Strings Class 11: Indexing, Slicing, Operations and String Methods in Python

What Strings is about

Weightage: This chapter is part of Unit II, Computational Thinking and Programming - 1, which carries 45 of the 70 theory marks in the school annual exam together with Introduction to Problem Solving, Getting Started with Python, Flow of Control, Lists, Tuples and Dictionaries and Introduction to Python Modules.Source: CBSE Computer Science (083) syllabus 2026-27, Class XI distribution of marks, Unit II Computational Thinking and Programming - 1

Key concepts

Creating strings and indexing

NCERT section NCERT 8.1, 8.2

A string is a sequence of characters (letters, digits, spaces, symbols, any Unicode character) enclosed in single quotes, double quotes or triple quotes. Triple quotes (''' or """) allow a string to run over several lines. Python has no separate character type: a single character is a string of length 1.

Each character has an index. Positive indices run 0, 1, 2, ... from the left; negative indices run -1, -2, ... from the right, so s[-1] is the last character. For a string of length n, valid indices are 0 to n - 1 and -n to -1. An index outside this range gives an IndexError. The index must be an integer (it may be an expression like s[2 + 1]).

Worked example

For s = 'PYTHON', give s[0], s[5], s[-1], s[-6], s[len(s) - 2], and say what s[6] does.

Try it first, then show the solutionSolution

1. Indices 0-5: P Y T H O N; negative -6 to -1.
2. s[0] = 'P', s[5] = 'N', s[-1] = 'N', s[-6] = 'P', s[4] = 'O'.
3. s[6] is beyond the last index 5.

Answer: 'P', 'N', 'N', 'P', 'O'; s[6] raises IndexError.

Strings are immutable

NCERT section NCERT 8.2.2

Once a string is created, its characters cannot be changed in place. s[0] = 'J' gives a TypeError ('str' object does not support item assignment). To 'change' a string, build a new one, for example by slicing and concatenation, and assign it to the variable: s = 'J' + s[1:]. Every string method that seems to modify a string, such as upper() or replace(), actually returns a new string; the original stays the same unless you assign the result back.

Worked example

Give the output:

s = 'mango'
s.upper()
print(s)
t = s.upper()
print(t, s)
s = 'T' + s[1:]
print(s)

Try it first, then show the solutionSolution

1. s.upper() returns 'MANGO' but the result is not stored; s is still 'mango'.
2. t stores 'MANGO'; s unchanged.
3. A new string 'T' + 'ango' is assigned to s.

Answer: mango
MANGO mango
Tango

Concatenation, repetition and membership

NCERT section NCERT 8.3.1-8.3.3

• Concatenation + joins two strings: 'Good' + 'Day' = 'GoodDay'. Both sides must be strings; 'Age' + 15 is a TypeError.
• Repetition * repeats a string an integer number of times: 'ab' * 3 = 'ababab'; a count of 0 or less gives the empty string ''.
• Membership in and not in test whether one string occurs as a substring of another: 'yth' in 'Python' is True; 'P' in 'python' is False (case-sensitive).

Strings can also be compared with ==, <, > and so on, character by character using character codes (uppercase letters come before lowercase).

Worked example

Give the output: print('Na' * 2 + ' ' + 'Batman', 'bat' in 'Batman', '' in 'abc', 'Z' < 'a')

Try it first, then show the solutionSolution

1. 'NaNa' + ' ' + 'Batman' = 'NaNa Batman'.
2. 'bat' is not in 'Batman' because of the capital B: False.
3. The empty string is a substring of every string: True.
4. 'Z' (90) < 'a' (97): True.

Answer: NaNa Batman False True True

Slicing

NCERT section NCERT 8.3.4

A slice s[start:stop:step] gives a new string from index start up to but not including stop, taking every step-th character.

• Missing start means from the beginning (or from the end if step is negative); missing stop means to the end (or the beginning if step is negative); step defaults to 1.
• Indices may be negative: s[-3:] gives the last three characters.
• Slicing never gives an IndexError: out-of-range indices are cut to the string's ends, and an impossible range gives ''.
• s[::-1] reverses the string.
• len(s[a:b]) = b - a when 0 ≤ a ≤ b ≤ len(s).

Worked example

For s = 'COMPUTER', find s[1:4], s[:3], s[4:], s[::2], s[-3:], s[::-1], s[5:2], s[2:100].

Try it first, then show the solutionSolution

1. Indices: C0 O1 M2 P3 U4 T5 E6 R7.
2. s[1:4] = 'OMP'; s[:3] = 'COM'; s[4:] = 'UTER'.
3. s[::2] = C, M, U, E = 'CMUE'; s[-3:] = 'TER'; s[::-1] = 'RETUPMOC'.
4. s[5:2] = '' (start after stop with step 1); s[2:100] = 'MPUTER' (stop cut to the end).

Answer: 'OMP', 'COM', 'UTER', 'CMUE', 'TER', 'RETUPMOC', '', 'MPUTER'

Traversing a string

NCERT section NCERT 8.4

Traversing means visiting every character, one at a time. Two ways:

• by character: for ch in s: ... (simplest when the position is not needed)
• by index: for i in range(len(s)): ... use s[i] (needed when the position matters, e.g. comparing s[i] with s[i + 1] or printing positions). A while loop with an index i from 0 to len(s) - 1 also works.

Traversal is the basis of counting vowels, digits or words, building a new string character by character, and checking palindromes.

Worked example

Write a program to count vowels, consonants, digits and other characters in 'CBSE 2027: Class XI'.

Try it first, then show the solutionSolution

1. s = 'CBSE 2027: Class XI'
   v = c = d = o = 0
   for ch in s:
       if ch.isalpha():
           if ch.lower() in 'aeiou':
               v += 1
           else:
               c += 1
       elif ch.isdigit():
           d += 1
       else:
           o += 1
   print(v, c, d, o)

2. Letters: C B S E C l a s s X I → vowels E, a, I = 3; consonants 8. Digits 2, 0, 2, 7 = 4. Others: 3 spaces and a colon = 4.

Answer: 3 8 4 4

Case and counting methods

NCERT section NCERT 8.5, Table 8.2

• len(s): number of characters (a built-in function, not a method).
• s.capitalize(): first character uppercase, all the rest lowercase: 'hELLO wORLD' → 'Hello world'.
• s.title(): first letter of every word uppercase, the rest lowercase: 'hELLO wORLD' → 'Hello World'.
• s.lower() / s.upper(): all letters to lowercase / uppercase.
• s.count(sub[, start[, end]]): number of non-overlapping occurrences of sub: 'banana'.count('an') = 2; 'aaaa'.count('aa') = 2.
• s.startswith(sub) / s.endswith(sub): True if s begins / ends with sub.

Worked example

s = 'india is my COUNTRY'. Give s.capitalize(), s.title(), s.upper().count('I'), s.endswith('try'), s.startswith('India').

Try it first, then show the solutionSolution

1. capitalize: 'India is my country' (rest lowered).
2. title: 'India Is My Country'.
3. Uppercase 'INDIA IS MY COUNTRY' has I at 'INDIA' (2) and 'IS' (1): 3.
4. s ends with 'TRY' in capitals, so endswith('try') is False; startswith('India') is False (s starts with 'india').

Answer: 'India is my country', 'India Is My Country', 3, False, False

Searching: find() and index()

NCERT section NCERT 8.5

• s.find(sub[, start[, end]]) returns the lowest index where sub begins, searching s[start:end]; it returns -1 if sub is not found.
• s.index(sub[, start[, end]]) does the same but raises a ValueError if sub is not found.

To find a later occurrence, start the search after the previous one: s.find(sub, previous + 1).

Worked example

s = 'abracadabra'. Find s.find('a'), s.find('a', 1), s.find('bra'), s.find('bra', 2), s.find('z'), and what s.index('z') does.

Try it first, then show the solutionSolution

1. First 'a' at 0; searching from 1, next 'a' at 3.
2. 'bra' at 1; searching from 2, next 'bra' at 8.
3. 'z' is absent: find gives -1, index raises ValueError.

Answer: 0, 3, 1, 8, -1; s.index('z') raises ValueError.

Testing methods (is...)

NCERT section NCERT 8.5

These return True or False. For an empty string all of them return False.

• isalnum(): every character is a letter or digit (no spaces or symbols).
• isalpha(): every character is a letter.
• isdigit(): every character is a digit.
• islower() / isupper(): there is at least one letter and all letters are lowercase / uppercase (digits and symbols are ignored): 'abc12'.islower() is True.
• isspace(): every character is whitespace (space, tab, newline).

Worked example

Give the results: 'Class11'.isalnum(), 'Class 11'.isalnum(), '2027'.isdigit(), '20.27'.isdigit(), 'HELLO 1'.isupper(), ' '.isspace(), ''.isalpha()

Try it first, then show the solutionSolution

1. 'Class11' has only letters and digits: True; the space in 'Class 11' makes it False.
2. '2027' True; '.' makes '20.27' False.
3. 'HELLO 1' has letters, all uppercase: True.
4. ' ' is all spaces: True; the empty string gives False.

Answer: True, False, True, False, True, True, False

Stripping, replacing, joining and splitting

NCERT section NCERT 8.5

• strip(), lstrip(), rstrip(): remove whitespace from both ends, the left end only, or the right end only. Given characters as an argument, they remove any of those characters from the ends: 'xxhixx'.strip('x') = 'hi'. Characters in the middle are never removed.
• s.replace(old, new): a new string with every occurrence of old replaced by new.
• sep.join(iterable): joins the strings of a list, tuple or string with sep between them: '-'.join(['10', '06', '2027']) = '10-06-2027'; '*'.join('abc') = 'a*b*c'. All items must be strings.
• s.partition(sep): splits at the first occurrence of sep into a tuple of three: (before, sep, after). If sep is not found: (s, '', '').
• s.split(sep): returns a list of the parts. With no argument it splits on any run of whitespace and drops empty parts; with a separator it splits at each occurrence and can give empty strings: 'a,,b'.split(',') = ['a', '', 'b'].

Worked example

Give the output:

s = '  roll:25:Asha  '
t = s.strip()
print(t.split(':'))
print(t.partition(':'))
print(t.replace(':', ' | '))
print('/'.join(t.split(':')[::-1]))

Try it first, then show the solutionSolution

1. t = 'roll:25:Asha' (spaces removed at both ends).
2. split(':') gives three parts in a list.
3. partition splits only at the first ':' into a 3-tuple.
4. Every ':' replaced by ' | '.
5. Reversed list ['Asha', '25', 'roll'] joined with '/'.

Answer: ['roll', '25', 'Asha']
('roll', ':', '25:Asha')
roll | 25 | Asha
Asha/25/roll

Strings in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• String Indexing, Slicing, Immutability and Traversal with Outputs (CBSE 11)
• String Methods: find, count, replace, strip, split, join and the is-Tests (CBSE 11)

Strings: syntax and key facts

Indices

0 to n-1 from the left; -1 to -n from the right; s[-k] = s[n-k]

s[n] is an IndexError.

Slice

s[start:stop:step], stop excluded; s[::-1] reverses

Slices never raise IndexError.

Operators

+ joins, * repeats (int count), in / not in test substrings

Comparison is by character codes; 'A' < 'a'.

Search

1. 1.find()
2. 2.index or -1; index()
3. 3.index or ValueError; count()
4. 4.non-overlapping occurrences

All take optional start and end.

Split and join

1. 1.s.split(sep)
2. 2.list; sep.join(list)
3. 3.str; s.partition(sep)
4. 4.3-tuple

split() with no argument splits on whitespace.

Case

capitalize(): first upper, rest lower; title(): each word capitalised; upper(); lower()

Methods return new strings.

How the school exam asks Strings

The question patterns to expect, and the method that works for each.

Tricks to remember Strings

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Stop is not served

  In a slice the stop index is never included, just as range() never reaches stop.

  Use it when: Every slicing question.

• find is friendly, index is strict

  find() quietly gives -1; index() complains with ValueError.

  Use it when: find versus index.

• Partition = 3 parts, split = list

  partition() always returns a 3-tuple; split() returns a list of any length.

  Use it when: Output type questions.

• Capitalise once, title every word

  capitalize() fixes the first letter of the whole string; title() fixes every word.

  Use it when: Case-method outputs.

• Methods give new strings

  Strings never change; catch the result in a variable.

  Use it when: Questions where a method's result is not stored.

Strings: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

s = '  learn PYTHON daily  '
print(s.strip().capitalize())
print(s.split())
print(s.strip().upper().count('N'))

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Strings

The slips examiners see most often, and what to write instead.

• ✗Mistake: Expecting a method to change the string in place

  Strings are immutable; methods return new strings.

  Wrong working

  s = 'hi'
  s.upper()
  print(s) → HI

  Right working

  s = 'hi'
  s.upper()
  print(s) → hi; use s = s.upper()

  ✓Do this: Assign the result back if you need it.

  Usually costs: 1 mark

• ✗Mistake: Including the stop character in a slice

  s[a:b] ends at index b - 1.

  Wrong working

  'PYTHON'[1:4] = 'YTHO'

  Right working

  'PYTHON'[1:4] = 'YTH'

  ✓Do this: Count b - a characters from index a.

  Usually costs: 1 mark

• ✗Mistake: Thinking capitalize() keeps the rest of the string

  capitalize() lowers every other character.

  Wrong working

  'python IS fun'.capitalize() = 'Python IS fun'

  Right working

  'Python is fun'

  ✓Do this: Remember the rest is lowered.

  Usually costs: 1 mark

• ✗Mistake: Writing the separator as an argument of join()

  join() is called on the separator and takes the list.

  Wrong working

  ['a', 'b'].join('-')

  Right working

  '-'.join(['a', 'b'])

  ✓Do this: separator.join(list).

  Usually costs: 1 mark

• ✗Mistake: Assuming strip() removes spaces inside the string

  strip() works only at the two ends.

  Wrong working

  ' a b '.strip() = 'ab'

  Right working

  ' a b '.strip() = 'a b'; use replace(' ', '') to remove all spaces

  ✓Do this: Ends only: strip; everywhere: replace.

  Usually costs: 1 mark

NCERT exercise guide: Strings

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Strings

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   What is the output of print('Python'[2:5])? (a) yth (b) tho (c) thon (d) ytho

   Hint

   Indices 2, 3, 4.

   Answer

   (b) tho

2. Q2 · 1 mark · Very short answerExam level

   Give the output of 'abc'.upper().isupper(), 'ABC1'.isalpha(), 'a-b-c'.split('-').

   Hint

   isalpha rejects digits.

   Answer

   True False ['a', 'b', 'c']

3. Q3 · 2 marks · Short answerHarder exam level

   Predict the output:

   s = 'GOOD MORNING'
   print(s.partition(' '), s.partition('X'))

   Hint

   Separator not found.

   Answer

   ('GOOD', ' ', 'MORNING') ('GOOD MORNING', '', '')

4. Q4 · 2 marks · Short answerExam level

   Why does the following give an error? Correct it.
   s = 'tool'
   s[0] = 'c'

   Hint

   Immutability.

   Answer

   Strings are immutable, so item assignment raises TypeError. Correct: s = 'c' + s[1:] (gives 'cool').

5. Q5 · 2 marks · Short answerHarder exam level

   Give the output of print('*'.join('2027'), '2027'.find('2', 1), '2027'.rstrip('7'))

   Hint

   join on a string uses each character.

   Answer

   2*0*2*7 2 202

6. Q6 · 3 marks · Short answerExam level

   Write a program to count the uppercase letters, lowercase letters and spaces in 'Digital India Mission'.

   Hint

   isupper(), islower(), isspace() on each character.

   Answer

   s = 'Digital India Mission'
   u = l = sp = 0
   for ch in s:
   if ch.isupper(): u += 1
   elif ch.islower(): l += 1
   elif ch.isspace(): sp += 1
   print(u, l, sp)
   Output: 3 16 2

7. Q7 · 1 mark · MCQHarder exam level

   Assertion (A): 'ab' in 'cab' and 'ab' in ['cab'] both give True. Reason (R): in checks substrings for a string and items for a list. Choose (a) both true, R explains A (b) both true, R does not explain A (c) A false, R true (d) both false.

   Hint

   Is 'ab' an item of ['cab']?

   Answer

   (c) A is false ('ab' in ['cab'] is False, since 'ab' is not an item of the list); R is true.

8. Q8 · 2 marks · Short answerHarder exam level

   Predict the output:

   w = 'level'
   print(w == w[::-1], w[1:-1], w.count('l', 1))

   Hint

   Search from index 1.

   Answer

   True eve 1

9. Q9 · 2 marks · Short answerHOTS / competency

   Write a program that prints the initials of a full name, e.g. 'Mohandas Karamchand Gandhi' → 'M.K.G.'

   Hint

   split() then take w[0].

   Answer

   name = 'Mohandas Karamchand Gandhi'
   print(''.join(w[0].upper() + '.' for w in name.split()))
   (or build the string in a loop) Output: M.K.G.

10. Q10 · 3 marks · Short answerHOTS / competency

    Write a program that removes all vowels from 'Education is power' and prints the result and the number of vowels removed.

    Hint

    Build a new string, count separately.

    Answer

    s = 'Education is power'
    r = ''
    c = 0
    for ch in s:
    if ch.lower() in 'aeiou':
    c += 1
    else:
    r += ch
    print(r, c)
    Output: dctn s pwr 8

11. Q11 · 2 marks · Short answerHOTS / competency

    s = 'banana'. Without running, give s.replace('a', 'o', 2) and explain the third argument.

    Hint

    Count of replacements.

    Answer

    'bonona'. The optional third argument limits how many occurrences (from the left) are replaced; here only the first two 'a's.

12. Q12 · 1 mark · Very short answerHarder exam level

    What is the output of print(len(' a b '.strip()), len('a,b,,c'.split(',')))?

    Hint

    Empty part between the commas.

    Answer

    3 4

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is s[-1] for s = 'CBSE'?Show answer

   'E'.

2. 2.What does 'abc' * 0 give?Show answer

   '' (an empty string).

3. 3.What does find() return when the substring is absent?Show answer

   -1.

4. 4.What type does partition() return?Show answer

   A tuple of three strings.

5. 5.What is 'hello world'.title()?Show answer

   'Hello World'.

6. 6.What error does s[0] = 'x' give?Show answer

   TypeError (strings are immutable).

7. 7.What is ''.isdigit()?Show answer

   False.

8. 8.What is 'a b c'.split()?Show answer

   ['a', 'b', 'c'].

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐String: immutable sequence of characters; indices 0..n-1 and -1..-n.
• ☐+ joins, * repeats, in tests substrings.
• ☐s[a:b:c] stops before b; s[::-1] reverses; slices never raise IndexError.
• ☐capitalize: first letter only (rest lower); title: every word.
• ☐find gives -1, index gives ValueError when absent.
• ☐is... methods are False for an empty string.
• ☐strip removes only from the ends; replace changes every occurrence.
• ☐split → list; partition → 3-tuple; sep.join(list) → string.

Strings: frequently asked questions