Concatenate strings in any order to get Maximum Number of “AB”

Given an array of strings of length N, it is allowed to concatenate them in any order. Find the maximum possible number of occurrences of ‘AB’ in the resulting string.

Examples:

Input : N = 4, arr={ “BCA”, “BGGGA”, “JKA”, “BALB” } 
Output : 3 
Concatenate them in the order JKA + BGGA + BCA + BALB and it will become JKABGGABCABALB and it has 3 occurrences of ‘AB’.

Input : N = 3, arr={ “ABCA”, “BOOK”, “BAND” } 
Output : 2

Approach:
Calculate the number of ABs within each string beforehand. Concentrate on the change of the number of ABs that extend over two strings when the strings are rearranged. The only characters that matter in each string are its first and last characters. 
The strings that can contribute to the answer are:  

1. A string that begins with B and ends with A.
2. A string that begins with B but does not end with A.
3. A string that does not begin with B but ends with A.

Let c1, c2, and c3 be the number of strings of categories 1, 2, and 3, respectively. 

• If c1 = 0, then the answer is min(c2, c3) as we can take both and concatenate as long as both are available.
• If c1 > 0 and c2 + c3 = 0, then the answer is c1 – 1 as we concatenate them in serial order one by one.
• If c1 > 0 and c2 + c3 > 0 and take min(c2, c3) = p, first concatenate the category 1 string one by one and the extra c1 – 1 ‘AB’ and then if both category 2 and 3 are available, then add category 3 at the beginning of present resulting string and add category 2 at the end of present resulting string.
• There are c1 – 1 + 2 = c1 + 1 extra ‘AB’ and now c2 and c3 decreased by one and p become p – 1 and now take both 
  category 2 and 3 and add them as long as both are available and now we get a total of c1 + 1 + (p – 1) = c1 + p extra ‘AB’. That means c1 + min(c2, c3) .

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find maximum number of ABs
int maxCountAB(string s[], int n)
{
    // variable A, B, AB for count strings that
    // end with 'A' but not end with 'B', 'B' but
    // does not end with 'A' and 'B' and ends
    // with 'A' respectively.
    int A = 0, B = 0, BA = 0, ans = 0;

    for (int i = 0; i < n; i++) {
        string S = s[i];
        int L = S.size();
        for (int j = 0; j < L - 1; j++) {

            // 'AB' is already present in string
            // before concatenate them
            if (S.at(j) == 'A' &&
                           S.at(j + 1) == 'B') {
                ans++;
            }
        }

        // count of strings that begins
        // with 'B' and ends with 'A
        if (S.at(0) == 'B' && S.at(L - 1) == 'A')
            BA++;

        // count of strings that begins
        // with 'B' but does not end with 'A'
        else if (S.at(0) == 'B')
            B++;

        // count of strings that ends with
        // 'A' but not end with 'B'
        else if (S.at(L - 1) == 'A')
            A++;
    }

    // updating the value of ans and
    // add extra count of 'AB'
    if (BA == 0)
        ans += min(B, A);
    else if (A + B == 0)
        ans += BA - 1;
    else
        ans += BA + min(B, A);

    return ans;
}

// Driver Code
int main()
{

    string s[] = { "ABCA", "BOOK", "BAND" };

    int n = sizeof(s) / sizeof(s[0]);

    cout << maxCountAB(s, n);

    return 0;
}

Java

// Java implementation of above approach
import java.util.*;

class GFG
{

// Function to find maximum number of ABs
static int maxCountAB(String s[], int n)
{
    // variable A, B, AB for count strings that
    // end with 'A' but not end with 'B', 'B' but
    // does not end with 'A' and 'B' and ends
    // with 'A' respectively.
    int A = 0, B = 0, BA = 0, ans = 0;

    for (int i = 0; i < n; i++)
    {
        String S = s[i];
        int L = S.length();
        for (int j = 0; j < L - 1; j++)
        {

            // 'AB' is already present in string
            // before concatenate them
            if (S.charAt(j) == 'A' &&
                        S.charAt(j + 1) == 'B')
            {
                ans++;
            }
        }

        // count of strings that begins
        // with 'B' and ends with 'A
        if (S.charAt(0) == 'B' && S.charAt(L - 1) == 'A')
            BA++;

        // count of strings that begins
        // with 'B' but does not end with 'A'
        else if (S.charAt(0) == 'B')
            B++;

        // count of strings that ends with
        // 'A' but not end with 'B'
        else if (S.charAt(L - 1) == 'A')
            A++;
    }

    // updating the value of ans and
    // add extra count of 'AB'
    if (BA == 0)
        ans += Math.min(B, A);
    else if (A + B == 0)
        ans += BA - 1;
    else
        ans += BA + Math.min(B, A);

    return ans;
}

// Driver Code
public static void main(String[] args)
{
    String s[] = { "ABCA", "BOOK", "BAND" };

    int n = s.length;

    System.out.println(maxCountAB(s, n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python 3 implementation of above approach

# Function to find maximum number of ABs
def maxCountAB(s,n):
    # variable A, B, AB for count strings that
    # end with 'A' but not end with 'B', 'B' but
    # does not end with 'A' and 'B' and ends
    # with 'A' respectively.
    A = 0
    B = 0
    BA = 0
    ans = 0

    for i in range(n):
        S = s[i]
        L = len(S)
        for j in range(L-1):
            # 'AB' is already present in string
            # before concatenate them
            if (S[j] == 'A' and S[j + 1] == 'B'):
                ans += 1

        # count of strings that begins
        # with 'B' and ends with 'A
        if (S[0] == 'B' and S[L - 1] == 'A'):
            BA += 1

        # count of strings that begins
        # with 'B' but does not end with 'A'
        elif (S[0] == 'B'):
            B += 1

        # count of strings that ends with
        # 'A' but not end with 'B'
        elif (S[L - 1] == 'A'):
            A += 1

    # updating the value of ans and
    # add extra count of 'AB'
    if (BA == 0):
        ans += min(B, A)
    elif (A + B == 0):
        ans += BA - 1
    else:
        ans += BA + min(B, A)
    return ans

# Driver Code
if __name__ == '__main__':
    s = ["ABCA", "BOOK", "BAND"]

    n = len(s)

    print(maxCountAB(s, n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of above approach
using System;

class GFG
{

    // Function to find maximum number of ABs
    static int maxCountAB(string []s, int n)
    {
        // variable A, B, AB for count strings that
        // end with 'A' but not end with 'B', 'B' but
        // does not end with 'A' and 'B' and ends
        // with 'A' respectively.
        int A = 0, B = 0, BA = 0, ans = 0;

        for (int i = 0; i < n; i++)
        {
            string S = s[i];
            int L = S.Length;
            for (int j = 0; j < L - 1; j++)
            {

                // 'AB' is already present in string
                // before concatenate them
                if (S[j] == 'A' &&
                            S[j + 1] == 'B')
                {
                    ans++;
                }
            }

            // count of strings that begins
            // with 'B' and ends with 'A
            if (S[0] == 'B' && S[L - 1] == 'A')
                BA++;

            // count of strings that begins
            // with 'B' but does not end with 'A'
            else if (S[0] == 'B')
                B++;

            // count of strings that ends with
            // 'A' but not end with 'B'
            else if (S[L - 1] == 'A')
                A++;
        }

        // updating the value of ans and
        // add extra count of 'AB'
        if (BA == 0)
            ans += Math.Min(B, A);

        else if (A + B == 0)
            ans += BA - 1;
        else
            ans += BA + Math.Min(B, A);

        return ans;
    }

    // Driver Code
    public static void Main()
    {
        string []s = { "ABCA", "BOOK", "BAND" };

        int n = s.Length;

        Console.WriteLine(maxCountAB(s, n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of above approach

// Function to find maximum number of ABs
function maxCountAB(s, n)
{
    // variable A, B, AB for count strings that
    // end with 'A' but not end with 'B', 'B' but
    // does not end with 'A' and 'B' and ends
    // with 'A' respectively.
    var A = 0, B = 0, BA = 0, ans = 0;

    for (var i = 0; i < n; i++) {
        var S = s[i];
        var L = S.length;
        for (var j = 0; j < L - 1; j++) {

            // 'AB' is already present in string
            // before concatenate them
            if (S[j] == 'A' &&
                           S[j + 1] == 'B') {
                ans++;
            }
        }

        // count of strings that begins
        // with 'B' and ends with 'A
        if (S[0] == 'B' && S[L - 1] == 'A')
            BA++;

        // count of strings that begins
        // with 'B' but does not end with 'A'
        else if (S[0] == 'B')
            B++;

        // count of strings that ends with
        // 'A' but not end with 'B'
        else if (S[L - 1] == 'A')
            A++;
    }

    // updating the value of ans and
    // add extra count of 'AB'
    if (BA == 0)
        ans += Math.min(B, A);
    else if (A + B == 0)
        ans += BA - 1;
    else
        ans += BA + Math.min(B, A);

    return ans;
}

// Driver Code
var s = ["ABCA", "BOOK", "BAND"];
var n = s.length;
document.write( maxCountAB(s, n));

</script>

Output

2

Time Complexity: O(N * L), where N is the size of the given string array and L is the maximum length of a string in the array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.