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 Below strings follow given constraints: aaa aab aac aba abc aca acb acc baa bac bca bcc caa cab cac cba cbc cca ccb Input : n = 4 Output : 39
Asked in Google Interview
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<bits/stdc++.h>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 codeint 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 withimport 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 codepublic 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 codeif __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// constraintsusing 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 codestatic public void Main (){ // Total number // of characters int n = 3; Console.WriteLine(countStr(n, 1, 2));}}// This code is contributed by aj_36 |
PHP
<?php// PHP 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.function countStr($n, $bCount, $cCount){ // Base cases if ($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. $res = countStr($n - 1, $bCount, $cCount); $res += countStr($n - 1, $bCount - 1, $cCount); $res += countStr($n - 1, $bCount, $cCount - 1); return $res;}// Driver code$n = 3; // Total number // of charactersecho countStr($n, 1, 2); // This code is contributed by ajit?> |
Javascript
<script>// JavaScript program for the above approach // n is total number of characters. // bCount and cCount are counts of 'b' // and 'c' respectively. function countStr(n, bCount, 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. let res = countStr(n - 1, bCount, cCount); res += countStr(n - 1, bCount - 1, cCount); res += countStr(n - 1, bCount, cCount - 1); return res; } // Driver Code let n = 3; // Total number of characters document.write(countStr(n, 1, 2)); // This code is contributed by splevel62.</script> |
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<bits/stdc++.h>using namespace std;// n is total number of characters.// bCount and cCount are counts of 'b'// and 'c' respectively.int countStrUtil(int dp[][2][3], 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][2][3]; memset(dp, -1, sizeof(dp)); return countStrUtil(dp, n);}// Driver codeint 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 codeif __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
<script>// javascript 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. function countStrUtil(dp , n, bCount , 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. var 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() function countStr(n , bCount , cCount) { dp = Array(n+1).fill(0).map (x => Array(2).fill(0).map (x => Array(3).fill(0))); for (i = 0; i < n + 1; i++) { for (j = 0; j < 2; j++) { for (k = 0; k < 3; k++) { dp[i][j][k] = -1; } } } return countStrUtil(dp, n,bCount,cCount); }// Driver code var n = 3; // Total number of characters var bCount = 1, cCount = 2;document.write(countStr(n,bCount,cCount));// This code contributed by shikhasingrajput </script> |
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<bits/stdc++.h>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
<?php// A O(1) PHP program to find // number of strings that can // be made under given constraints.function countStr($n){ return 1 + ($n * 2) + ($n * (($n * $n) - 1) / 2);}// Driver code $n = 3;echo countStr($n);// This code is contributed by aj_36?> |
Javascript
<script>// A O(1) javascript program to // find number of strings// that can be made under// given constraints. function countStr(n) { return 1 + (n * 2) + (n * ((n * n) - 1) / 2); } // Driver code var n = 3; document.write(countStr(n));// This code is contributed by Princi Singh</script> |
Output
19
Time Complexity : O(1)
Auxiliary Space : O(1)
Thanks to Niharika Sahai for providing above solution.
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