Bagaimana Anda membalikkan karakter k terakhir dengan python?

Misalkan kita memiliki string dan bilangan bulat k, kita harus membalik k karakter pertama untuk setiap 2k karakter dihitung dari awal string. Jika tidak ada karakter yang tersisa, balikkan semuanya. Jika ada kurang dari 2k karakter tetapi lebih besar dari atau sama dengan k karakter, maka balikkan k karakter pertama dan biarkan yang lain seperti aslinya

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Bagaimana Anda membalikkan n karakter terakhir dengan Python?
Bagaimana Anda membalikkan karakter dengan Python?
Bagaimana Anda mengubah karakter terakhir dengan Python?
Bagaimana Anda menyingkirkan karakter terakhir dalam sebuah string dengan Python?

Jadi, jika inputnya seperti "abcdefgh" dan k = 3, maka outputnya adalah "cbadefhg"

Untuk mengatasi ini, kami akan mengikuti langkah-langkah ini −

l. = membuat daftar karakter dari s
saya. = k-1
sedangkan i < ukuran l + k −
- A. = l[dari indeks 0 sampai i-k+1]
- B. = l[dari indeks i-k+1 sampai i+1]
- c. = l[dari indeks i+1 sampai akhir]
- l. = a concatenate b[dari indeks 0 sampai akhir] concatenate c
- Saya. = i + 2*k
mengembalikan string dengan menggabungkan setiap karakter dalam l

Contoh

Mari kita lihat implementasi berikut untuk mendapatkan pemahaman yang lebih baik −

Demo Langsung

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

Memasukkan

"abcdefg", 3

Keluaran

cbadefg

GeeksFor

GeeksFor

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// C++ implementation of the approach0

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90______________________

eegksfgroeeks

__________________________________________________________________________________913

// C++ implementation of the approach_0

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_13_______99

// C++ implementation of the approach0

eegksfgroeeks

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15 // C++ implementation of the approach05 // C++ implementation of the approach06

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5using1 using2

// C++ implementation of the approach_11

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55 namespace0

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61// C++ implementation of the approach18

// C++ implementation of the approach_06

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// C++ implementation of the approach22// C++ implementation of the approach23

eegksfgroeeks

51// C++ implementation of the approach25

// C++ implementation of the approach_26

C#

// C++ implementation of the approach_27

using // C++ implementation of the approach29

std;5 std;6

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// Function to return the string after

// reversing the alternate k characters

// Function to return the string after0 // Function to return the string after1

// Function to return the string after2

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0 // C++ implementation of the approach47

eegksfgroeeks

eegksfgroeeks

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_191_______1

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_191_______3

// C++ implementation of the approach0// C++ implementation of the approach5 // C++ implementation of the approach6

________191

GeeksFor

_191_______8// C++ implementation of the approach9

eegksfgroeeks

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______1

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_191_______64

________191

GeeksFor

_191_______66

________191

GeeksFor

_191_______68

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______5

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______7

eegksfgroeeks

5#include <bits/stdc++.h>9

eegksfgroeeks

5using1 using2

#include <bits/stdc++.h>_9

// Function to return the string after0 // C++ implementation of the approach80// C++ implementation of the approach81 // C++ implementation of the approach82

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0 // C++ implementation of the approach84

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0 // C++ implementation of the approach86

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5// C++ implementation of the approach89

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5// C++ implementation of the approach81 // C++ implementation of the approach92

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59 // C++ implementation of the approach95

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________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_191_______99

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______01

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______03

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______05

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______07

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______09

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5#include <bits/stdc++.h>9

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5using1 #include <bits/stdc++.h>14#include <bits/stdc++.h>15#include <bits/stdc++.h>16

#include <bits/stdc++.h>_9

using_4

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24 // Function to return the string after0

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26 #include <bits/stdc++.h>22

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30namespace0// C++ implementation of the approach9

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0 #include <bits/stdc++.h>30

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0 namespace7

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5#include <bits/stdc++.h>35

#include <bits/stdc++.h>_9

#include <bits/stdc++.h>_38

Javascript

#include <bits/stdc++.h>_39

#include <bits/stdc++.h>_40

// Function to return the string after

// reversing the alternate k characters

#include <bits/stdc++.h>43 #include <bits/stdc++.h>44

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6 #include <bits/stdc++.h>48

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#include <bits/stdc++.h>_51

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_191_______1

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_191_______3

// C++ implementation of the approach0// C++ implementation of the approach5 // C++ implementation of the approach6

________191

GeeksFor

_191_______8// C++ implementation of the approach9

#include <bits/stdc++.h>_51

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______1

// C++ implementation of the approach0#include <bits/stdc++.h>66#include <bits/stdc++.h>15#include <bits/stdc++.h>68#include <bits/stdc++.h>15#include <bits/stdc++.h>70

________191

GeeksFor

_13_______10

#include <bits/stdc++.h>_51

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______5

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______7

eegksfgroeeks

5#include <bits/stdc++.h>9

eegksfgroeeks

5using1 using2

#include <bits/stdc++.h>_9

using_4

#include <bits/stdc++.h>85namespace0// C++ implementation of the approach9

#include <bits/stdc++.h>_88

#include <bits/stdc++.h>_89

#include <bits/stdc++.h>_90

#include <bits/stdc++.h>_91

#include <bits/stdc++.h>_92

Keluaran

eegksfgroeeks

Kompleksitas Waktu. O(N) di mana N adalah panjang string “s” yang diberikan

Ruang Bantu. O(1)

Pendekatan 2.

Ide untuk memecahkan masalah ini adalah untuk melintasi string, dan saat melintasi mundur karakter K pertama, lalu lewati karakter K berikutnya, lalu balik lagi K karakter berikutnya dan seterusnya hingga string lengkap dimodifikasi

Ikuti langkah-langkah untuk memecahkan masalah

Lintasi sampai akhir string asli
- Lintasi mundur dari i+k ke i dan simpan karakter dalam string yang dihasilkan
- Perbarui i ke i+k
- Lintasi dari i ke i + k sekarang, dan simpan karakter dalam string yang dihasilkan
Kembalikan string asli

Di bawah ini adalah implementasi dari pendekatan di atas

C++

// C++ implementation of the approach

#include <bits/stdc++.h>

using namespace std;

// Function to return the string after

#include <bits/stdc++.h>_99

string revAlternateK(string s,

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

0 using04

eegksfgroeeks

eegksfgroeeks

5using07#include <bits/stdc++.h>15// C++ implementation of the approach9

eegksfgroeeks

eegksfgroeeks

0 using12

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5using14

eegksfgroeeks

5using16

eegksfgroeeks

eegksfgroeeks

59 using19

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______21

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______23

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______25

// C++ implementation of the approach0

eegksfgroeeks

6 using28

________191

GeeksFor

_197_______30

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______32

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______34

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______36

// C++ implementation of the approach0

eegksfgroeeks

6 using39

________191

GeeksFor

_197_______30

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______43

eegksfgroeeks

5#include <bits/stdc++.h>9

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5using47

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5using1 using50

#include <bits/stdc++.h>_9

using_4

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0 using6

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5using57namespace0// C++ implementation of the approach9

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0 using62

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0 using65

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5using67

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5using1 std;2

#include <bits/stdc++.h>_9

Jawa

std;_4

using_73 using74

std;5 using76

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5// Function to return the string after

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5#include <bits/stdc++.h>99

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5// Function to return the string after0 using83

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

0 using04

eegksfgroeeks

eegksfgroeeks

// C++ implementation of the approach0using91#include <bits/stdc++.h>15// C++ implementation of the approach9

// C++ implementation of the approach0

eegksfgroeeks

0 // reversing the alternate k characters2// reversing the alternate k characters3using98// reversing the alternate k characters3// C++ implementation of the approach9

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______14

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______16

// C++ implementation of the approach0

eegksfgroeeks

59 using19

________191

GeeksFor

_197_______21

________191

GeeksFor

_197_______23

________191

GeeksFor

_197_______25

// C++ implementation of the approach7

eegksfgroeeks

6 namespace16

eegksfgroeeks

91namespace18

namespace19namespace20

________191

GeeksFor

_229_______9

// C++ implementation of the approach7________200______24

________191

GeeksFor

_197_______34

________191

GeeksFor

_197_______36

________191

GeeksFor

_13_______6 namespace31

namespace19namespace20

________191

GeeksFor

_229_______9

________191

GeeksFor

_197_______43

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_229_______9

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_197_______47

// C++ implementation of the approach0using1 using50

eegksfgroeeks

5#include <bits/stdc++.h>9

eegksfgroeeks

eegksfgroeeks

24 // Function to return the string after0

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

// C++ implementation of the approach0namespace55namespace0// C++ implementation of the approach9

// C++ implementation of the approach0

eegksfgroeeks

0 using62

// C++ implementation of the approach0

eegksfgroeeks

0 namespace63

eegksfgroeeks

39// C++ implementation of the approach9

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_200_______67

eegksfgroeeks

5#include <bits/stdc++.h>9

#include <bits/stdc++.h>_9

namespace_71

Python3

eegksfgroeeks

_46

eegksfgroeeks

_47

namespace_74

eegksfgroeeks

49 namespace76

eegksfgroeeks

5namespace78

eegksfgroeeks

55 #include <bits/stdc++.h>15

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

55 // reversing the alternate k characters3

eegksfgroeeks

5namespace87

eegksfgroeeks

55 // reversing the alternate k characters3

eegksfgroeeks

eegksfgroeeks

5namespace92

eegksfgroeeks

5namespace94

eegksfgroeeks

eegksfgroeeks

59 namespace97

// C++ implementation of the approach_0

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_210_______00

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_210_______02

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_210_______04

// C++ implementation of the approach0

eegksfgroeeks

6 namespace87std;08 std;09

eegksfgroeeks

7std;11

eegksfgroeeks

eegksfgroeeks

70std;14

eegksfgroeeks

eegksfgroeeks

91std;17

eegksfgroeeks

eegksfgroeeks

91std;20

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

// C++ implementation of the approach7namespace78

eegksfgroeeks

eegksfgroeeks

55std;28

// C++ implementation of the approach_7

// C++ implementation of the approach0

eegksfgroeeks

eegksfgroeeks

55 std;11

eegksfgroeeks

eegksfgroeeks

70 std;36

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_210_______38

________191

class Solution:
   def reverseStr(self, s, k):
      l = list(s)
      i = k-1
      while i < len(l)+k:
         a = l[:i-k+1]
         b = l[i-k+1:i+1]
         c = l[i+1:]
         l = a + b[::-1] + c
         i += 2*k
      return ''.join(l)

ob = Solution()
print(ob.reverseStr("abcdefg", 3))

_210_______40

// C++ implementation of the approach0

eegksfgroeeks

6 namespace87std;08 std;09std;46std;11

eegksfgroeeks

eegksfgroeeks

70std;50

// C++ implementation of the approach7namespace78

eegksfgroeeks

eegksfgroeeks

55std;28

// C++ implementation of the approach0

eegksfgroeeks

eegksfgroeeks

55 namespace87

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

5std;63

eegksfgroeeks

5using1 namespace78

// C++ implementation of the approach_11

std;_68

eegksfgroeeks

55 namespace0

std;71

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

7std;68std;76

std;_77

eegksfgroeeks

eegksfgroeeks

// C++ implementation of the approach22std;81std;68std;83

std;_84

C#

// C++ implementation of the approach_27

using // C++ implementation of the approach29

eegksfgroeeks

24 std;5 using76

eegksfgroeeks

5// Function to return the string after

eegksfgroeeks

5#include <bits/stdc++.h>99

eegksfgroeeks

5// Function to return the string after0 std;97 std;98std;97 // Function to return the string after00

eegksfgroeeks

eegksfgroeeks

eegksfgroeeks

0 using04

Bagaimana Anda membalikkan n karakter terakhir dengan Python?

Metode 2. Menggunakan Membalikkan String .

Menggunakan pengirisan String. string = "C# Corner" print ("String aslinya adalah. ",end="") print (string) print ("String terbalik (menggunakan loop) adalah. ",akhir="").

Menggunakan Loop. def terbalik. _str = "" untuk saya di s. .

Menggunakan Rekursi. def terbalik. jika len(s) == 0. kembali s. .

Menggunakan "terbalik"

Bagaimana Anda membalikkan karakter dengan Python?

Fungsi Bawaan terbalik() bergabung() untuk membuat string terbalik . Namun, maksud utama dan kasus penggunaan reversed() adalah untuk mendukung iterasi terbalik pada iterables Python. Dengan string sebagai argumen, reversed() mengembalikan iterator yang menghasilkan karakter dari input string dalam urutan terbalik.

Bagaimana Anda mengubah karakter terakhir dengan Python?

Python - ganti karakter terakhir dalam string .

teks = "ABC"

ukuran = len(teks) # panjang teks

penggantian = "X" # ganti dengan ini

teks = teks. ganti(teks[ukuran - 1. ], penggantian)

print("Rangkaian setelah. ", teks) # ABX

Bagaimana Anda menyingkirkan karakter terakhir dalam sebuah string dengan Python?

Metode 1. Menggunakan Pengiris daftar untuk Menghapus Elemen Terakhir dari string . Teknik mengiris juga dapat menghapus elemen terakhir dari string. str[. -1] akan menghapus elemen terakhir kecuali untuk semua elemen.