Count Palindromic Substrings

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Steps

setup

Initialize variables

We start by setting the string length n to 3 and initializing the palindrome count to 0.

💡 Initializing count and length is essential before starting expansions to track palindromes found.

Line:

n = len(s)
count = 0

💡 The algorithm prepares to iterate over each character as a center for palindrome expansion.

expand

Start first odd-length expansion at center 0

Expand around center index 0 for odd-length palindromes: left=0, right=0. Characters match ('a' == 'a'), so count increments to 1.

💡 Odd-length palindromes always start with a single character, which is a palindrome by definition.

Line:

expand(i, i)  # i=0

💡 Single characters are palindromes; this step confirms the base case.

expand

Attempt to expand odd-length palindrome at center 0 further

Try expanding left to -1 and right to 1, but left is out of bounds, so expansion stops.

💡 Expansion stops when pointers go out of string bounds.

Line:

while left >= 0 and right < n and s[left] == s[right]: ...

💡 Boundaries limit palindrome expansion.

expand

Start first even-length expansion at center between 0 and 1

Expand around center between indices 0 and 1: left=0, right=1. Characters 'a' and 'b' do not match, so no palindrome found.

💡 Even-length palindromes require two matching characters at the center.

Line:

expand(i, i + 1)  # i=0

💡 No even-length palindrome starts at this center.

expand

Start odd-length expansion at center 1

Expand around center index 1: left=1, right=1. Characters match ('b' == 'b'), count increments to 2.

💡 Each single character is a palindrome, so count increases.

Line:

expand(i, i)  # i=1

💡 Single character palindrome found at index 1.

expand

Attempt to expand odd-length palindrome at center 1 further

Expand left=0, right=2: characters 'a' and 'c' do not match, stop expansion.

💡 Expansion stops when characters differ.

Line:

while left >= 0 and right < n and s[left] == s[right]: ...

💡 Mismatch ends palindrome expansion.

expand

Start even-length expansion at center between 1 and 2

Expand around center between indices 1 and 2: left=1, right=2. Characters 'b' and 'c' do not match, no palindrome found.

💡 Even-length palindrome requires matching adjacent characters.

Line:

expand(i, i + 1)  # i=1

💡 No even-length palindrome here.

expand

Start odd-length expansion at center 2

Expand around center index 2: left=2, right=2. Characters match ('c' == 'c'), count increments to 3.

💡 Single character palindrome found at last index.

Line:

expand(i, i)  # i=2

💡 Counting all single character palindromes.

expand

Attempt to expand odd-length palindrome at center 2 further

Expand left=1, right=3, but right is out of bounds, stop expansion.

💡 Expansion stops at string boundary.

Line:

while left >= 0 and right < n and s[left] == s[right]: ...

💡 Boundaries limit palindrome expansion.

expand

Start even-length expansion at center between 2 and 3

Expand around center between indices 2 and 3: right index 3 is out of bounds, no palindrome found.

💡 Even-length expansions require valid right index.

Line:

expand(i, i + 1)  # i=2

💡 No palindrome possible beyond string end.

reconstruct

Return total palindrome count

All expansions done, return the total count of palindromic substrings found: 3.

💡 Final count reflects all palindromes found by expansions.

Line:

return count

💡 The algorithm successfully counted all palindromic substrings.

def countSubstrings(s: str) -> int:
    n = len(s)  # STEP 1
    count = 0  # STEP 1

    def expand(left, right):
        nonlocal count
        while left >= 0 and right < n and s[left] == s[right]:  # STEP 2-10
            count += 1  # STEP 2,5,8
            left -= 1  # STEP 3,6,9
            right += 1  # STEP 3,6,9

    for i in range(n):  # STEP 2,4,5,7,8,10
        expand(i, i)      # Odd length palindromes
        expand(i, i + 1)  # Even length palindromes

    return count  # STEP 11

if __name__ == '__main__':
    print(countSubstrings('abc'))  # Expected output: 3

📊

Count Palindromic Substrings - Watch the Algorithm Execute, Step by Step

Watching the expansion around centers visually reveals how palindromes grow and overlap, making the counting process intuitive and clear.

Initializing dp array

i\w	0	1	2
i=0	?	?	?
i=1	?	?	?
i=2	?	?	?

Item 0 - wt:1 val:1

i\w	0	1	2
i=0	1	?	?
i=1	?	?	?
i=2	?	?	?

palindrome found

Item 0 - wt:1 val:1

i\w	0	1	2
i=0	1	?	?
i=1	?	?	?
i=2	?	?	?

palindrome confirmed

Item 0 - wt:0 val:1

i\w	0	1	2
i=0	1	?	?
i=1	?	?	?
i=2	?	?	?

no palindrome

Item 1 - wt:1 val:2

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	?

palindrome found

Item 1 - wt:0 val:2

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	?

no palindrome

Item 1 - wt:0 val:2

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	?

no palindrome

Item 2 - wt:1 val:3

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	1

palindrome found

Item 2 - wt:0 val:3

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	1

Item 2 - wt:0 val:3

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	1

Initializing dp array

i\w	0	1	2
i=0	1	?	?
i=1	?	1	?
i=2	?	?	1

palindrome

Key Takeaways

✓ Every single character in the string is a palindrome by itself.

This is the base case that the algorithm counts first before expanding.

✓ Palindrome expansions stop immediately when characters at the expanding pointers differ or boundaries are exceeded.

This shows how the algorithm efficiently avoids unnecessary checks.

✓ Even-length palindromes require two adjacent matching characters as a center, which may not always exist.

This explains why some expansions yield zero palindromes and helps understand the difference between odd and even expansions.

Practice

1. What is the time complexity of the dynamic programming solution for the longest palindromic substring problem, and why?

medium

A. O(n^2) because the DP table of size nxn is filled once

B. O(n) because each character is processed once

C. O(n^3) because each substring is checked with a nested loop and palindrome check

D. O(n^2) but with extra O(n) recursion stack space

Solution

1. Step 1: Identify loops in DP solution

   Two nested loops: outer over substring length (up to n), inner over start index (up to n), total O(n^2).

2. Step 2: Confirm no extra recursion or nested palindrome checks

   Palindrome check is O(1) via DP table lookup, no recursion stack used.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP table size nxn filled once -> O(n^2) time [OK]

Hint: DP fills nxn table once, no recursion [OK]

Common Mistakes:

• Confusing brute force O(n^3) with DP
• Assuming recursion stack adds O(n) space
• Thinking each substring palindrome check is O(n)

2. The following code attempts to find the longest palindromic substring using DP. Identify the line containing the subtle bug that causes incorrect results for some inputs.

medium

A. Line: if length < 3:

B. Line: dp[i][i] = True

C. Line: dp[i][j] = dp[i + 1][j - 1]

D. Line: for i in range(n - length + 1):

Solution

1. Step 1: Compare condition with correct code

   The correct condition is length <= 3, but here it is length < 3, so length=3 substrings are not handled correctly.

2. Step 2: Understand impact

   For length=3, dp[i][j] should be True if s[i]==s[j], but this code skips that, causing missed palindromes of length 3.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing < to <= fixes the bug [OK]

Hint: Check boundary conditions for length 3 palindromes [OK]

Common Mistakes:

• Off-by-one in length condition
• Confusing substring indices
• Incorrect DP table initialization

3. Suppose you want to count the number of distinct subsequences of t in s, but now characters in s can be reused multiple times (i.e., you can use the same character in s multiple times to form subsequences). Which modification to the DP approach correctly solves this variant?

hard

A. Use a 2D DP where dp[i][j] counts subsequences using first i chars of s and j chars of t, but allow reusing s[i] by not incrementing i on match

B. Change the recurrence to allow staying at the same i index when matching characters

C. Use the same DP but iterate j forward instead of backward in the inner loop

D. No change needed; original DP already counts subsequences with reuse

Solution

1. Step 1: Understand reuse means characters in s can be used multiple times

   Original DP assumes each character in s can be used at most once per subsequence.

2. Step 2: Modify DP to allow reusing s[i]

   Use 2D DP where dp[i][j] counts subsequences using first i chars of s for first j chars of t, and when s[i] matches t[j], allow staying at i (not incrementing) to reuse character.

3. Step 3: Recognize that iterating j forward or no change won't handle reuse correctly

   Only explicit state allowing reuse by not advancing i on match works.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Allowing reuse requires 2D DP with careful state transitions [OK]

Hint: Reuse means allowing same s[i] multiple times, so must not always increment i

Common Mistakes:

• Assuming original DP handles reuse
• Trying to fix by changing iteration order only

4. Suppose you want to find the longest palindromic subsequence in a string where characters can be reused unlimited times. Which modification to the DP approach correctly handles this variant?

hard

A. Modify DP to consider infinite reuse by allowing i to stay fixed while j moves, and vice versa

B. Use a greedy approach selecting matching characters repeatedly from both ends

C. Use the same DP but allow indices i and j to move independently without restriction

D. This problem reduces to longest palindromic substring, so use Manacher's algorithm

Solution

1. Step 1: Understand reuse means characters can be counted multiple times

   Standard LPS counts each character once; reuse requires relaxing index movement constraints.

2. Step 2: Modify DP transitions to allow staying at i or j to reuse characters

   DP must consider cases where i or j does not advance to reuse characters, effectively allowing repeated use.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Allowing indices to stay or move independently models reuse correctly [OK]

Hint: Reuse means relaxing index movement in DP transitions [OK]

Common Mistakes:

• Using greedy approach
• Confusing substring with subsequence
• Applying substring algorithms incorrectly

5. Suppose the wildcard pattern matching problem is extended so that the '*' wildcard can match any sequence of characters, but now the input string can be infinitely long (streaming input). Which modification to the DP approach is necessary to handle this scenario efficiently?

hard

A. Switch to a greedy two-pointer approach that backtracks on '*' positions to avoid DP overhead

B. Use recursion with memoization caching all substring pairs to handle infinite input

C. Precompute all possible matches for '*' in advance and store in a lookup table

D. Use the same space-optimized DP with arrays sized by pattern length, processing input character by character

Solution

1. Step 1: Understand infinite input constraints

   DP requires O(m*n) time and space, not feasible for infinite input length.

2. Step 2: Identify suitable approach

   Greedy two-pointer with backtracking on '*' positions can process streaming input efficiently without full DP table.

3. Step 3: Why other options fail

   Precomputing or recursion with memoization is impossible with infinite input; DP arrays cannot grow infinitely.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Greedy with backtracking handles streaming input efficiently [OK]

Hint: Infinite input forbids DP; greedy with backtracking is needed [OK]

Common Mistakes:

• Trying DP on infinite input
• Assuming memoization works with infinite data
• Ignoring streaming constraints