# Count of strings that can be formed using a, b and c under given constraints

Given a length n, count the number of strings of length n that can be made using ‘a’, ‘b’ and ‘c’ with at most one ‘b’ and two ‘c’s allowed.

Examples :

```Input : n = 3
Output : 19
aaa aab aac aba abc aca acb acc baa
bac bca bcc caa cab cac cba cbc cca ccb

Input  : n = 4
Output : 39```

A simple solution is to recursively count all possible combinations of strings that can be made up to latter ‘a’, ‘b’, and ‘c’.

Below is the implementation of the above idea

## C++

 `// C++ program to count number of strings` `// of n characters with` `#include` `using` `namespace` `std;`   `// n is total number of characters.` `// bCount and cCount are counts of 'b'` `// and 'c' respectively.` `int` `countStr(``int` `n, ``int` `bCount, ``int` `cCount)` `{` `    ``// Base cases` `    ``if` `(bCount < 0 || cCount < 0) ``return` `0;` `    ``if` `(n == 0) ``return` `1;` `    ``if` `(bCount == 0 && cCount == 0) ``return` `1;`   `    ``// Three cases, we choose, a or b or c` `    ``// In all three cases n decreases by 1.` `    ``int` `res = countStr(n-1, bCount, cCount);` `    ``res += countStr(n-1, bCount-1, cCount);` `    ``res += countStr(n-1, bCount, cCount-1);`   `    ``return` `res;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 3;  ``// Total number of characters` `    ``cout << countStr(n, 1, 2);` `    ``return` `0;` `}`

## Java

 `// Java program to count number ` `// of strings of n characters with` `import` `java.io.*;`   `class` `GFG ` `{` `    `  `// n is total number of characters.` `// bCount and cCount are counts of 'b'` `// and 'c' respectively.` `static` `int` `countStr(``int` `n, ` `                    ``int` `bCount, ` `                    ``int` `cCount)` `{` `    ``// Base cases` `    ``if` `(bCount < ``0` `|| cCount < ``0``) ``return` `0``;` `    ``if` `(n == ``0``) ``return` `1``;` `    ``if` `(bCount == ``0` `&& cCount == ``0``) ``return` `1``;`   `    ``// Three cases, we choose, a or b or c` `    ``// In all three cases n decreases by 1.` `    ``int` `res = countStr(n - ``1``, bCount, cCount);` `    ``res += countStr(n - ``1``, bCount - ``1``, cCount);` `    ``res += countStr(n - ``1``, bCount, cCount - ``1``);`   `    ``return` `res;` `}`   `// Driver code` `public` `static` `void` `main (String[] args)` `{` `    ``int` `n = ``3``; ``// Total number of characters` `    ``System.out.println(countStr(n, ``1``, ``2``));` `}` `}`   `// This code is contributed by akt_mit`

## Python 3

 `# Python 3 program to ` `# count number of strings` `# of n characters with`   `# n is total number of characters.` `# bCount and cCount are counts ` `# of 'b' and 'c' respectively.` `def` `countStr(n, bCount, cCount):`   `    ``# Base cases` `    ``if` `(bCount < ``0` `or` `cCount < ``0``):` `        ``return` `0` `    ``if` `(n ``=``=` `0``) :` `        ``return` `1` `    ``if` `(bCount ``=``=` `0` `and` `cCount ``=``=` `0``):` `        ``return` `1`   `    ``# Three cases, we choose, a or b or c` `    ``# In all three cases n decreases by 1.` `    ``res ``=` `countStr(n ``-` `1``, bCount, cCount)` `    ``res ``+``=` `countStr(n ``-` `1``, bCount ``-` `1``, cCount)` `    ``res ``+``=` `countStr(n ``-` `1``, bCount, cCount ``-` `1``)`   `    ``return` `res`   `# Driver code` `if` `__name__ ``=``=``"__main__"``:` `    ``n ``=` `3` `# Total number of characters` `    ``print``(countStr(n, ``1``, ``2``))`   `# This code is contributed ` `# by ChitraNayal`

## C#

 `// C# program to count number ` `// of strings of n characters ` `// with a, b and c under given` `// constraints` `using` `System;`   `class` `GFG` `{` `    `  `// n is total number of ` `// characters. bCount and` `// cCount are counts of ` `// 'b' and 'c' respectively.` `static` `int` `countStr(``int` `n, ` `                    ``int` `bCount, ` `                    ``int` `cCount)` `{` `    ``// Base cases` `    ``if` `(bCount < 0 || cCount < 0) ` `        ``return` `0;` `    ``if` `(n == 0) ``return` `1;` `    ``if` `(bCount == 0 && cCount == 0) ` `        ``return` `1;`   `    ``// Three cases, we choose, ` `    ``// a or b or c. In all three` `    ``// cases n decreases by 1.` `    ``int` `res = countStr(n - 1, ` `                       ``bCount, cCount);` `    ``res += countStr(n - 1, ` `                    ``bCount - 1, cCount);` `    ``res += countStr(n - 1, ` `                    ``bCount, cCount - 1);`   `    ``return` `res;` `}`   `// Driver code` `static` `public` `void` `Main ()` `{` `    ``// Total number ` `    ``// of characters` `    ``int` `n = 3; ` `    ``Console.WriteLine(countStr(n, 1, 2));` `}` `}`   `// This code is contributed by aj_36`

## PHP

 ``

## Javascript

 ``

Output

`19`

Time Complexity: O(3^N).
Auxiliary Space: O(1).

Efficient Solution:

If we drown a recursion tree of the above code, we can notice that the same values appear multiple times. So we store results that are used later if repeated.

## C++

 `// C++ program to count number of strings` `// of n characters with` `#include` `using` `namespace` `std;`   `// n is total number of characters.` `// bCount and cCount are counts of 'b'` `// and 'c' respectively.` `int` `countStrUtil(``int` `dp[], ``int` `n, ``int` `bCount=1,` `                 ``int` `cCount=2)` `{` `    ``// Base cases` `    ``if` `(bCount < 0 || cCount < 0) ``return` `0;` `    ``if` `(n == 0) ``return` `1;` `    ``if` `(bCount == 0 && cCount == 0) ``return` `1;`   `    ``// if we had saw this combination previously` `    ``if` `(dp[n][bCount][cCount] != -1)` `        ``return` `dp[n][bCount][cCount];`   `    ``// Three cases, we choose, a or b or c` `    ``// In all three cases n decreases by 1.` `    ``int` `res = countStrUtil(dp, n-1, bCount, cCount);` `    ``res += countStrUtil(dp, n-1, bCount-1, cCount);` `    ``res += countStrUtil(dp, n-1, bCount, cCount-1);`   `    ``return` `(dp[n][bCount][cCount] = res);` `}`   `// A wrapper over countStrUtil()` `int` `countStr(``int` `n)` `{` `    ``int` `dp[n+1];` `    ``memset``(dp, -1, ``sizeof``(dp));` `    ``return` `countStrUtil(dp, n);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 3; ``// Total number of characters` `    ``cout << countStr(n);` `    ``return` `0;` `}`

## Java

 `// Java program to count number of strings ` `// of n characters with `   `class` `GFG ` `{` `    ``// n is total number of characters. ` `    ``// bCount and cCount are counts of 'b' ` `    ``// and 'c' respectively. `   `    ``static` `int` `countStrUtil(``int``[][][] dp, ``int` `n,` `                            ``int` `bCount, ``int` `cCount) ` `    ``{`   `        ``// Base cases ` `        ``if` `(bCount < ``0` `|| cCount < ``0``) ` `        ``{` `            ``return` `0``;` `        ``}` `        ``if` `(n == ``0``)` `        ``{` `            ``return` `1``;` `        ``}` `        ``if` `(bCount == ``0` `&& cCount == ``0``) ` `        ``{` `            ``return` `1``;` `        ``}`   `        ``// if we had saw this combination previously ` `        ``if` `(dp[n][bCount][cCount] != -``1``) ` `        ``{` `            ``return` `dp[n][bCount][cCount];` `        ``}`   `        ``// Three cases, we choose, a or b or c ` `        ``// In all three cases n decreases by 1. ` `        ``int` `res = countStrUtil(dp, n - ``1``, bCount, cCount);` `        ``res += countStrUtil(dp, n - ``1``, bCount - ``1``, cCount);` `        ``res += countStrUtil(dp, n - ``1``, bCount, cCount - ``1``);`   `        ``return` `(dp[n][bCount][cCount] = res);` `    ``}`   `    ``// A wrapper over countStrUtil() ` `    ``static` `int` `countStr(``int` `n, ``int` `bCount, ``int` `cCount) ` `    ``{` `        ``int``[][][] dp = ``new` `int``[n + ``1``][``2``][``3``];` `        ``for` `(``int` `i = ``0``; i < n + ``1``; i++) ` `        ``{` `            ``for` `(``int` `j = ``0``; j < ``2``; j++) ` `            ``{` `                ``for` `(``int` `k = ``0``; k < ``3``; k++) ` `                ``{` `                    ``dp[i][j][k] = -``1``;` `                ``}` `            ``}` `        ``}` `        ``return` `countStrUtil(dp, n,bCount,cCount);` `    ``}`   `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``int` `n = ``3``; ``// Total number of characters ` `        ``int` `bCount = ``1``, cCount = ``2``;` `        ``System.out.println(countStr(n,bCount,cCount));` `    ``}` `}`   `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python 3 program to count number of strings` `# of n characters with`   `# n is total number of characters.` `# bCount and cCount are counts of 'b'` `# and 'c' respectively.` `def` `countStrUtil(dp, n, bCount``=``1``,cCount``=``2``):`   `    ``# Base cases` `    ``if` `(bCount < ``0` `or` `cCount < ``0``):` `        ``return` `0` `    ``if` `(n ``=``=` `0``):` `        ``return` `1` `    ``if` `(bCount ``=``=` `0` `and` `cCount ``=``=` `0``):` `        ``return` `1`   `    ``# if we had saw this combination previously` `    ``if` `(dp[n][bCount][cCount] !``=` `-``1``):` `        ``return` `dp[n][bCount][cCount]`   `    ``# Three cases, we choose, a or b or c` `    ``# In all three cases n decreases by 1.` `    ``res ``=` `countStrUtil(dp, n``-``1``, bCount, cCount)` `    ``res ``+``=` `countStrUtil(dp, n``-``1``, bCount``-``1``, cCount)` `    ``res ``+``=` `countStrUtil(dp, n``-``1``, bCount, cCount``-``1``)`   `    ``dp[n][bCount][cCount] ``=` `res` `    ``return` `dp[n][bCount][cCount]`   `# A wrapper over countStrUtil()` `def` `countStr(n):`   `    ``dp ``=` `[ [ [``-``1` `for` `x ``in` `range``(``2``+``1``)] ``for` `y ``in` `range``(``1``+``1``)]``for` `z ``in` `range``(n``+``1``)]` `    ``return` `countStrUtil(dp, n)`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``n ``=` `3` `# Total number of characters` `    ``print``(countStr(n))` `    `  `# This code is contributed by chitranayal    `

## C#

 `// C# program to count number of strings ` `// of n characters with ` `using` `System; `   `class` `GFG ` `{ ` `    ``// n is total number of characters. ` `    ``// bCount and cCount are counts of 'b' ` `    ``// and 'c' respectively. ` `    ``static` `int` `countStrUtil(``int``[,,] dp, ``int` `n, ` `                    ``int` `bCount=1, ``int` `cCount=2) ` `    ``{ ` `        ``// Base cases ` `        ``if` `(bCount < 0 || cCount < 0)` `            ``return` `0; ` `        ``if` `(n == 0) ` `            ``return` `1; ` `        ``if` `(bCount == 0 && cCount == 0) ` `            ``return` `1; ` `    `  `        ``// if we had saw this combination previously ` `        ``if` `(dp[n,bCount,cCount] != -1) ` `            ``return` `dp[n,bCount,cCount]; ` `    `  `        ``// Three cases, we choose, a or b or c ` `        ``// In all three cases n decreases by 1. ` `        ``int` `res = countStrUtil(dp, n - 1, bCount, cCount); ` `        ``res += countStrUtil(dp, n - 1, bCount - 1, cCount); ` `        ``res += countStrUtil(dp, n - 1, bCount, cCount - 1); ` `    `  `        ``return` `(dp[n, bCount, cCount] = res); ` `    ``} ` `    `  `    ``// A wrapper over countStrUtil() ` `    ``static` `int` `countStr(``int` `n) ` `    ``{ ` `        ``int``[,,] dp = ``new` `int``[n + 1, 2, 3]; ` `        ``for``(``int` `i = 0; i < n + 1; i++)` `            ``for``(``int` `j = 0; j < 2; j++)` `                ``for``(``int` `k = 0; k < 3; k++)` `                    ``dp[i, j, k] = -1;` `        ``return` `countStrUtil(dp, n); ` `    ``} ` `    `  `    ``// Driver code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``int` `n = 3; ``// Total number of characters ` `        `  `        ``Console.Write(countStr(n)); ` `    ``}` `}`   `// This code is contributed by DrRoot_`

## Javascript

 ``

Output

`19`

Time Complexity : O(n)
Auxiliary Space : O(n)

Thanks to Mr. Lazy for suggesting above solutions.

A solution that works in O(1) time :

We can apply the concepts of combinatorics to solve this problem in constant time. we may recall the formula that the number of ways we can arrange a total of n objects, out of which p number of objects are of one type, q objects are of another type, and r objects are of the third type is n!/(p!q!r!)

Let us proceed towards the solution step by step.

How many strings we can form with no ‘b’ and ‘c’? The answer is 1 because we can arrange a string consisting of only ‘a’ in one way only and the string would be aaaa….(n times).

How many strings we can form with one ‘b’? The answer is n because we can arrange a string consisting (n-1) ‘a’s and 1 ‘b’ is n!/(n-1)! = n . The same goes for ‘c’ .

How many strings we can form with 2 places, filled up by ‘b’ and/or ‘c’ ?  Answer is n*(n-1) + n*(n-1)/2 . Because that 2 places can be either 1 ‘b’ and 1 ‘c’  or 2 ‘c’ according to our given constraints. For the first case, total number of arrangements is n!/(n-2)! = n*(n-1) and for second case that is n!/(2!(n-2)!) = n*(n-1)/2 .

Finally, how many strings we can form with 3 places, filled up by ‘b’ and/or ‘c’ ?  Answer is (n-2)*(n-1)*n/2 . Because those 3 places can only be consisting of 1 ‘b’ and 2’c’  according to our given constraints. So, total number of arrangements is n!/(2!(n-3)!) = (n-2)*(n-1)*n/2 .

Implementation:

## C++

 `// A O(1) CPP program to find number of strings` `// that can be made under given constraints.` `#include` `using` `namespace` `std;` `int` `countStr(``int` `n){` `    `  `    ``int` `count = 0;` `    `  `    ``if``(n>=1){` `        ``//aaa...` `        ``count += 1;` `        ``//b...aaa...` `          ``count += n;` `        ``//c...aaa...` `        ``count += n;` `        `  `        ``if``(n>=2){` `          ``//bc...aaa...` `          ``count += n*(n-1);` `          ``//cc...aaa...` `          ``count += n*(n-1)/2;` `          `  `          ``if``(n>=3){` `            ``//bcc...aaa...` `            ``count += (n-2)*(n-1)*n/2;` `          ``}` `        ``}` `    `  `    ``}` `    `  `    ``return` `count;` `    `  `}`   `// Driver code ` `int` `main()` `{` `  ``int` `n = 3;` `  ``cout << countStr(n);` `  ``return` `0;` `} `

## Java

 `// A O(1) Java program to ` `// find number of strings` `// that can be made under` `// given constraints.` `import` `java.io.*;`   `class` `GFG` `{` `    ``static` `int` `countStr(``int` `n)` `    ``{` `    ``return` `1` `+ (n * ``2``) + ` `           ``(n * ((n * n) - ``1``) / ``2``);` `    ``}`   `// Driver code ` `public` `static` `void` `main (String[] args)` `{` `    ``int` `n = ``3``;` `    ``System.out.println( countStr(n));` `}` `}`   `// This code is contributed by ajit`

## Python 3

 `# A O(1) Python3 program to find ` `# number of strings that can be` `# made under given constraints.`   `def` `countStr(n):` `    ``return` `(``1` `+` `(n ``*` `2``) ``+` `                ``(n ``*` `((n ``*` `n) ``-` `1``) ``/``/` `2``))`   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `3` `    ``print``(countStr(n))`   `# This code is contributed` `# by ChitraNayal`

## C#

 `// A O(1) C# program to ` `// find number of strings` `// that can be made under` `// given constraints.` `using` `System;`   `class` `GFG` `{` `    ``static` `int` `countStr(``int` `n)` `    ``{` `    ``return` `1 + (n * 2) + ` `          ``(n * ((n * n) - 1) / 2);` `    ``}`   `// Driver code ` `static` `public` `void` `Main ()` `{` `    ``int` `n = 3;` `    ``Console.WriteLine(countStr(n));` `}` `}`   `// This code is contributed by m_kit`

## PHP

 ``

## Javascript

 ``

Output

`19`

Time Complexity : O(1)
Auxiliary Space : O(1)

Thanks to Niharika Sahai for providing above solution.

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!