DSA Bite #5 - Strings

Welcome back, coding explorer!
Today’s DSA Bite takes you into the world of Strings - one of the most fundamental and interview-heavy data structures. Whether you’re reversing names, validating palindromes, or parsing user inputs, strings are everywhere.

We’ll break it all down - from basics to advanced patterns - just like we did for arrays.

Part 1: What is a String?

A string is a sequence of characters - letters, digits, symbols - enclosed in quotes.
Think of it like a necklace of characters, where each bead (character) has an index.

Example

name = "Fahim"
print(name[0])  # F

let name = "Fahim";
console.log(name[0]); // F

Key String Terms

Term	Description	Example
Character	A single symbol	`'A'`, `'5'`, `'$'`
Index	Position of a character (starts at 0)	`"Code"[1] → 'o'`
Immutable	Strings can’t be changed directly	You must create a new one
Length	Number of characters	`len(s)` / `s.length`

Part 2: Common String Operations

1️⃣ Accessing Characters

s = "hello"
print(s[1])  # e

let s = "hello";
console.log(s[1]); // e

2️⃣ Concatenation

greet = "Hi " + "there"
print(greet)  # Hi there

let greet = "Hi " + "there";
console.log(greet); // Hi there

3️⃣ Substring

s = "CodeAlignr"
print(s[0:4])  # Code

let s = "CodeAlignr";
console.log(s.substring(0,4)); // Code

4️⃣ Searching

if "Align" in s:
    print("Found!")

if (s.includes("Align")) console.log("Found!");

5️⃣ Replace & Split

text = "I love Python"
print(text.replace("Python", "AI"))
print(text.split(" "))

let text = "I love JavaScript";
console.log(text.replace("JavaScript", "AI"));
console.log(text.split(" "));

Part 3: Common Interview Patterns

1️⃣ Reverse a String

s = "hello"
print(s[::-1])  # olleh

let s = "hello";
console.log(s.split('').reverse().join('')); // olleh

2️⃣ Check Palindrome

s = "madam"
print(s == s[::-1])  # True

let s = "madam";
console.log(s === s.split('').reverse().join('')); // true

3️⃣ Count Vowels

s = "developer"
count = sum(1 for c in s if c in "aeiou")
print(count)

let s = "developer";
let count = [...s].filter(c => "aeiou".includes(c)).length;
console.log(count);

4️⃣ Frequency of Characters

from collections import Counter
print(Counter("banana"))

let s = "banana";
let freq = {};
for (let ch of s) freq[ch] = (freq[ch] || 0) + 1;
console.log(freq);

Part 4: Advanced Patterns

Two Pointers – Remove Duplicates or Check Palindrome Efficiently

s = "level"
left, right = 0, len(s)-1
is_pal = True
while left < right:
    if s[left] != s[right]:
        is_pal = False
        break
    left += 1
    right -= 1
print(is_pal)

let s = "level";
let left = 0, right = s.length - 1, isPal = true;
while (left < right) {
  if (s[left] !== s[right]) { isPal = false; break; }
  left++; right--;
}
console.log(isPal);

Sliding Window – Longest Substring Without Repeating Characters

s = "abcabcbb"
left = 0
seen = set()
max_len = 0

for right in range(len(s)):
    while s[right] in seen:
        seen.remove(s[left])
        left += 1
    seen.add(s[right])
    max_len = max(max_len, right - left + 1)

print(max_len)

let s = "abcabcbb";
let left = 0, seen = new Set(), maxLen = 0;

for (let right = 0; right < s.length; right++) {
  while (seen.has(s[right])) {
    seen.delete(s[left]);
    left++;
  }
  seen.add(s[right]);
  maxLen = Math.max(maxLen, right - left + 1);
}
console.log(maxLen);

Anagram Check

s1, s2 = "listen", "silent"
print(sorted(s1) == sorted(s2))

let s1 = "listen", s2 = "silent";
console.log([...s1].sort().join('') === [...s2].sort().join(''));

Part 5: Mini Challenges

1️⃣ Reverse words in a sentence
Input: "I love Python" → Output: "Python love I"

2️⃣ Find the first non-repeating character
Input: "aabbcd" → Output: "c"

3️⃣ Check if two strings are anagrams.
Input: "angel", "glean" → Output: True

4️⃣ Find the longest common prefix.
Input: ["flower", "flow", "flight"] → Output: "fl"

Expert Challenge – String Compression

Compress a string like "aaabbcc" → "a3b2c2"

s = "aaabbcc"
res, count = "", 1
for i in range(1, len(s)):
    if s[i] == s[i-1]:
        count += 1
    else:
        res += s[i-1] + str(count)
        count = 1
res += s[-1] + str(count)
print(res)

let s = "aaabbcc";
let res = "", count = 1;
for (let i = 1; i < s.length; i++) {
  if (s[i] === s[i - 1]) count++;
  else { res += s[i - 1] + count; count = 1; }
}
res += s[s.length - 1] + count;
console.log(res);

🤖 AI Tip

Ask AI:

❝

“Explain and generate problems using Two Pointer and Sliding Window techniques on strings.”

Use it to practice substring, palindrome, and frequency problems efficiently.

✅ Sign-off
That’s your Strings Bite – Beginner to Expert! 🎉
You’ve explored everything from basic manipulation to advanced string algorithms.

Next up: Linked Lists - where data gets dynamic and connected!