Remove Palindromic Subsequences - Practice Coding Problems

Remove Palindromic Subsequences - Problem

You're given a string s consisting only of the letters 'a' and 'b'. Your task is to completely empty this string by removing palindromic subsequences in the minimum number of steps possible.

In each step, you can remove one palindromic subsequence from the string. A subsequence doesn't need to be contiguous - you can pick characters from anywhere in the string as long as you maintain their relative order.

Key insight: Since we only have two characters ('a' and 'b'), this problem has a surprisingly elegant solution! Think about what palindromic subsequences you can always form with just these two letters.

Goal: Return the minimum number of steps needed to make the string empty.

Input & Output

example_1.py — Basic Palindrome

$ Input: s = "ababa"

› Output: 1

💡 Note: The entire string "ababa" is already a palindrome, so we can remove it in just 1 step.

example_2.py — Non-Palindrome

$ Input: s = "abb"

› Output: 2

💡 Note: "abb" is not a palindrome. We can remove all 'a's in step 1 (subsequence "a"), then remove all 'b's in step 2 (subsequence "bb"), requiring 2 steps total.

example_3.py — Single Character

$ Input: s = "baabb"

› Output: 2

💡 Note: "baabb" is not a palindrome (first char 'b' ≠ last char 'b'... wait, it starts and ends with 'b'! Let's check: b-a-a-b-b. Comparing positions: b≠b? Actually b=b, a≠b, so not palindrome). We need 2 steps: remove all 'b's, then all 'a's.

Visualization

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Understanding the Visualization

Check if Already Palindrome

If the entire string reads same forwards/backwards, remove in 1 step

Otherwise, Use 2 Steps

Step 1: Remove all 'a's (forms 'aaa...' palindrome)

Clean Up Remaining

Step 2: Remove all 'b's (forms 'bbb...' palindrome)

Key Takeaway

🎯 Key Insight: With only 2 characters, we can always group identical characters together to form palindromic subsequences, requiring at most 2 steps!

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n)

Generate 2^n subsequences, check each for palindrome in O(n) time

⚠ Quadratic Growth

Space Complexity

O(2^n)

Store all possible subsequences

⚠ Quadratic Space

Constraints

1 ≤ s.length ≤ 1000
s[i] is either 'a' or 'b'
Key insight: With only 2 distinct characters, answer is always 0, 1, or 2

Asked in

f Facebook 15 G Google 12 a Amazon 8 ⊞ Microsoft 6

The key insight is that with only 2 characters ('a' and 'b'), we can always solve this in at most 2 steps. If the string is already a palindrome, return 1. Otherwise, return 2 (remove all 'a's, then all 'b's). Use two pointers for O(n) palindrome checking.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Generate All Subsequences)	O(2^n * n)	O(2^n)	Try all possible palindromic subsequences and find minimum steps
Two Pointers (Palindrome Check + Logic)	O(n)	O(1)	Use two pointers to check if string is palindrome, then apply optimal logic

Brute Force (Generate All Subsequences) — Algorithm Steps

Generate all possible subsequences of the string
Filter subsequences that are palindromes
Use backtracking to find minimum set of palindromic subsequences
Return the size of the minimum set

Visualization

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Step-by-Step Walkthrough

Generate Subsequences

Create all 2^n possible subsequences

Filter Palindromes

Keep only palindromic subsequences

Find Minimum Cover

Use backtracking to find minimum set

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int isPalindrome(char* seq) {
    int len = strlen(seq);
    int left = 0, right = len - 1;
    while (left < right) {
        if (seq[left] != seq[right]) return 0;
        left++;
        right--;
    }
    return 1;
}

int minPalindromicSubsequences(char* s) {
    int len = strlen(s);
    if (len == 0) return 0;
    
    // Simplified brute force - just check if entire string is palindrome
    // If yes, return 1, otherwise return 2 (since we only have 'a' and 'b')
    if (isPalindrome(s)) return 1;
    return 2;
}

int removePalindromeSub(char* s) {
    return minPalindromicSubsequences(s);
}

int main() {
    char s[1001];
    scanf("%s", s);
    
    // Remove quotes if present
    int len = strlen(s);
    if (s[0] == '"' && s[len-1] == '"') {
        s[len-1] = '\0';
        memmove(s, s+1, len-1);
    }
    
    printf("%d\n", removePalindromeSub(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n)

Generate 2^n subsequences, check each for palindrome in O(n) time

⚠ Quadratic Growth

Space Complexity

O(2^n)

Store all possible subsequences

⚠ Quadratic Space

Constraints

1 ≤ s.length ≤ 1000
s[i] is either 'a' or 'b'
Key insight: With only 2 distinct characters, answer is always 0, 1, or 2

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Generate All Subsequences) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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