Find number of Strictly Increasing Sequences with ‘a’ and ‘b’ in a String

Given a string S of length N (1 <= N <= 10⁵) consisting of lower case alphabets. The task is to find the number of strictly increasing sequences P₁,P₂, …, P_k such that:

• For each i (1 ≤ i ≤ k), S_pi = ‘a’
• For each i (1 ≤ i < k), there exists j such that p_i < j < p_i+1 and S_j = ‘b’

The result should be calculated modulo 10^9 + 7.

Example:

Input: s = “abbaa”
Output: 5

Input: s = “baaaa”
Output: 4

Approach:

The idea is to use a greedy approach to solve this problem. Maintains two variables, result and temp, where result stores the current number of strictly increasing sequences ending with ‘a’ and temp stores the number of strictly increasing sequences ending with ‘a’ before the last ‘b’ was encountered.

Iterates over each character in the string

• Check if the character is ‘a’, then increments result by temp + 1. The +1 is for the sequence ending at the current ‘a’, and temp adds the sequences that can be formed by appending the current ‘a’ to all sequences ending with ‘a’ before the last ‘b’.
• If the character is ‘b’, it updates temp to result, preparing for the next ‘a’ characters.

Finally, the result is the total number of strictly increasing sequences ending with ‘a’.

Steps-by-step apprach:

• Set result and temp to 0. These variables will be used to keep track of the final result and a temporary value.
• Loop through each character in the given string (s).
  • Check for ‘a’: Inside the loop, if the current character is ‘a’, update the result using the formula (result + temp + 1) % 1000000007.
  • Check for ‘b’: If the current character is ‘b’, update the temp value using the result.
• After iterating through the string, return the final result

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

int solve(string & s)
{
    // Variables to store the result and a temporary value
    int result = 0, temp = 0;

    // Iterate over each character in the string
    for (int i = 0; i < s.size(); i++) {
        // If the character is 'a', update the result
        if (s[i] == 'a')
            result = (result + temp + 1) % 1000000007;
        // If the character is 'b', update the temporary
        // value
        if (s[i] == 'b')
            temp = result;
    }

    // Return the result
    return result;
}

int main()
{
    // Input string
    string s = "abbaa";

    cout << solve(s);

    return 0;
}

Java

public class LexicographicallyLargest {

    public static int solve(String s)
    {
        // Variables to store the result and a temporary
        // value
        int result = 0, temp = 0;

        // Iterate over each character in the string
        for (int i = 0; i < s.length(); i++) {
            // If the character is 'a', update the result
            if (s.charAt(i) == 'a')
                result = (result + temp + 1) % 1000000007;
            // If the character is 'b', update the temporary
            // value
            if (s.charAt(i) == 'b')
                temp = result;
        }

        // Return the result
        return result;
    }

    public static void main(String[] args)
    {
        // Input string
        String s = "abbaa";

        System.out.println(solve(s));
    }
}

Python3

def solve(s):
    # Variables to store the result and a temporary value
    result = 0
    temp = 0

    # Iterate over each character in the string
    for char in s:
        # If the character is 'a', update the result
        if char == 'a':
            result = (result + temp + 1) % 1000000007
        # If the character is 'b', update the temporary value
        if char == 'b':
            temp = result

    # Return the result
    return result

#Driver code
s = "abbaa"

# Print the result of the solve function
print(solve(s))

C#

using System;

class Program {
    static int Solve(string s)
    {
        // Variables to store the result and a temporary
        // value
        int result = 0, temp = 0;

        // Iterate over each character in the string
        for (int i = 0; i < s.Length; i++) {
            // If the character is 'a', update the result
            if (s[i] == 'a')
                result = (result + temp + 1) % 1000000007;

            // If the character is 'b', update the temporary
            // value
            if (s[i] == 'b')
                temp = result;
        }

        // Return the result
        return result;
    }

    static void Main()
    {
        // Input string
        string s = "abbaa";

        Console.WriteLine(Solve(s));
    }
}

Javascript

// JavaScript Implementation

function solve(s) {
  let result = 0;
  let temp = 0;

  for (let i = 0; i < s.length; i++) {
    if (s[i] === 'a') {
      result = (result + temp + 1) % 1000000007;
    }
    if (s[i] === 'b') {
      temp = result;
    }
  }

  return result;
}

const s = "abbaa";
console.log(solve(s));

// This code is contributed by Tapesh(tapeshdu420)

Output

5

Time Complexity: O(N), where N is the length of the input string.
Auxiliary Space: O(1)