Longest Palindromic Substring in DSA Python - Execution Trace

Concept Flow - Longest Palindromic Substring

Start at each index i in string

↓

Expand around center i (odd length)

↓

Expand around center i and i+1 (even length)

↓

Check palindrome length

↓

Update longest palindrome if longer

↓

Move to next index i+1

↓

End

We check each position as a center and expand outwards to find palindromes, updating the longest found.

Execution Sample

DSA Python

def longest_palindrome(s):
    start = 0
    max_len = 1
    for i in range(len(s)):
        # Odd length palindrome
        l, r = i, i
        while l >= 0 and r < len(s) and s[l] == s[r]:
            if (r - l + 1) > max_len:
                start = l
                max_len = r - l + 1
            l -= 1
            r += 1
        # Even length palindrome
        l, r = i, i + 1
        while l >= 0 and r < len(s) and s[l] == s[r]:
            if (r - l + 1) > max_len:
                start = l
                max_len = r - l + 1
            l -= 1
            r += 1
    return s[start:start+max_len]

This code finds the longest palindrome by expanding around each character for odd length and around each pair for even length palindromes.

Execution Table

Step	Operation	Center Indices (l,r)	Palindrome Checked	Longest Palindrome	Visual State
1	Start at index 0	l=0, r=0	"a"	"a"	"a"
2	Expand around center 0	l=-1, r=1	Stop (l<0)	"a"	"a"
3	Start at index 1	l=1, r=1	"b"	"a"	"a"
4	Expand around center 1	l=0, r=2	"aba"	"aba"	"aba"
5	Expand around center 1	l=-1, r=3	Stop (l<0)	"aba"	"aba"
6	Start at index 2	l=2, r=2	"a"	"aba"	"aba"
7	Expand around center 2	l=1, r=3	"bab"	"aba"	"aba"
8	Expand around center 2	l=0, r=4	"ababa"	"ababa"	"ababa"
9	Expand around center 2	l=-1, r=5	Stop (l<0)	"ababa"	"ababa"
10	Start at index 3	l=3, r=3	"b"	"ababa"	"ababa"
11	Expand around center 3	l=2, r=4	"aba"	"ababa"	"ababa"
12	Expand around center 3	l=1, r=5	Stop (r>=len)	"ababa"	"ababa"
13	Start at index 4	l=4, r=4	"a"	"ababa"	"ababa"
14	Expand around center 4	l=3, r=5	Stop (r>=len)	"ababa"	"ababa"
15	End of string reached	-	-	"ababa"	"ababa"

💡 All centers checked, longest palindrome found is "ababa"

Variable Tracker

Variable	Start	After Step 4	After Step 8	After Step 15
start	0	0	0	0
max_len	1	3	5	5
l	-	0	0	-
r	-	2	4	-

Key Moments - 3 Insights

Why do we expand around two centers (i and i+1) for even length palindromes?

For even length palindromes, we expand around i and i+1 to catch palindromes like "abba". The original code only expands around one center, so even length palindromes are missed. The updated code includes expansion around both centers.

Why does the expansion stop when l < 0 or r >= len(s)?

Because l and r represent indices expanding outwards. If l < 0 or r >= len(s), we are outside the string bounds, so we cannot check further. This is shown in steps 2, 5, 9, 12, and 14 where expansion stops.

How is the longest palindrome updated during expansion?

At each expansion, if the current palindrome length (r - l + 1) is greater than max_len, we update start and max_len. For example, at step 4, max_len changes from 1 to 3 and start to 0, and at step 8, max_len updates to 5 and start to 0.

Visual Quiz - 3 Questions

Look at the execution table, what is the longest palindrome after step 4?

A "aba"

B "b"

C "a"

D "ababa"

💡 Hint

Check the 'Longest Palindrome' column at step 4 in the execution_table.

At which step does the palindrome length first update to 5?

A Step 12

B Step 5

C Step 8

D Step 15

💡 Hint

Look at the 'Longest Palindrome' column and find when it changes to "ababa".

If we added even length palindrome checks, which step would change?

A Step 1

B Step 6

C Step 3

D Step 14

💡 Hint

Even length checks happen around centers i and i+1, which is done at each index like step 6.

Concept Snapshot

Longest Palindromic Substring:
- Check each index as palindrome center
- Expand left and right while chars match
- Update longest substring if longer found
- Repeat for all indices
- Return substring from start with max length

Full Transcript

The longest palindromic substring is found by checking each character as a center and expanding outwards to see if the characters on both sides match. We keep track of the longest palindrome found so far by updating the start index and maximum length. Expansion stops when indices go out of string bounds or characters don't match. This example checks both odd length palindromes by expanding around one center and even length palindromes by expanding around two centers. The final longest palindrome substring is returned after checking all centers.

Practice

1. What is the main goal of the Longest Palindromic Substring problem?

easy

A. Count the total number of palindromic substrings

B. Reverse the string and compare with original

C. Check if the entire string is a palindrome

D. Find the longest substring that reads the same forwards and backwards

Solution

1. Step 1: Understand the problem statement

   The problem asks to find the longest continuous part of the string that reads the same forwards and backwards.

2. Step 2: Differentiate from related problems

   Counting palindromic substrings or checking the whole string is different from finding the longest palindromic substring.

3. Final Answer:

   Find the longest substring that reads the same forwards and backwards -> Option D

4. Quick Check:

   Longest palindromic substring = Find longest palindrome [OK]

Hint: Longest palindrome means longest continuous mirrored substring [OK]

Common Mistakes:

• Confusing substring with subsequence
• Counting all palindromes instead of longest
• Checking whole string palindrome only

2. Which of the following Python code snippets correctly initializes variables to find the longest palindromic substring?

easy

A. start, max_len = 0, 1

B. start = max_len = None

C. start, max_len = 1, 0

D. start = max_len = 0

Solution

1. Step 1: Understand variable roles

   We need to track the start index and maximum length of the palindrome found. Starting with 0 and 1 is common because a single character is a palindrome.

2. Step 2: Check initialization correctness

   start, max_len = 0, 1 sets start=0 and max_len=1, which is correct. start, max_len = 1, 0 reverses values incorrectly. Options A, B, and C do not set max_len properly.

3. Final Answer:

   start, max_len = 0, 1 -> Option A

4. Quick Check:

   Initialize start=0 and max_len=1 for single char palindrome [OK]

Hint: Start at 0, max length at least 1 for single char palindrome [OK]

Common Mistakes:

• Setting max_len to 0 initially
• Using None instead of integers
• Swapping start and max_len values

3. What will be the output of the following code snippet?

def expand_around_center(s, left, right):
    while left >= 0 and right < len(s) and s[left] == s[right]:
        left -= 1
        right += 1
    return right - left - 1

s = "babad"
length = expand_around_center(s, 1, 1)
print(length)

medium

A. 1

B. 5

C. 3

D. 0

Solution

1. Step 1: Understand the function behavior

   The function expands around the center indices (left and right) while characters match, then returns the length of the palindrome found.

2. Step 2: Trace the expansion for s='babad' at center (1,1)

   Center at index 1 ('a'). Expand left=1, right=1: 'a'=='a' true, expand to left=0, right=2: 'b'=='b' true, expand to left=-1 (stop). Length = right - left - 1 = 3 - (-1) - 1 = 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Expand around center returns palindrome length 3 [OK]

Hint: Expand while chars match, length = right - left - 1 [OK]

Common Mistakes:

• Not subtracting 1 in length calculation
• Confusing indices after expansion
• Assuming length equals right - left

4. Identify the error in the following code snippet for finding the longest palindromic substring:

def longest_palindrome(s):
    start, max_len = 0, 1
    for i in range(len(s)):
        len1 = expand_around_center(s, i, i)
        len2 = expand_around_center(s, i, i+1)
        length = max(len1, len2)
        if length > max_len:
            start = i - length // 2
            max_len = length
    return s[start:start+max_len]

medium

A. The loop should run till len(s)-1 to avoid index error

B. The start index calculation is incorrect for even length palindromes

C. max_len should be initialized to 0 instead of 1

D. expand_around_center function is missing

Solution

1. Step 1: Analyze start index update logic

   For even length palindromes, start index calculation differs because center is between two characters. Using i - length // 2 works for odd length but not always for even length.

2. Step 2: Identify correct start calculation

   For even length, start should be i - (length // 2) + 1 to correctly locate the palindrome start.

3. Final Answer:

   The start index calculation is incorrect for even length palindromes -> Option B

4. Quick Check:

   Start index must adjust differently for even palindromes [OK]

Hint: Adjust start index differently for even-length palindromes [OK]

Common Mistakes:

• Using same start formula for odd and even palindromes
• Ignoring even length palindrome cases
• Not updating max_len properly

5. Given the string "forgeeksskeegfor", which substring is the longest palindromic substring?

hard

A. "geeksskeeg"

B. "forgeeksskeegfor"

C. "geeksskeegf"

D. "eksskeeg"

Solution

1. Step 1: Identify palindromic substrings in the string

   Check substrings that read the same forwards and backwards. "geeksskeeg" is a palindrome of length 10 inside the string.

2. Step 2: Compare lengths of candidate substrings

   "forgeeksskeegfor" is not a palindrome as a whole. "geeksskeegf" and "eksskeeg" are shorter palindromes.

3. Final Answer:

   "geeksskeeg" -> Option A

4. Quick Check:

   Longest palindrome in "forgeeksskeegfor" is "geeksskeeg" [OK]

Hint: Look for longest mirrored substring inside the string [OK]

Common Mistakes:

• Assuming whole string is palindrome
• Choosing substring with extra characters
• Ignoring palindrome length