i];\n }\n return helper(strs,1,0);\n }\n};\n```
1
1
['Dynamic Programming', 'Recursion', 'Memoization', 'C', 'C++']
1
delete-columns-to-make-sorted-iii
Delete Columns to Make Sorted III Solution Java
delete-columns-to-make-sorted-iii-soluti-ef2v
class Solution {\n public int minDeletionSize(String[] A) {\n final int n = A[0].length();\n // dp[i] := LIS ending at A[*][i]\n int[] dp = new int[n]
bhupendra786
NORMAL
2022-09-26T08:20:20.889886+00:00
2022-09-26T08:20:20.889924+00:00
270
false
class Solution {\n public int minDeletionSize(String[] A) {\n final int n = A[0].length();\n // dp[i] := LIS ending at A[*][i]\n int[] dp = new int[n];\n Arrays.fill(dp, 1);\n\n for (int i = 1; i < n; ++i)\n for (int j = 0; j < i; ++j)\n if (isSorted(A, j, i))\n dp[i] = Math.max(dp[i], dp[j] + 1);\n\n return n - Arrays.stream(dp).max().getAsInt();\n }\n\n private boolean isSorted(String[] A, int j, int i) {\n for (final String a : A)\n if (a.charAt(j) > a.charAt(i))\n return false;\n return true;\n }\n}\n
1
0
['Array', 'String', 'Dynamic Programming']
0
delete-columns-to-make-sorted-iii
Java Basic Dp
java-basic-dp-by-mi1-ap3k
\nclass Solution {\n public int minDeletionSize(String[] strs) {\n for(int i=0;i<strs.length;i++){\n strs[i] = "0"+strs[i];\n }\n
mi1
NORMAL
2021-11-06T05:38:28.000759+00:00
2021-11-06T05:38:28.000805+00:00
99
false
```\nclass Solution {\n public int minDeletionSize(String[] strs) {\n for(int i=0;i<strs.length;i++){\n strs[i] = "0"+strs[i];\n }\n int[][]dp= new int[105][105];\n for(int[]row:dp) Arrays.fill(row,-1);\n return solve(strs,0,1,dp);\n }\n \n private int solve(String [] strs,int prev,int idx,int[][]dp){\n if(idx>=strs[0].length()){\n return 0;\n }\n if(dp[prev][idx]!=-1){\n return dp[prev][idx];\n }\n if(!isValid(strs,prev,idx)){\n int count = 1+solve(strs,prev,idx+1,dp);\n return dp[prev][idx]=count;\n }else{\n int count = 1+solve(strs,prev,idx+1,dp);\n count = Math.min(count,solve(strs,idx,idx+1,dp));\n return dp[prev][idx]=count;\n }\n }\n \n private boolean isValid(String[] strs,int prev,int idx){\n for(int i=0;i<strs.length;i++){\n String curr = strs[i];\n if(curr.charAt(idx)<curr.charAt(prev)){\n return false;\n }\n }\n return true;\n }\n \n}\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
Java DP
java-dp-by-prathihaspodduturi-37q8
\nclass Solution {\n public boolean check(String s[],int i,int j)\n {\n for(int p=0;p<s.length;p++)\n {\n if(s[p].charAt(i)>s[p].
prathihaspodduturi
NORMAL
2021-10-16T08:51:37.837432+00:00
2021-10-16T08:51:37.837462+00:00
83
false
```\nclass Solution {\n public boolean check(String s[],int i,int j)\n {\n for(int p=0;p<s.length;p++)\n {\n if(s[p].charAt(i)>s[p].charAt(j))\n return false;\n }\n return true;\n }\n public int minDeletionSize(String[] s) {\n int n=s.length,m=s[0].length(),ans=1;\n int dp[]=new int[m];\n for(int i=m-1;i>=0;i--)\n {\n dp[i]=1;\n for(int j=i+1;j<m;j++)\n {\n if(check(s,i,j)==true)\n dp[i]=Math.max(1+dp[j],dp[i]);\n }\n ans=Math.max(ans,dp[i]);\n }\n return m-ans;\n }\n}\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
Python3 maximum increasing sequence with explanation (Runtime: 96 ms, faster than 100.00%)
python3-maximum-increasing-sequence-with-kcnh
Explanation:\nSave to dp[i] max len of possible string can be made out of all strings of str[0:i] when all satisfy the requirement.\nSo, first dp[0] is 1, becau
MVR11
NORMAL
2021-10-06T21:47:34.077788+00:00
2021-10-06T21:47:34.079699+00:00
124
false
**Explanation:**\nSave to dp[i] max len of possible string can be made out of all strings of str[0:i] when all satisfy the requirement.\nSo, first dp[0] is 1, because if all strs are 1 char, then all satisfy the requirements.\nFor dp[1]:\n- check if all str[0] and [1] satisfy requirements. If so, save to dp[1] max of (already saved, dp[0]+1)\nFor dp[2]\n- check if all str[0] and [2] satisfy requirements. If so, save to dp[2] max of (already saved, dp[0]+1)\n- check if all str[1] and [2] satisfy requirements. If so, save to dp[2] max of (already saved, dp[1]+1)\nFor dp[3] - same, compare with 0,1,2.\nAnd so on.\n\nNow we have dp that has information about longest string can be made before i, like [1,2,3,2,2]\nTo know minimum possible deletion of chars we get max(dp) and substract from len(str)\n\nTo speed up and exit from subiteration we added break (instead of more beautiful realisation with all() )\n\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n n = len(strs[0])\n dp = [1] * n\n for i in range(1, n):\n for j in range(i):\n ok = True\n for s in strs:\n if s[i] < s[j]:\n ok = False\n break\n if ok:\n dp[i] = max(dp[i], dp[j] + 1)\n return n - max(dp)\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
C++ | Solution | Fast and short
c-solution-fast-and-short-by-paraslaul-gl9d
\n int minDeletionSize(vector<string>& strs) {\n// so what question demands is internal sorting of strings (not among each other)\n// so what
ParasLaul
NORMAL
2021-06-12T22:48:47.307330+00:00
2021-06-12T22:48:47.307372+00:00
214
false
```\n int minDeletionSize(vector<string>& strs) {\n// so what question demands is internal sorting of strings (not among each other)\n// so what this is a good variation of LIS and truly a hard question\n// so let me draw what i want to explain\n// 1. k-> b a b c a\n// 2. k-> b b a z b\n// i\n// j\n// so at every i th and jth column we make comparison in each kth row if(strs[k][j]>strs[k][i]) then we have to remove this column\n// optimal way is to just count columns which does not violate above conditions because that are columns we want to keep ;)\n// at last our answer -----> total columns(n) - columns to keep(should max so that we can minimise no of columns to be deleted)\n// Lets code it;\n \n int i,j, k;\n int n=strs[0].size();\n int m=strs.size();\n vector<int>dp(n,1);\n for(i=0;i<n;i++){\n for(j=0;j<i;j++){\n for(k=0;k<m;k++){\n if(strs[k][j]>strs[k][i]) break;\n }\n if(k==m&&dp[i]<1+dp[j]) dp[i]=1+dp[j];\n }\n \n }\n return n - *max_element(dp.begin(), dp.end());\n }\n};\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
Solution in Go + Explanation
solution-in-go-explanation-by-lutraman-0y53
After a long time and some pain, I figured that this problem can be seen as a graph problem:\n\nlet\'s look at this example: ["aaaaabaa", "abcdefgb"]\nrearrange
lutraman
NORMAL
2021-04-09T11:05:27.907263+00:00
2021-04-26T09:30:42.175793+00:00
107
false
After a long time and some pain, I figured that this problem can be seen as a graph problem:\n\nlet\'s look at this example: ["aaaaabaa", "abcdefgb"]\nrearranged to a matrix shape:\n```\n\ta a a a a b a a\n\ta b c d e f g b\n```\nWe can think of each column in the matrix as a node in the graph\n\nFrom each node to another node there is a path if and only if:\n1. the source node is a column from the left of the destination node (i.e. we can only move from left to right)\n2. all the rows in the column of the source node are smaller or equal to their corrosponding row in the destination node\n\nfor example, there is a path from column 0 to column 7 (last one) because for the first string [0] and [7] are both \'a\' and for the second [0] is a and [7] is b\n\n```\n\ta ----------> a\n\ta ----------> b\n```\nThe cost of which is 6, since we have to delete 6 characters to get there directly. The cost can be calculated `nextColumn - column - 1` where `column` and `nextColumn` are index numbers.\n\nHowever, there is no route from 6 to 7:\n```\n\ta ----------> a\n\tg ----X-----> b\n```\n\nMoving from one column to the next is free (if possible), no deletions.\n\nLet\'s also add a Start and an End node to our graph (they are placeholders)\n```\nS a a a a a b a a E\nS\ta b c d e f g b E\n```\nand rephrase our question as follows:\n**What\'s the cheapest route to get from S to E?**\n\nNow the question looks much simpler. At first I thought of Dijkstra, but it was just too complex to impement in this scenario (setting up a priority queue in go is a bit too much work for what we need). Then I tried DFS, but that was too slow. So I added some memory to the solution in the form of an array keeping the cheapest route to evey node at any point of the search, and now I get to call it dynamic programming :-)\n\nSo, here\'s the solution in Go. A DFS with memory:\n\n```go\nfunc minDeletionSize(strs []string) int {\n deletionsToColumn := make([]int, len(strs[0])+1) // deletionsToColumn is 1 place longer, to accomodate for the very important End node\n\t// this loop simulates going from our placeholder start node to every other node (the cost is just the column index)\n\t// also there will always be a path from S to any other node, it translates to where we choose to start\n for i := 0; i < len(deletionsToColumn); i++ {\n deletionsToColumn[i] = i\n }\n return minDeletionsRoute(strs, deletionsToColumn)\n}\n\n// I guess there was no real need for me to split to another function, just a remainder of my previous DFS attempt\nfunc minDeletionsRoute(strs []string, deletionsToColumn []int) int {\n for column := 0; column < len(strs[0]); column++ {\n for nextColumn := column + 1; nextColumn < len(deletionsToColumn); nextColumn++ {\n\t\t\t// it\'s cheaper (computationally) to check if the cost of the edge is cheaper than the previous best way to arrive to it, so that condition comes before the call to hasRoute\n if deletionsToColumn[nextColumn] >= deletionsToColumn[column] + nextColumn - column - 1 && hasRoute(strs, column, nextColumn) {\n deletionsToColumn[nextColumn] = deletionsToColumn[column] + nextColumn - column - 1\n }\n }\n }\n return deletionsToColumn[len(deletionsToColumn)-1]\n}\n\n// hasRoute checks that all the chars in the column (c) are in order with the chars in the next column (nc)\nfunc hasRoute(strs []string, c, nc int) bool {\n if c == -1 || nc == len(strs[0]) { return true } // if nc is out of bounds, that means it\'s the end node, to which all nodes are free to reach (at a cost!)\n for si := 0; si < len(strs); si++ {\n if strs[si][c] > strs[si][nc] {\n return false\n }\n } \n return true\n}\n```\n
1
0
['Dynamic Programming', 'Depth-First Search', 'Graph', 'Go']
0
delete-columns-to-make-sorted-iii
[Rust] 100%
rust-100-by-nickx720-n651
\nimpl Solution {\n pub fn min_deletion_size(a: Vec<String>) -> i32 {\n let len = a[0].len();\n let mut dp: Vec<i32> = vec![1; len];\n let mut r
nickx720
NORMAL
2020-12-01T11:03:05.113883+00:00
2020-12-01T11:03:05.115136+00:00
84
false
```\nimpl Solution {\n pub fn min_deletion_size(a: Vec<String>) -> i32 {\n let len = a[0].len();\n let mut dp: Vec<i32> = vec![1; len];\n let mut res = 1;\n for i in 0..len {\n for j in 0..i {\n let mut flag: bool = false;\n for item in a.iter() {\n let item = item.chars().collect::<Vec<char>>();\n if item[i] >= item[j] {\n flag = true;\n continue;\n }\n flag = false;\n break;\n }\n if flag {\n dp[i] = std::cmp::max(dp[i], 1 + dp[j]);\n res = std::cmp::max(res, dp[i]);\n }\n }\n }\n let len = len as i32;\n len - res\n }\n}\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
Longest Increasing Subsequence - 1D DP
longest-increasing-subsequence-1d-dp-by-vsq0f
For general LIS problem, there is a nice O(nlogn) solution https://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/LongestIncreasingSubsequence.pd
huangdachuan
NORMAL
2020-10-15T22:58:42.201100+00:00
2020-10-15T22:58:42.201144+00:00
178
false
For general LIS problem, there is a nice O(nlogn) solution https://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/LongestIncreasingSubsequence.pdf\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& A) {\n int n = A[0].size();\n vector<int> dp(n, 1);\n int lis = 1;\n for (int i = 1; i < n; ++i) {\n for (int j = 0; j < i; ++j) {\n if (lexiOrder(A, j, i)) {\n dp[i] = max(dp[i], dp[j] + 1);\n }\n }\n lis = max(lis, dp[i]);\n }\n return n - lis;\n }\n\nprivate:\n bool lexiOrder(vector<string>& A, int j, int i) {\n for (int r = 0; r < A.size(); ++r) {\n if (A[r][i] < A[r][j]) {\n return false;\n }\n }\n return true;\n }\n};\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
Simple Java sol
simple-java-sol-by-cuny-66brother-yx49
\nclass Solution {\n public int minDeletionSize(String[] A) {\n int dp[]=new int[A[0].length()];\n int res=1;\n Arrays.fill(dp,1);\n
CUNY-66brother
NORMAL
2020-05-01T22:48:26.404305+00:00
2020-05-01T22:48:26.404360+00:00
128
false
```\nclass Solution {\n public int minDeletionSize(String[] A) {\n int dp[]=new int[A[0].length()];\n int res=1;\n Arrays.fill(dp,1);\n for(int i=1;i<A[0].length();i++){\n int longest=1;\n for(int j=i-1;j>=0;j--){\n boolean all=true;\n for(int k=0;k<A.length;k++){\n if(A[k].charAt(i)<A[k].charAt(j))all=false;\n }\n if(all)longest=Math.max(longest,1+dp[j]);\n }\n dp[i]=longest;\n }\n for(int n:dp)res=Math.max(res,n);\n return A[0].length()-res;\n }\n}\n```
1
0
[]
0
delete-columns-to-make-sorted-iii
Accepted C# DP Solution
accepted-c-dp-solution-by-maxpushkarev-hya0
\n public class Solution\n {\n public int MinDeletionSize(string[] rows)\n {\n int length = rows[0].Length;\n int [] d
maxpushkarev
NORMAL
2020-03-18T02:55:49.301123+00:00
2020-03-18T02:55:49.301177+00:00
90
false
```\n public class Solution\n {\n public int MinDeletionSize(string[] rows)\n {\n int length = rows[0].Length;\n int [] dp = new int[length];\n\n for (int i = length - 1; i >= 0; i--)\n {\n dp[i] = length - i - 1;\n\n for (int j = i + 1; j < length; j++)\n {\n int inner = (j - i) - 1;\n bool valid = true;\n\n foreach (var row in rows)\n {\n if (row[i] > row[j])\n {\n valid = false;\n break;\n }\n }\n\n if (valid)\n {\n dp[i] = Math.Min(dp[i], dp[j] + inner);\n }\n }\n }\n\n int res = int.MaxValue;\n for (int i = 0; i < length; i++)\n {\n res = Math.Min(res, dp[i] + i);\n }\n\n return res;\n }\n }\n```
1
1
['Dynamic Programming']
0
delete-columns-to-make-sorted-iii
[C++] DP with Memo
c-dp-with-memo-by-shaktirajpandey1996-4xcw
Assumptions:-\nDP[i][j] will give information about minimum number of deletion required to make array index upto i lexicographically sorted where j is the previ
shaktirajpandey1996
NORMAL
2019-08-28T18:09:36.432235+00:00
2019-08-28T18:09:36.432272+00:00
303
false
Assumptions:-\n**DP[i][j]** will give information about ***minimum number of deletion*** required to make array index upto **i** lexicographically sorted where j is the previously used value.\n\n```\nconst int inf = 1e8;\nclass Solution {\npublic:\n int N, M;\n vector<string > P;\n vector<vector<int> > DP;\n int solve(int id, int prev){\n if(id >= N){\n return 0;\n }\n if(DP[id][prev] != -1){\n return DP[id][prev];\n }\n int ans = inf;\n int f = 0;\n for(int i=0;i<P.size();i++){\n if ( P[i][id] < P[i][prev]){\n f = 1;\n break;\n }\n }\n if(f == 0){\n ans = solve(id+1,id);\n }\n ans = min(ans,1 + solve(id+1, prev));\n return DP[id][prev]=ans; \n }\n\n int minDeletionSize(vector<string>& A) {\n for(int i=0;i<A.size();i++){\n P.push_back("_"+A[i]);\n }\n N = P[0].size();\n M = P.size();\n DP.resize(N+5,vector<int > (N+5,-1));\n return solve(1,0);\n }\n};\n\n\n```
1
0
[]
1
delete-columns-to-make-sorted-iii
Java 10ms DP solution
java-10ms-dp-solution-by-brianjr-8dg9
\nclass Solution {\n public int minDeletionSize(String[] A) {\n int len = A[0].length();\n int min = len;\n int[] dp = new int[len];\n
brianjr
NORMAL
2018-12-16T18:44:01.095058+00:00
2018-12-16T18:44:01.095146+00:00
180
false
```\nclass Solution {\n public int minDeletionSize(String[] A) {\n int len = A[0].length();\n int min = len;\n int[] dp = new int[len];\n for(int i=1; i<len; i++) {\n dp[i] = i;\n for(int j=i-1; j>=0; j--) {\n if(canCompare(A, i, j)) {\n dp[i] = Math.min(dp[i], dp[j]+(i-j-1));\n }\n }\n min = Math.min(min, dp[i]+(len-i-1));\n }\n return min;\n }\n \n private boolean canCompare(String[] A, int x, int y) {\n for(int i=0; i<A.length; i++) {\n if(A[i].charAt(x)<A[i].charAt(y)) {\n return false;\n }\n }\n return true;\n }\n}\n```
1
1
[]
0
delete-columns-to-make-sorted-iii
[Java] DP | Longest Increasing Subsequence
java-dp-longest-increasing-subsequence-b-ihss
My solution is O(kn^2) \n\n\n public int minDeletionSize(String[] A) {\n int[] dp = new int[A[0].length()];\n int max = 1;\n dp[0] = 1;\
gcarrillo
NORMAL
2018-12-16T04:00:53.003447+00:00
2018-12-16T04:00:53.003519+00:00
346
false
My solution is ```O(kn^2)``` \n\n```\n public int minDeletionSize(String[] A) {\n int[] dp = new int[A[0].length()];\n int max = 1;\n dp[0] = 1;\n for(int i = 1; i < dp.length; i++){\n dp[i] = 1;\n for(int j = 0; j < i; j++){\n boolean mark = true;\n\t\t// if all characters in column i are greater than j then its a valid subsequence\n for(String k : A){\n if(k.charAt(j) > k.charAt(i)){\n mark = false;\n break;\n }\n \n }\n \n if(mark)\n dp[i] = Math.max(dp[j] + 1, dp[i]);\n }\n \n max = Math.max(max,dp[i]);\n }\n \n return A[0].length() - max;\n \n }\n```
1
1
[]
1
delete-columns-to-make-sorted-iii
C++ DP Solution based on LIS
c-dp-solution-based-on-lis-by-gengar33-ic61
IntuitionInstead of caring for what to delete, think of what to keep. Since the questions says lexicographic, we can think of LIS approach.ApproachComplexity T
gengar33
NORMAL
2025-03-14T04:48:10.111023+00:00
2025-03-14T04:48:10.111023+00:00
5
false
# Intuition Instead of caring for what to delete, think of what to keep. Since the questions says lexicographic, we can think of LIS approach. # Approach Idea here is to use LIS. If you see the first 2 loops they are pretty much standard LIS way. Now for each string, we check that if the order is non-decreasing for s[i] and s[j] where j < i If for any string, we see that s[i] >= s[j] does not hold true, that means index i characters need to be removed for all strs The k == strs.size check ensures that we have not come out of the loop due to above reason we increase the LIS length by 1 At the end, we find the max LIS length. This represents the no. of characters we are going to keep in all strings. So n-MaxLISLength is the total deleted characters. # Complexity - Time complexity: O(StringLength * StringLength * strs.length) - Space complexity: O(StringLength) # Code ```cpp [] class Solution { public: int minDeletionSize(vector<string>& strs) { int n = strs[0].length(); vector<int> dp(n, 1); for(int i = 1; i<n; i++) { for(int j=0; j<i; j++) { int k =0 ; while(k < strs.size()) { if(strs[k][i] < strs[k][j]) { break; } k++; } if(k == strs.size()) { dp[i] = max(dp[i], 1 + dp[j]); } } } int res = 0; for(int i: dp) res = max(res, i); return n-res; } }; ```
0
0
['Dynamic Programming', 'C++']
0
delete-columns-to-make-sorted-iii
C++ DP Solution based on LIS
c-dp-solution-based-on-lis-by-gengar33-ug2j
IntuitionInstead of caring for what to delete, think of what to keep. Since the questions says lexicographic, we can think of LIS approach.ApproachComplexity T
gengar33
NORMAL
2025-03-14T04:45:24.671422+00:00
2025-03-14T04:45:24.671422+00:00
7
false
# Intuition Instead of caring for what to delete, think of what to keep. Since the questions says lexicographic, we can think of LIS approach. # Approach Idea here is to use LIS. If you see the first 2 loops they are pretty much standard LIS way. Now for each string, we check that if the order is non-decreasing for s[i] and s[j] where j < i If for any string, we see that s[i] >= s[j] does not hold true, that means index i characters need to be removed for all strs The k == strs.size check ensures that we have not come out of the loop due to above reason we increase the LIS length by 1 At the end, we find the max LIS length. This represents the no. of characters we are going to keep in all strings. So n-MaxLISLength is the total deleted characters. # Complexity - Time complexity: O(StringLength * StringLength * strs.length) - Space complexity: O(StringLength) # Code ```cpp [] class Solution { public: int minDeletionSize(vector<string>& strs) { int n = strs[0].length(); vector<int> dp(n, 1); for(int i=n-2; i>=0; i--) { for(int j = i+1; j < n; j++) { int c=0; for(string s: strs) { if(s[i] > s[j]) break; c++; } if(c == strs.size()) dp[i] = max(dp[i], 1+dp[j]); } } int res = 0; for(int i: dp) res = max(res, i); return n-res; } }; ```
0
0
['Dynamic Programming', 'C++']
0
delete-columns-to-make-sorted-iii
960. Delete Columns to Make Sorted III
960-delete-columns-to-make-sorted-iii-by-hl4u
IntuitionApproachComplexity Time complexity: Space complexity: Code
G8xd0QPqTy
NORMAL
2025-01-02T03:48:18.851028+00:00
2025-01-02T03:48:18.851028+00:00
6
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code ```python [] class Solution: def minDeletionSize(self, strs): n = len(strs) m = len(strs[0]) dp = [1] * m for j in range(1, m): for i in range(j): if all(strs[k][i] <= strs[k][j] for k in range(n)): dp[j] = max(dp[j], dp[i] + 1) return m - max(dp) ```
0
0
['Python']
0
delete-columns-to-make-sorted-iii
EASY PICK NOT PICK SOLUTION
easy-pick-not-pick-solution-by-uppalaaka-dhuq
IntuitionApproachComplexity Time complexity: Space complexity: Code
UppalaAkash
NORMAL
2025-01-01T13:31:08.627559+00:00
2025-01-01T13:31:08.627559+00:00
14
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code ```cpp [] class Solution { public: int n; int dp[105][105]; int solve(int ind, int prev, vector<string>& strs) { if (ind >= n) return 0; if (dp[ind][prev + 1] != -1) return dp[ind][prev + 1]; int flag = 0; for (int i = 0; i < strs.size(); i++) { if (prev != -1 && strs[i][prev] > strs[i][ind]) { flag = 1; break; } } int minval = INT_MAX; // We don't need to remove the index if (!flag) minval = solve(ind + 1, ind, strs); // Consider deleting the current index minval = min(minval, 1 + solve(ind + 1, prev, strs)); return dp[ind][prev + 1] = minval; } int minDeletionSize(vector<string>& strs) { memset(dp, -1, sizeof dp); n = strs[0].size(); return solve(0, -1, strs); } }; ```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
100% Memory || 100% Runtime || DP || Memoized || with witout approach
100-memory-100-runtime-dp-memoized-with-cw2hy
IntuitionApproachComplexity Time complexity: Space complexity: Code
twinwal
NORMAL
2024-12-17T12:03:14.711414+00:00
2024-12-17T12:03:14.711414+00:00
4
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```kotlin []\nclass Solution {\n fun minDeletionSize(strs: Array<String>): Int {\n val dp = Array(strs[0].length) { Array<Int?>(strs[0].length) { -1 } }\n return minDeletionSize1(strs, 0, -1, dp) ?: -1\n }\n\n fun minDeletionSize1(strs: Array<String>, index: Int, pre: Int, dp: Array<Array<Int?>>): Int? {\n if (index == strs[0].length) {\n return 0\n }\n if(dp[index][pre + 1] != -1) return dp[index][pre + 1]\n val without = minDeletionSize1(strs, index + 1, pre, dp)?.let { 1 + it }\n var with: Int? = null\n var isWithValid = true\n for (i in 0..strs.lastIndex) {\n if (pre != -1 && strs[i][index] < strs[i][pre]) {\n isWithValid = false\n break\n }\n }\n if (isWithValid) {\n with = minDeletionSize1(strs, index + 1, index, dp)\n }\n dp[index][pre + 1] = listOfNotNull(with, without).minOrNull()\n return dp[index][pre + 1]\n }\n}\n```
0
0
['Kotlin']
0
delete-columns-to-make-sorted-iii
Python (Simple DP)
python-simple-dp-by-rnotappl-lkh2
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
rnotappl
NORMAL
2024-11-28T06:28:46.681586+00:00
2024-11-28T06:28:46.681622+00:00
3
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n def minDeletionSize(self, strs):\n n, m = len(strs), len(strs[0])\n\n @lru_cache(None)\n def function(cur,prev):\n if cur == m:\n return 0 \n\n min_val = 1 + function(cur+1,prev)\n\n if prev == -1 or all(strs[i][prev] <= strs[i][cur] for i in range(n)):\n min_val = min(min_val,function(cur+1,cur))\n\n return min_val \n\n return function(0,-1)\n```
0
0
['Python3']
0
delete-columns-to-make-sorted-iii
python3 recursive dp
python3-recursive-dp-by-0icy-n0vh
\n# Code\npython3 []\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n\n @cache\n def dp(prev,i):\n if i>=len(
0icy
NORMAL
2024-11-22T05:26:49.390307+00:00
2024-11-22T05:26:49.390357+00:00
3
false
\n# Code\n```python3 []\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n\n @cache\n def dp(prev,i):\n if i>=len(strs[0]):\n return 0\n t1 = 1+dp(prev,i+1)\n t2 = inf\n if prev == -1:\n t2 = dp(i,i+1)\n else:\n flag = True\n for e in strs:\n if e[i] >= e[prev]:\n continue\n else:\n flag = False\n break\n if flag:\n t2 = dp(i,i+1)\n return min(t1,t2)\n \n return dp(-1,0)\n```
0
0
['Python3']
0
delete-columns-to-make-sorted-iii
DP
dp-by-linda2024-2vy0
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
linda2024
NORMAL
2024-10-19T22:33:48.700246+00:00
2024-10-19T22:33:48.700262+00:00
3
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```csharp []\npublic class Solution {\n private bool ValidOrder(string[] strs, int idx1, int idx2)\n {\n int size = strs.Length;\n for(int i = 0; i < size; i++)\n {\n if(idx2 >= strs[i].Length)\n return false;\n\n if(strs[i][idx1]>strs[i][idx2])\n return false;\n }\n\n return true;\n }\n\n public int MinDeletionSize(string[] strs) {\n int row = strs.Length, col = strs[0].Length;\n int[] dp = Enumerable.Repeat(1, col).ToArray();\n\n int minDel = col;\n for(int i = 0; i < col; i++)\n {\n for(int j = 0; j < i; j++)\n {\n if(ValidOrder(strs, j, i))\n dp[i] = Math.Max(dp[i], dp[j]+1);\n }\n \n minDel = Math.Min(minDel, col-dp[i]);\n }\n\n return minDel;\n }\n}\n```
0
0
['C#']
0
delete-columns-to-make-sorted-iii
pyhthon
pyhthon-by-hassam_472-q4cb
\n# Code\npython3 []\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n if not strs:\n return 0\n \n n =
hassam_472
NORMAL
2024-10-19T13:15:46.149910+00:00
2024-10-19T13:15:46.149944+00:00
5
false
\n# Code\n```python3 []\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n if not strs:\n return 0\n \n n = len(strs)\n m = len(strs[0])\n \n # dp[i] represents the maximum length of the sorted subsequence ending at index i\n dp = [1] * m\n \n for i in range(1, m):\n for j in range(i):\n if all(s[j] <= s[i] for s in strs):\n dp[i] = max(dp[i], dp[j] + 1)\n \n return m - max(dp)\n```
0
0
['Python3']
0
delete-columns-to-make-sorted-iii
Pick max indices
pick-max-indices-by-theabbie-9nn8
\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n m = len(strs[0])\n @lru_cache(maxsize = None)\n def dp(i, prev)
theabbie
NORMAL
2024-10-17T11:48:04.191704+00:00
2024-10-17T11:48:04.191727+00:00
0
false
```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n m = len(strs[0])\n @lru_cache(maxsize = None)\n def dp(i, prev):\n if i >= m:\n return 0\n take = True\n for s in strs:\n if prev != -1 and s[prev] > s[i]:\n take = False\n break\n res = dp(i + 1, prev)\n if take:\n res = max(res, 1 + dp(i + 1, i))\n return res\n return m - dp(0, -1)\n```
0
0
['Python']
0
delete-columns-to-make-sorted-iii
delete-columns-to-make-sorted-iii - TypeScript Solution
delete-columns-to-make-sorted-iii-typesc-uckr
\n\n# Code\ntypescript []\nfunction minDeletionSize(strs: string[]): number {\n const numRows = strs.length;\n const numCols = strs[0].length;\n const
himashusharma
NORMAL
2024-10-10T04:59:28.876193+00:00
2024-10-10T04:59:28.876229+00:00
2
false
\n\n# Code\n```typescript []\nfunction minDeletionSize(strs: string[]): number {\n const numRows = strs.length;\n const numCols = strs[0].length;\n const dp: number[] = new Array(numCols).fill(1);\n let minDeletions = numCols;\n\n for(let i = 0; i < numCols; ++i){\n for(let j = 0; j<i; ++j){\n let valid = true;\n for(let k = 0; k < numRows; ++k){\n if(strs[k][j] > strs[k][i]){\n valid = false;\n break;\n }\n }\n if(valid){\n dp[i] = Math.max(dp[i], dp[j] + 1)\n }\n }\n minDeletions = Math.min(minDeletions, numCols - dp[i]);\n }\n return minDeletions;\n \n};\n```
0
0
['Array', 'String', 'Dynamic Programming', 'TypeScript', 'JavaScript']
0
delete-columns-to-make-sorted-iii
Delete Column With DP
delete-column-with-dp-by-ansh1707-jt1l
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
Ansh1707
NORMAL
2024-10-04T01:09:28.093089+00:00
2024-10-04T01:09:28.093112+00:00
1
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python []\nclass Solution(object):\n def minDeletionSize(self, strs):\n n = len(strs[0])\n \n def checkRest(idx1, idx2):\n for i in range(1, len(strs)):\n if not strs[i][idx1] <= strs[i][idx2]: \n return False\n return True\n \n def lis(idx, memo):\n if idx in memo:\n return memo[idx]\n \n longest = 1\n for i in range(idx + 1, n):\n if strs[0][idx] <= strs[0][i] and checkRest(idx, i):\n longest = max(longest, 1 + lis(i, memo))\n \n memo[idx] = longest\n return longest\n\n longest = 0\n memo = {}\n for i in range(n):\n longest = max(longest, lis(i, memo))\n \n return n - longest\n\n\n```
0
0
['Python']
0
delete-columns-to-make-sorted-iii
Delete Columns to Make Sorted III
delete-columns-to-make-sorted-iii-by-sha-zmkk
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
Shaludroid
NORMAL
2024-09-06T21:24:23.664368+00:00
2024-09-06T21:24:23.664399+00:00
19
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int minDeletionSize(String[] strs) {\n int n = strs.length;\n int m = strs[0].length();\n \n // dp[j] represents the length of the longest increasing subsequence of columns ending at column j\n int[] dp = new int[m];\n Arrays.fill(dp, 1); // Each column itself can be a subsequence\n \n for (int j = 1; j < m; j++) {\n for (int i = 0; i < j; i++) {\n // Check if column i can come before column j in the sequence\n boolean valid = true;\n for (int k = 0; k < n; k++) {\n if (strs[k].charAt(i) > strs[k].charAt(j)) {\n valid = false;\n break;\n }\n }\n if (valid) {\n dp[j] = Math.max(dp[j], dp[i] + 1);\n }\n }\n }\n \n // The length of the longest subsequence is the maximum value in dp\n int lis = 0;\n for (int len : dp) {\n lis = Math.max(lis, len);\n }\n \n // Minimum deletions = total columns - length of longest increasing subsequence\n return m - lis;\n }\n}\n\n```
0
0
['Java']
0
delete-columns-to-make-sorted-iii
960. Delete Columns to Make Sorted III.cpp
960-delete-columns-to-make-sorted-iiicpp-p6d9
Code\n\nclass Solution {\npublic:\n vector<bitset<100>>t; \n void fill_table(vector<string>& strs){\n for(auto &w: strs){\n vector<vector<int>>v(26,
202021ganesh
NORMAL
2024-08-22T10:09:53.330250+00:00
2024-08-22T10:09:53.330281+00:00
1
false
**Code**\n```\nclass Solution {\npublic:\n vector<bitset<100>>t; \n void fill_table(vector<string>& strs){\n for(auto &w: strs){\n vector<vector<int>>v(26, vector<int>());\n v[w.back() - \'a\'].push_back(w.size()-1);\n for(int j = w.size()-2; j >= 0; j--){\n bitset<100>b;\n int id = w[j] - \'a\'; \n for(int k = id; k != 26; k++)\n for(auto id: v[k]) b[id] = 1; \n v[id].push_back(j); \n t[j] &= b;\n }\n }\n } \n int minDeletionSize(vector<string>& strs){\n int n = strs[0].size();\n t.resize(n-1);\n for(auto &b : t) b.set();\n fill_table(strs); \n vector<int>dp(n, 0); \n queue<pair<int, int>>; \n for(int i = 0; i != t.size(); i++)\n for(int j = 0; t[i].any(); j++)\n if(t[i][j]) dp[j] = max(dp[j], dp[i] + 1), t[i][j] = 0; \n return n - ( *max_element(dp.begin(), dp.end()) + 1);\n }\n};\n```
0
0
['C']
0
delete-columns-to-make-sorted-iii
In C++
in-c-by-ranjithgopinath-0mfu
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
RanjithGopinath
NORMAL
2024-08-10T13:56:29.960076+00:00
2024-08-10T13:56:29.960103+00:00
13
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n = strs.size();\n int m = strs[0].size();\n vector<int> dp(m, 1); // dp[i] stores the length of the longest increasing subsequence ending at column i\n \n for (int i = 1; i < m; ++i) {\n for (int j = 0; j < i; ++j) {\n bool valid = true;\n for (int k = 0; k < n; ++k) {\n if (strs[k][j] > strs[k][i]) {\n valid = false;\n break;\n }\n }\n if (valid) {\n dp[i] = max(dp[i], dp[j] + 1);\n }\n }\n }\n \n return m - *max_element(dp.begin(), dp.end());\n }\n};\n```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
very easy
very-easy-by-omsonekar4-tuhv
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
omsonekar4
NORMAL
2024-08-06T20:51:54.270521+00:00
2024-08-06T20:51:54.270545+00:00
4
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n = strs.size();\n int m = strs[0].size();\n vector<int> ways(m,0);\n for(int i=m-1;i>=0;i--){\n int ans = 0;\n for(int j=i+1;j<m;j++){\n int k;\n for(k=0;k<n && strs[k][i] <= strs[k][j];k++);\n if (k == n)\n ans = max(ans, ways[j]+1);\n }\n ways[i] = ans;\n }\n return m-*max_element(ways.begin(),ways.end())-1;\n }\n};\n```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
DP: Maximal increasing subsequence(all strings through)
dp-maximal-increasing-subsequenceall-str-fpkf
Approach\nSimilar to the search for the maximum length of an increasing subsequence.\n\u041Enly the comparison goes through all strings in a special function\n#
sav20011962
NORMAL
2024-08-01T09:25:14.834287+00:00
2024-08-01T11:57:56.844867+00:00
9
false
# Approach\nSimilar to the search for the maximum length of an increasing subsequence.\n\u041Enly the comparison goes through all strings in a special function\n# Complexity\n![image.png](https://assets.leetcode.com/users/images/4818b8f4-da36-4ac5-a494-7e36c21e8327_1722504079.8276875.png)\n# Code\n```\n// similar to the search for the maximum length of an increasing subsequence\n// only the comparison goes through all strings \n// in a special function\nclass Solution {\n fun minDeletionSize(strs: Array<String>): Int {\n val length = strs[0].length\n val dp = IntArray(length, {1})\n var max = 0\n fun checkAllStrs(j: Int, i: Int): Boolean {\n for (s in strs) \n if (s[j] > s[i]) return false\n return true\n }\n for (i in 0 until length) {\n for (j in 0 until i) {\n if (checkAllStrs(j, i)) {\n dp[i] = Math.max(dp[i], dp[j] + 1)\n }\n }\n max = Math.max(max, dp[i])\n }\n return length - max\n }\n\n}\n```
0
0
['Dynamic Programming', 'Kotlin']
0
delete-columns-to-make-sorted-iii
simple lis
simple-lis-by-harsh99429-ucd4
\n# Complexity\n- Time complexity:\nO(n^2*m)\nm is the size of each string \n\n- Space complexity:\nO(n)\n\n# Code\n\nclass Solution:\n def minDeletionSize(s
harsh99429
NORMAL
2024-07-25T19:45:48.394421+00:00
2024-07-25T19:45:48.394457+00:00
8
false
\n# Complexity\n- Time complexity:\n$$O(n^2*m)$$\nm is the size of each string \n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution:\n def minDeletionSize(self, l: List[str]) -> int:\n n=len(l[0]) #len of each str\n m=len(l) #len of list\n dp=[1 for _ in range(n)]\n\n for i in range(n):\n for k in range(i):\n ct=0\n for j in range(m):\n if l[j][k]<=l[j][i]:\n ct+=1\n else:\n break\n if(ct==m):\n dp[i]=max(dp[i],1+dp[k])\n \n return n-max(dp)\n\n```
0
0
['Python3']
0
delete-columns-to-make-sorted-iii
C++ || DP || Commented
c-dp-commented-by-saiteja_balla0413-mk3c
\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n=strs[0].size();\n vector<vector<int>> dp(n+1,vector<int>(n+1,
saiteja_balla0413
NORMAL
2024-04-19T15:03:30.879577+00:00
2024-04-19T15:03:30.879614+00:00
19
false
```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n=strs[0].size();\n vector<vector<int>> dp(n+1,vector<int>(n+1,-1));\n return dfs(0,1,strs,dp);\n }\n //returns true if every string\'s char at ind is >= char at last, since it should be in lexicographical order\n bool isValid(vector<string>& strs,int ind,int last)\n {\n if(last==0)\n return true;\n int n=strs.size();\n ind-=1;\n last-=1;\n for(int i=0;i<n;i++)\n {\n if(strs[i][ind] < strs[i][last])\n return false;\n }\n return true;\n }\n int dfs(int last,int ind,vector<string>& strs,vector<vector<int>>& dp)\n {\n //we have calculated all columns and reached end \n if(ind==strs[0].size()+1)\n return 0;\n \n //if pre computed then return the ans\n if(dp[ind][last]!=-1)\n return dp[ind][last];\n \n int curr=INT_MAX;\n \n //we choose to delete it and find ans\n curr=min(curr, 1+ dfs(last,ind+1,strs,dp));\n \n //Choose not to delete only if the last char and ind char are in lexicographical order\n if(isValid(strs,ind,last))\n curr=min(curr, dfs(ind,ind+1,strs,dp));\n return dp[ind][last]=curr;\n }\n};\n```
0
0
[]
0
delete-columns-to-make-sorted-iii
JS/TS Solution, beats 100%
jsts-solution-beats-100-by-jamauss-pgts
Able to determine the deletion size by using .charAt to compare with what\'s still left.\n\n# Code\n\nconst minDeletionSize = (strs: string[]): number => {\n
jamauss
NORMAL
2023-12-21T18:33:48.869478+00:00
2023-12-21T18:33:48.869516+00:00
19
false
Able to determine the deletion size by using `.charAt` to compare with what\'s still left.\n\n# Code\n```\nconst minDeletionSize = (strs: string[]): number => {\n \n const totalStrs = strs.length;\n const strSize = strs[0].length;\n let result:number = strSize - 1;\n let k:number = 0;\n\n let dp:number[] = new Array<number>(strSize).fill(1); // using fill so compile with tsc --lib "es2015,dom"\n\n for(let i = 0; i < strSize; ++i) {\n for(let j = 0; j < i; ++j) {\n for(k = 0; k < totalStrs; ++k) {\n if(strs[k].charAt(j) > strs[k].charAt(i)) break;\n }\n if(k === totalStrs && dp[j] + 1 > dp[i]) {\n dp[i] = dp[j] + 1;\n }\n }\n result = Math.min(result, strSize - dp[i]);\n }\n\n return result;\n\n};\n```\nSee more `TypeScript`/`JavaScript` (and other language) LeetCode solutions at https://github.com/jasonmauss/LeetCode
0
0
['TypeScript', 'JavaScript']
0
delete-columns-to-make-sorted-iii
Solution to 960. Delete Columns to Make Sorted III
solution-to-960-delete-columns-to-make-s-vh1e
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this problem is to find the longest increasing subsequence of colu
Askalany
NORMAL
2023-12-15T09:14:11.070911+00:00
2023-12-15T09:14:11.070934+00:00
44
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this problem is to find the longest increasing subsequence of columns. This is because the columns in the longest increasing subsequence do not need to be deleted to make the rows sorted. The remaining columns, which are not part of this subsequence, need to be deleted.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach is to use dynamic programming. We iterate over the columns of the input from right to left. For each column, we check if it is lexicographically less than or equal to all other columns to its right. If it is, we update the dynamic programming array to keep track of the maximum increasing subsequence of columns. The final answer is the total number of columns minus the length of the maximum increasing subsequence.\n# Pseudocode\n```\nProcedure minDeletionSize(strs: List of strings) -> int:\n Initialize W as the width of the input (length of any string in strs)\n Initialize dp as an array of size W with all elements as 1\n\n For i from W-2 to 0 (inclusive) in reverse order:\n For j from i+1 to W (exclusive):\n If all characters at index j in strs are greater than or equal to characters at index i:\n Update dp[i] as the maximum of dp[i] and 1 + dp[j]\n\n Return W - maximum value in dp\nEnd Procedure\n\n```\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity is $$O(W2\u22C5N)$$, where $$W$$ is the width of the input and $$N$$ is the number of strings. This is because for each column, we compare it with all other columns to its right, and for each comparison, we check all the strings.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is $$O(W)$$, where $$W$$ is the width of the input. This is because we use a dynamic programming array of size $$W$$ to keep track of the maximum increasing subsequence of columns.\n# Code\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n W = len(strs[0])\n dp = [1] * W\n for i in range(W-2, -1, -1):\n for j in range(i+1, W):\n if all(row[i] <= row[j] for row in strs):\n dp[i] = max(dp[i], 1 + dp[j])\n\n return W - max(dp)\n\n```
0
0
['Array', 'String', 'Dynamic Programming', 'Python3']
0
delete-columns-to-make-sorted-iii
python super easy to understand (dp top down)
python-super-easy-to-understand-dp-top-d-ce41
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
harrychen1995
NORMAL
2023-11-06T15:04:41.802657+00:00
2023-11-06T15:04:41.802687+00:00
25
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n \n @functools.lru_cache(None)\n def dp(i, j): # i -> current index , j -> prev index after deletion\n\n if i == len(strs[0]):\n return 0\n ans = float("inf")\n\n if j == -1: # -> delete or not delete beginning of characters\n ans = min(ans, dp(i+1, i), 1 + dp(i+1, j))\n return ans\n\n for x in range(len(strs)):\n if strs[x][i] < strs[x][j]: # if current index is less than prev index, must delete current index\n ans = min(ans, 1 + dp(i+1, j))\n return ans\n # delete current index or not delete current index\n ans = min(ans, dp(i+1, i), 1 + dp(i+1, j))\n return ans\n\n return dp(0, -1)\n \n```
0
0
['Dynamic Programming', 'Python3']
0
delete-columns-to-make-sorted-iii
Solution
solution-by-user0352v-85l4
\n\n# Code\n\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n \n int n = strs.size();\n int m = strs[0].length
user0352V
NORMAL
2023-10-30T15:13:21.014742+00:00
2023-10-30T15:13:21.014762+00:00
18
false
\n\n# Code\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n \n int n = strs.size();\n int m = strs[0].length();\n\n vector<int> dp(m, 1);\n int res = 1;\n for(int i = 0; i < m; i++) {\n for(int j = 0; j < i; j++) {\n for(int k = 0; k <= n; k++) {\n if(k == n) {\n dp[i] = max(dp[i], dp[j] + 1);\n res = max(res, dp[i]);\n } else if (strs[k][j] > strs[k][i]) {\n break;\n }\n }\n }\n }\n return m - res;\n }\n};\n```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
CPP || INCLUDE - EXCLUDE CALL ONLY || DP || memoization
cpp-include-exclude-call-only-dp-memoiza-ciks
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
abhi_bittu2525
NORMAL
2023-10-12T17:39:33.803279+00:00
2023-10-12T17:39:33.803312+00:00
36
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int solve(vector<string>& strs,int len,int index,int prev,vector<vector<int> >&dp){\n if(index >= len){\n return 0;\n }\n if(dp[index][prev+1] != -1){\n return dp[index][prev+1];\n }\n bool isD = true;\n for(int i =0;i<strs.size();i++){\n if(prev == -1){\n break;\n }\n if(strs[i][prev] > strs[i][index]){\n isD = false;\n break;\n }\n }\n int include = INT_MAX;\n if(isD){//AGR AB TK DIKKAT NHI H TOH AAGE BDA \n include = solve(strs,len,index+1,index,dp);\n }\n int exclude = 1+ solve(strs,len,index+1,prev,dp); //ek ko exclude kr rhe h mtlb ek colom dlt krna pdega \n\n return dp[index][prev+1] = min(include,exclude);\n }\n int minDeletionSize(vector<string>& strs) {\n int len = strs[0].length();\n vector<vector<int> >dp(len+1,vector<int>(len+1,-1));\n return solve(strs,len,0,-1,dp);\n }\n};\n```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
Efficient JS Solution (Beat over 80% time and 100% memory)
efficient-js-solution-beat-over-80-time-q9wr2
\n\n# Approach\nClassic LIS.\n\n# Complexity\n- let n = strs.length, m = max(strs[i].length)\n- Time complexity: O(nm^2)\n- Space complexity: O(m)\n\n# Code\njs
CuteTN
NORMAL
2023-10-03T19:13:31.043576+00:00
2023-10-03T19:13:31.043606+00:00
15
false
![image.png](https://assets.leetcode.com/users/images/5ab32d9d-c295-4d17-93b2-409168613b28_1696360254.4706516.png)\n\n# Approach\nClassic LIS.\n\n# Complexity\n- let `n = strs.length`, `m = max(strs[i].length)`\n- Time complexity: $$O(nm^2)$$\n- Space complexity: $$O(m)$$\n\n# Code\n```js\nlet dp = new Uint8Array(100);\n/**\n * @param {string[]} strs\n * @return {number}\n */\nvar minDeletionSize = function(strs) {\n let m = strs.length;\n let n = strs[0].length;\n\n dp.fill(1, 0, n);\n let res = 1;\n\n for (let r = 1; r < n; r++) {\n for (let l = r - 1; l >= 0; l--) {\n let t = dp[l] + 1;\n if (t <= dp[r]) continue;\n\n for (let i = 0; i < m; i++) {\n if (strs[i].charCodeAt(l) > strs[i].charCodeAt(r)) {\n t = 0;\n break;\n }\n }\n if (t) dp[r] = t;\n res = Math.max(dp[r], res);\n }\n }\n\n return n - res;\n};\n```
0
0
['Dynamic Programming', 'JavaScript']
0
delete-columns-to-make-sorted-iii
C++, JavaScript, Java Solution.
c-javascript-java-solution-by-samuel-akt-w12n
\n\n# C++\n\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n = strs.size();\n int m = strs[0].size();\n
Samuel-Aktar-Laskar
NORMAL
2023-09-12T16:25:19.747819+00:00
2023-09-12T16:41:39.518380+00:00
13
false
\n\n# C++\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n = strs.size();\n int m = strs[0].size();\n vector<int> ways(m,0);\n for(int i=m-1;i>=0;i--){\n int ans = 0;\n for(int j=i+1;j<m;j++){\n int k;\n for(k=0;k<n && strs[k][i] <= strs[k][j];k++);\n if (k == n)\n ans = max(ans, ways[j]+1);\n }\n ways[i] = ans;\n }\n return m-*max_element(ways.begin(),ways.end())-1;\n }\n};\n```\n\n# JavaScript\n```\n/**\n * @param {string[]} strs\n * @return {number}\n */\nvar minDeletionSize = function(strs) {\n const n = strs.length;\n const m = strs[0].length;\n const ways = new Array(m).fill(0);\n for(let i=m-1;i>=0;i--){\n let ans = 0;\n for(let j=i+1;j<m;j++){\n let k;\n for(k=0;k<n && strs[k][i] <= strs[k][j];k++);\n if (k == n) ans = Math.max(ans, ways[j]+1);\n }\n ways[i] = ans;\n }\n return m - Math.max(...ways)-1;\n};\n```\n\n# Java\n```\nclass Solution {\n public int minDeletionSize(String[] strs) {\n int n = strs.length;\n int m = strs[0].length();\n int ways[] = new int[m];\n int res = 0;\n for(int i=m-1;i>=0;i--){\n int ans = 0;\n for(int j=i+1;j<m;j++){\n int k;\n for(k=0;k<n && strs[k].charAt(i) <= strs[k].charAt(j);k++);\n if (k == n)\n ans = Math.max(ans, ways[j]+1);\n }\n ways[i] = ans;\n res = Math.max(res,ans);\n }\n return m-res-1; \n }\n}\n```
0
0
['C++', 'JavaScript']
0
delete-columns-to-make-sorted-iii
C++ Easy Solution
c-easy-solution-by-md_aziz_ali-mg71
Code\n\nclass Solution {\npublic:\n vector<vector<int>>dp;\n int solve(vector<string>& strs,int i,int n,int prev){\n if(i == n)\n return
Md_Aziz_Ali
NORMAL
2023-09-09T02:32:39.001297+00:00
2023-09-09T02:32:39.001318+00:00
23
false
# Code\n```\nclass Solution {\npublic:\n vector<vector<int>>dp;\n int solve(vector<string>& strs,int i,int n,int prev){\n if(i == n)\n return 0;\n if(dp[i][prev+1] != -1)\n return dp[i][prev + 1];\n bool flag = true;\n for(int j = 0;j < strs.size();j++){\n if(prev == -1)\n break;\n if(strs[j][prev] > strs[j][i]){\n flag = false;\n break;\n }\n }\n int take = INT_MAX;\n if(flag)\n take = solve(strs,i+1,n,i);\n return dp[i][prev + 1] = min(take,1 + solve(strs,i+1,n,prev));\n }\n int minDeletionSize(vector<string>& strs) {\n dp = vector<vector<int>>(strs[0].length(),vector<int>(strs[0].length(),-1));\n return solve(strs,0,strs[0].length(),-1);\n }\n};\n```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
Easy to Understand DP solution
easy-to-understand-dp-solution-by-raj_ti-yaco
Explanation\nHere dp[i] means the solution for 1...i and taking the ith index in our solution. Similarly dp1[i] means the solution for i...n taking in the ith
raj_tirtha
NORMAL
2023-08-30T08:09:56.963591+00:00
2023-08-30T08:09:56.963621+00:00
19
false
# Explanation\nHere dp[i] means the solution for 1...i and taking the ith index in our solution. Similarly dp1[i] means the solution for i...n taking in the ith index. But for dp11 we will maintain the reverse order as after i the string should be increasing. Hence our final answer will be dp[i]+dp1[i]. Time complexity is O(m*n^2).\n\n# Code\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n=strs[0].size();\n vector<int> dp(n);\n for(int i=0;i<n;i++){\n dp[i]=i;\n for(int j=i-1;j>=0;j--){\n bool ok=true;\n for(int k=0;k<strs.size();k++){\n if(strs[k][j]>strs[k][i]){\n ok=false;\n break;\n }\n }\n if(ok){\n dp[i]=min(dp[i],dp[j]+(i-j)-1);\n }\n }\n }\n vector<int> dp1(n);\n for(int i=n-1;i>=0;i--){\n dp1[i]=n-i-1;\n for(int j=i+1;j<n;j++){\n bool ok=true;\n for(int k=0;k<strs.size();k++){\n if(strs[k][j]<strs[k][i]){\n ok=false;\n break;\n }\n }\n if(ok){\n dp1[i]=min(dp1[i],dp1[j]+(j-i)-1);\n }\n }\n }\n int ans=n;\n for(int i=0;i<n;i++){\n ans=min(ans,dp[i]+dp1[i]);\n }\n return ans;\n }\n};\n```
0
0
['C++']
0
delete-columns-to-make-sorted-iii
Python 3: LIS DP Solution with comments; TC - 99.37%
python-3-lis-dp-solution-with-comments-t-62l0
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
hootielander
NORMAL
2023-08-28T14:50:07.200436+00:00
2023-08-28T14:50:07.200468+00:00
23
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n\n # dp[i], including i for every strs, the length of the LIS\n N = len(strs[0])\n dp = [1] * N\n\n for i in range(N - 1, -1, -1):\n for j in range(i + 1, N):\n canIncrease = True\n\n # check against every strings whether including current\n # can still mean it is LIS\n for s in strs:\n if s[i] > s[j]:\n canIncrease = False\n break\n \n # if including current char, for all string, does not go against LIS\n if canIncrease:\n dp[i] = max(dp[i], dp[j] + 1)\n\n return N - max(dp)\n\n```
0
0
['Python3']
0
delete-columns-to-make-sorted-iii
Python: pick or skip DP: top-down to 1D bottom-up optimized
python-pick-or-skip-dp-top-down-to-1d-bo-8xnh
"Pick or Skip DP" (Knapsack)\n\nYou can pick a column or skip it. Then you need to figure out when you can pick a column.\nYou can pick it when it\'s either the
vokasik
NORMAL
2023-08-24T02:26:31.750267+00:00
2023-08-24T02:44:18.415820+00:00
26
false
**"Pick or Skip DP" (Knapsack)**\n\nYou can pick a column or skip it. Then you need to figure out when you can pick a column.\nYou can pick it when it\'s either the first time you pick any column or `all_rows[prev_picked_col] < all_rows[trying_to_pick_col]`\nReturn `COLS - max(picked_columns)`\n\nLooks like medium LIS #300\n\n```\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n # 1. top-down\n @cache\n def dfs(i, prev):\n if i == C:\n return 0\n pick = 0\n can_pick = True\n for r in range(R):\n if prev != -1 and strs[r][prev] > strs[r][i]:\n can_pick = False\n break\n if can_pick:\n pick = 1 + dfs(i + 1, i)\n skip = dfs(i + 1, prev)\n return max(pick, skip)\n C,R = len(strs[0]), len(strs) \n return C - dfs(0, -1)\n\n # 2. bottom-up\n C,R = len(strs[0]), len(strs) \n dp = defaultdict(int)\n for c in reversed(range(C)):\n for prev in range(-1, c):\n dp[c, prev] = dp[c + 1, prev]\n if not any(prev != -1 and strs[r][prev] > strs[r][c] for r in range(R)):\n dp[c, prev] = max(1 + dp[c + 1, c], dp[c + 1, prev])\n return C - dp[0, -1]\n\n # 3. shift indices coz you can\'t have negative\n C,R = len(strs[0]), len(strs)\n dp = defaultdict(int)\n for c in reversed(range(1, C + 1)):\n for prev in range(c + 1):\n if not any(prev and strs[r][prev - 1] > strs[r][c - 1] for r in range(R)):\n dp[c, prev] = max(1 + dp[c + 1, c], dp[c + 1, prev])\n else:\n dp[c, prev] = dp[c + 1, prev]\n return C - dp[1, 0]\n\n # 4. defaultdict to 2D\n C,R = len(strs[0]), len(strs)\n dp = [[0] * (C + 2) for _ in range(C + 2)]\n for c in reversed(range(1, C + 1)):\n for prev in range(c + 1):\n dp[c][prev] = dp[c + 1][prev]\n if not any(prev and strs[r][prev - 1] > strs[r][c - 1] for r in range(R)):\n dp[c][prev] = max(dp[c][prev], 1 + dp[c + 1][c], dp[c + 1][prev])\n return C - dp[1][0]\n\n # 5. 2D to 1D\n C,R = len(strs[0]), len(strs)\n dp = [0] * (C + 1)\n for c in reversed(range(1, C + 1)):\n for prev in range(c):\n if not any(prev and strs[r][prev - 1] > strs[r][c - 1] for r in range(R)):\n dp[prev] = max(dp[prev], 1 + dp[c])\n return C - dp[0]\n```\n```
0
0
['Dynamic Programming', 'Python']
0
delete-columns-to-make-sorted-iii
Easy C# 100%
easy-c-100-by-kuramshin34-y9b4
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
kuramshin34
NORMAL
2023-08-23T08:02:08.901371+00:00
2023-08-23T08:02:08.901393+00:00
9
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\npublic class Solution {\n public int MinDeletionSize(string[] strs) {\n int n = strs.Length;\n int m = strs[0].Length;\n\n int[] dp = Enumerable.Repeat(1, m).ToArray();\n int res = 1;\n for(int i = 0; i < m; i++)\n {\n for (int j = 0; j < i; j++) \n {\n for (int k = 0; k <= n; k++) \n {\n if (k == n) \n {\n dp[i] = Math.Max(dp[i], dp[j] + 1);\n res = Math.Max(res, dp[i]);\n }\n else if(strs[k][j] > strs[k][i]) \n {\n break;\n }\n } \n }\n }\n return m - res;\n }\n}\n```
0
0
['C#']
0
delete-columns-to-make-sorted-iii
Simple C++ solution Based on LIS
simple-c-solution-based-on-lis-by-tag_98-1j1d
Intuition\n Describe your first thoughts on how to solve this problem. \nSame intuition as LIS\n\nIn this Problem we find the maximum length so our answer is (
Tag_989
NORMAL
2023-07-10T14:22:18.688158+00:00
2023-07-10T14:22:18.688177+00:00
57
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSame intuition as LIS\n\nIn this Problem we find the maximum length so our answer is **( strs[0].size() - maxlength )**\n\n# Code\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int n=strs.size();\n int m=strs[0].size();\n\n if(m==1)return 0;\n vector<int> dp(m,1);\n\n int ans=0;\n for(int i=1;i<m;i++)\n {\n int j=0;\n while(j<i)\n {\n bool flag=1;\n for(int k=0;k<n;k++)\n {\n if(strs[k][j]>strs[k][i])\n {\n flag=0;\n }\n }\n if(flag==1)\n {\n if(dp[i]<dp[j]+1)\n {\n dp[i]=dp[j]+1;\n }\n }\n\n j++;\n }\n\n ans=max(dp[i],ans);\n }\n return m-ans;\n }\n};\n```
0
0
['Dynamic Programming', 'C++']
0
delete-columns-to-make-sorted-iii
My Solution
my-solution-by-hope_ma-5ovj
\n/**\n * Time Complexity: O(rows * cols * cols)\n * Space Complexity: O(cols)\n * where `rows` is the length of the vector `strs`\n * `cols` is the lengt
hope_ma
NORMAL
2023-07-06T07:57:58.876254+00:00
2023-07-06T07:57:58.876284+00:00
18
false
```\n/**\n * Time Complexity: O(rows * cols * cols)\n * Space Complexity: O(cols)\n * where `rows` is the length of the vector `strs`\n * `cols` is the length of the string `strs.front()`\n */\nclass Solution {\n public:\n int minDeletionSize(const vector<string> &strs) {\n const int cols = static_cast<int>(strs.front().size());\n int dp[cols];\n fill(dp, dp + cols, 1);\n for (int c = 0; c < cols; ++c) {\n for (int pc = 0; pc < c; ++pc) {\n if (less_than(strs, pc, c)) {\n dp[c] = max(dp[c], dp[pc] + 1);\n }\n }\n }\n return cols - *max_element(dp, dp + cols);\n }\n \n private:\n bool less_than(const vector<string> &strs, const int c1, const int c2) {\n const int n = static_cast<int>(strs.front().size());\n bool ret = true;\n for (const string &str : strs) {\n if (str[c1] > str[c2]) {\n ret = false;\n break;\n }\n }\n return ret;\n }\n};\n```
0
0
[]
0
delete-columns-to-make-sorted-iii
(Python) Delete Columns to Make Sorted III
python-delete-columns-to-make-sorted-iii-28ef
Intuition\nThe intuition behind the approach is to use dynamic programming to find the length of the longest increasing subsequence (LIS) for each position in t
codewithsom
NORMAL
2023-07-01T17:45:47.832552+00:00
2023-07-01T17:45:47.832571+00:00
32
false
# Intuition\nThe intuition behind the approach is to use dynamic programming to find the length of the longest increasing subsequence (LIS) for each position in the strings. By keeping track of this information, we can determine the minimum number of deletions required to make all the strings lexicographically sorted.\n\n# Approach\n1. Initialize a 1D array dp of length `n` (number of columns) with all elements set to 1. This array will keep track of the length of the LIS at each position.\n2. Iterate through each column (j) from 1 to n.\n3. For each column (j), iterate through all the previous columns (i) from 0 to j-1.\n4. Check if all elements in the current column (j) are greater than or equal to the corresponding elements in the previous column (i) for all rows. If this condition is true, it means we can extend the LIS by including the current column (j).\n5. If the condition in step 4 is met, update `dp[j]` to be the maximum of `dp[j]` and `dp[i] + 1`, indicating the extended length of the LIS at column j.\n6. After iterating through all previous columns for the current column (j), we will have the length of the LIS at position j in the dp array.\n7. The minimum number of deletions required to make all the strings lexicographically sorted will be the total number of columns (n) minus the length of the longest increasing subsequence, which is `n - max(dp)`.\n# Complexity\n- Time complexity:\nThe time complexity of the solution is `O(n^2)`, where n is the number of columns in the strings. The dynamic programming approach involves two nested loops to fill the dp array.\n- Space complexity:\nThe space complexity is `O(n)`, where n is the number of columns. We use a 1D array dp of length n to store the LIS at each position.\n# Code\n```\nfrom typing import List\n\nclass Solution:\n def minDeletionSize(self, strs: List[str]) -> int:\n m, n = len(strs), len(strs[0])\n dp = [1] * n\n\n for j in range(1, n):\n for i in range(j):\n if all(strs[k][i] <= strs[k][j] for k in range(m)):\n dp[j] = max(dp[j], dp[i] + 1)\n\n return n - max(dp)\n\n```
0
0
['Array', 'String', 'Dynamic Programming', 'Sorting', 'Python3']
0
delete-columns-to-make-sorted-iii
Scala solution
scala-solution-by-malovig-uy1m
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
malovig
NORMAL
2023-06-16T20:02:10.773556+00:00
2023-06-16T20:02:10.773576+00:00
24
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(m^2 * n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(m)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nobject Solution {\n def minDeletionSize(strs: Array[String]): Int = {\n val m = strs(0).length\n val dp = Array.fill(m)(1)\n var overallMax = 0\n for (i <- 0 until m) {\n for (j <- 0 until i)\n if (strs.indices.forall(k => strs(k).charAt(j) <= strs(k).charAt(i))) dp(i) = math.max(dp(i), dp(j) + 1)\n overallMax = math.max(dp(i), overallMax)\n }\n m - overallMax\n }\n\n}\n```
0
0
['Scala']
0
delete-columns-to-make-sorted-iii
O(n*m*m) DP solution
onmm-dp-solution-by-xjpig-bjpf
Intuition\n Describe your first thoughts on how to solve this problem. \ndp[i] memorize the longest increasing subsequnce up to ith position in the string.\n\n-
XJPIG
NORMAL
2023-05-01T12:14:22.632308+00:00
2023-05-01T12:14:36.792077+00:00
46
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ndp[i] memorize the longest increasing subsequnce up to ith position in the string.\n\n- Time complexity: $$O(n*m*m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minDeletionSize(vector<string>& strs) {\n int result=INT_MAX;\n vector<int> dp(strs[0].size(),1);\n for(int i=0;i<strs[0].size();i++){\n for(int j=0;j<i;j++){\n int k=0;\n for(;k<strs.size();k++)\n if(strs[k][i]<strs[k][j]) break;\n if(k==strs.size()) dp[i]=max(dp[i],dp[j]+1);\n }\n result=min(result,(int)strs[0].size()-dp[i]);\n }\n return result;\n }\n};\n```
0
0
['C++']
0