Problem Overview

You can view the full problem statement here: Codeforces 1146A - Love "A"

In this problem, you are given a string consisting of lowercase English letters. You need to find the maximum possible length of a subsequence (not necessarily contiguous) such that the number of letter 'a' in it is strictly more than the number of all other letters combined.

Key Points:

  • A subsequence allows picking or skipping characters, but order must be preserved.
  • We must maximize the length while satisfying the condition:
    • Number of 'a' > Number of non-'a'.

C++ Code

#include<bits/stdc++.h>

using namespace std;

int main()
{
    char a[105];
    cin >> a;
    int x = 0, y = 0;
    for(int i = 0; i < strlen(a); i++){
        if(a[i] == 'a'){
            x++;
        }
        else{
            y++;
        }
    }
    if(x > y){
        cout << x + y << endl;
    }
    else{
        cout << 2 * x - 1 << endl;
    }

    return 0;
}

Explanation of the Approach

  1. Input the String:
    Read the input string a.

  2. Count Characters:
    Traverse the string:
    • x = count of character 'a'.
    • y = count of characters other than 'a'.

  3. Check Condition:

    • If the number of 'a's (x) is already greater than other characters (y), the entire string can be taken as it satisfies the condition.
      • Otherwise, we can only take some non-'a' letters to maintain the condition.
      • In that case, the maximum length possible is 2 * x - 1:
      • We can take all 'a's and only enough non-'a's so that 'a's still remain more.

  4. Print the Answer: Output the maximum possible subsequence length according to the condition.

Step-by-Step Dry Run Example

Input:

baa

Process:

  • Characters: 'b', 'a', 'a'
  • x = 2 (count of 'a')
  • y = 1 (count of other letters)
  • Since x > y (2 > 1), we can take the whole string.
  • Output: x + y = 3

Input:

abcde

Process:

  • Characters: 'a', 'b', 'c', 'd', 'e'
  • x = 1 (only one 'a')
  • y = 4 (other letters)
  • Since x <= y, we take as many characters as possible without violating the rule.
  • Output: 2 * x - 1 = 2 * 1 - 1 = 1

Time and Space Complexity

  • Time Complexity:

    • Traversing the string to count characters: O(n)

  • Space Complexity:

    • Using only a few integer variables: O(1)

Common Mistakes to Avoid

  • Incorrect Subsequence Understanding:
    Remember, a subsequence allows skipping characters but keeps the order.
  • Forgetting the Strictly Greater Condition:
    The number of 'a's must be strictly more, not equal.
  • Off-by-One Errors:
    When calculating 2 * x - 1, ensure you correctly subtract 1 to maintain "strictly more" condition.
  • Not Resetting Counters (if multiple test cases):
    If extended to multiple test cases, be careful to reset x and y for each case.

Related Problems for Practice

Conclusion

The "Love 'A'" problem is an excellent exercise for beginners to practice string traversal, counting characters, and building conditions based on character frequencies.
It teaches how to optimize subsequences to meet a specific condition.
With practice, you can develop a strong intuition for problems that involve counting and basic string manipulation.

Keep practicing such problems to sharpen your problem-solving skills!