Ways to arrange Balls such that adjacent balls are of different types

There are ‘p’ balls of type P, ‘q’ balls of type Q and ‘r’ balls of type R. Using the balls we want to create a straight line such that no two balls of same type are adjacent.

Examples :

Input  : p = 1, q = 1, r = 0
Output : 2
There are only two arrangements PQ and QP

Input  : p = 1, q = 1, r = 1
Output : 6
There are only six arrangements PQR, QPR,
QRP, RQP, PRQ and RPQ

Input  : p = 2, q = 1, r = 1
Output : 6
There are only six arrangements PQRP, QPRP,
PRQP, RPQP, PRPQ and PQPR

We strongly recommend that you click here and practice it, before moving on to the solution.

Naive Solution:
The naive solution to this problem is a recursive solution. We recursively call for three cases
1) Last ball to be placed is of type P
2) Last ball to be placed is of type Q
3) Last ball to be placed is of type R

Below is the implementation of above idea.

C++

 // C++ program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other #include using namespace std;    // Returns count of arrangements where last placed ball is // 'last'.  'last' is 0 for 'p', 1 for 'q' and 2 for 'r' int countWays(int p, int q, int r, int last) {     // if number of balls of any color becomes less     // than 0 the number of ways arrangements is 0.     if (p<0 || q<0 || r<0)         return 0;        // If last ball required is of type P and the number     // of balls of P type is 1 while number of balls of     // other color is 0 the number of ways is 1.     if (p==1 && q==0 && r==0 && last==0)         return 1;        // Same case as above for 'q' and 'r'     if (p==0 && q==1 && r==0 && last==1)         return 1;     if (p==0 && q==0 && r==1 && last==2)         return 1;        // if last ball required is P and the number of ways is     // the sum of number of ways to form sequence with 'p-1' P     // balls, q Q Balls and r R balls ending with Q and R.     if (last==0)         return countWays(p-1,q,r,1) + countWays(p-1,q,r,2);        // Same as above case for 'q' and 'r'     if (last==1)         return countWays(p,q-1,r,0) + countWays(p,q-1,r,2);     if (last==2)         return countWays(p,q,r-1,0) + countWays(p,q,r-1,1); }    // Returns count of required arrangements int countUtil(int p, int q, int r) {    // Three cases arise:    return countWays(p, q, r, 0) +  // Last required balls is type P           countWays(p, q, r, 1) +  // Last required balls is type Q           countWays(p, q, r, 2); // Last required balls is type R }    // Driver code to test above int main() {     int p = 1, q = 1, r = 1;     printf("%d", countUtil(p, q, r));     return 0; }

Java

 // Java program to count number  // of ways to arrange three types of // balls such that no two balls of  // same color are adjacent to each other class GFG {            // Returns count of arrangements     // where last placed ball is     // 'last'. 'last' is 0 for 'p',      // 1 for 'q' and 2 for 'r'     static int countWays(int p, int q, int r, int last)      {         // if number of balls of any          // color becomes less than 0          // the number of ways arrangements is 0.         if (p < 0 || q < 0 || r < 0)         return 0;                // If last ball required is         // of type P and the number         // of balls of P type is 1         // while number of balls of         // other color is 0 the number         // of ways is 1.         if (p == 1 && q == 0 && r == 0 && last == 0)             return 1;                // Same case as above for 'q' and 'r'         if (p == 0 && q == 1 && r == 0 && last == 1)             return 1;         if (p == 0 && q == 0 && r == 1 && last == 2)             return 1;                // if last ball required is P         // and the number of ways is         // the sum of number of ways          // to form sequence with 'p-1' P         // balls, q Q Balls and r R balls         // ending with Q and R.         if (last == 0)         return countWays(p - 1, q, r, 1) +                countWays(p - 1, q, r, 2);                // Same as above case for 'q' and 'r'         if (last == 1)             return countWays(p, q - 1, r, 0) +                    countWays(p, q - 1, r, 2);                    if (last == 2)         return countWays(p, q, r - 1, 0) +                countWays(p, q, r - 1, 1);                return 0;     }            // Returns count of required arrangements     static int countUtil(int p, int q, int r) {         // Three cases arise:         return countWays(p, q, r, 0) + // Last required balls is type P                countWays(p, q, r, 1) + // Last required balls is type Q                countWays(p, q, r, 2);  // Last required balls is type R     }            // Driver code     public static void main(String[] args)      {         int p = 1, q = 1, r = 1;         System.out.print(countUtil(p, q, r));     } }    // This code is contributed by Anant Agarwal.

Python3

 # Python3 program to count  # number of ways to arrange  # three types of balls such   # that no two balls of same  # color are adjacent to each  # other    # Returns count of arrangements # where last placed ball is # 'last'. 'last' is 0 for 'p', # 1 for 'q' and 2 for 'r' def countWays(p, q, r, last):            # if number of balls of      # any color becomes less     # than 0 the number of      # ways arrangements is 0.     if (p < 0 or q < 0 or r < 0):         return 0;        # If last ball required is     # of type P and the number     # of balls of P type is 1     # while number of balls of     # other color is 0 the number     # of ways is 1.     if (p == 1 and q == 0 and         r == 0 and last == 0):         return 1;        # Same case as above      # for 'q' and 'r'     if (p == 0 and q == 1 and         r == 0 and last == 1):         return 1;                if (p == 0 and q == 0 and          r == 1 and last == 2):         return 1;        # if last ball required is P      # and the number of ways is     # the sum of number of ways      # to form sequence with 'p-1' P     # balls, q Q Balls and r R      # balls ending with Q and R.     if (last == 0):         return (countWays(p - 1, q, r, 1) +                  countWays(p - 1, q, r, 2));        # Same as above case     # for 'q' and 'r'     if (last == 1):         return (countWays(p, q - 1, r, 0) +                  countWays(p, q - 1, r, 2));     if (last == 2):         return (countWays(p, q, r - 1, 0) +                  countWays(p, q, r - 1, 1));    # Returns count of  # required arrangements def countUtil(p, q, r):            # Three cases arise:     # Last required balls is type P     # Last required balls is type Q     # Last required balls is type R     return (countWays(p, q, r, 0) +              countWays(p, q, r, 1) +              countWays(p, q, r, 2));     # Driver Code p = 1;  q = 1; r = 1; print(countUtil(p, q, r));        # This code is contributed by mits

C#

 // C# program to count number  // of ways to arrange three types of // balls such that no two balls of  // same color are adjacent to each other using System;    class GFG {            // Returns count of arrangements     // where last placed ball is     // 'last'. 'last' is 0 for 'p',      // 1 for 'q' and 2 for 'r'     static int countWays(int p, int q,                             int r, int last)      {                    // if number of balls of any          // color becomes less than 0          // the number of ways         // arrangements is 0.         if (p < 0 || q < 0 || r < 0)             return 0;                // If last ball required is         // of type P and the number         // of balls of P type is 1         // while number of balls of         // other color is 0 the number         // of ways is 1.         if (p == 1 && q == 0 && r == 0                                && last == 0)             return 1;                // Same case as above for 'q' and 'r'         if (p == 0 && q == 1 && r == 0                                 && last == 1)             return 1;         if (p == 0 && q == 0 && r == 1                                 && last == 2)             return 1;                // if last ball required is P         // and the number of ways is         // the sum of number of ways          // to form sequence with 'p-1' P         // balls, q Q Balls and r R balls         // ending with Q and R.         if (last == 0)             return countWays(p - 1, q, r, 1) +                     countWays(p - 1, q, r, 2);                // Same as above case for 'q' and 'r'         if (last == 1)             return countWays(p, q - 1, r, 0) +                 countWays(p, q - 1, r, 2);                    if (last == 2)             return countWays(p, q, r - 1, 0) +                     countWays(p, q, r - 1, 1);                return 0;     }            // Returns count of required arrangements     static int countUtil(int p, int q, int r)     {                    // Three cases arise:         // 1. Last required balls is type P         // 2. Last required balls is type Q         // 3. Last required balls is type R         return countWays(p, q, r, 0) +                 countWays(p, q, r, 1) +                 countWays(p, q, r, 2);      }            // Driver code     public static void Main()      {         int p = 1, q = 1, r = 1;                    Console.Write(countUtil(p, q, r));     } }    // This code is contributed by nitin mittal.

PHP



Output:

6

Time Complexity of this solution is exponential.

We can observe that there are many subproblems being solved again and again so the problem can be solved using Dynamic Programming (DP). We can easily make memoization solution to this problem.

C++

 // C++ program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other #include using namespace std; #define MAX 100    // table to store to store results of subproblems int dp[MAX][MAX][MAX];    // Returns count of arrangements where last placed ball is // 'last'.  'last' is 0 for 'p', 1 for 'q' and 2 for 'r' int countWays(int p, int q, int r, int last) {     // if number of balls of any color becomes less     // than 0 the number of ways arrangements is 0.     if (p<0 || q<0 || r<0)         return 0;        // If last ball required is of type P and the number     // of balls of P type is 1 while number of balls of     // other color is 0 the number of ways is 1.     if (p==1 && q==0 && r==0 && last==0)         return 1;        // Same case as above for 'q' and 'r'     if (p==0 && q==1 && r==0 && last==1)         return 1;     if (p==0 && q==0 && r==1 && last==2)         return 1;        // If this subproblem is already evaluated     if (dp[p][q][r][last] != -1)         return dp[p][q][r][last];        // if last ball required is P and the number of ways is     // the sum of number of ways to form sequence with 'p-1' P     // balls, q Q Balls and r R balls ending with Q and R.     if (last==0)        dp[p][q][r][last] = countWays(p-1,q,r,1) + countWays(p-1,q,r,2);        // Same as above case for 'q' and 'r'     else if (last==1)        dp[p][q][r][last] = countWays(p,q-1,r,0) + countWays(p,q-1,r,2);     else //(last==2)        dp[p][q][r][last] =  countWays(p,q,r-1,0) + countWays(p,q,r-1,1);        return dp[p][q][r][last]; }    // Returns count of required arrangements int countUtil(int p, int q, int r) {    // Initialize 'dp' array    memset(dp, -1, sizeof(dp));       // Three cases arise:    return countWays(p, q, r, 0) +  // Last required balls is type P           countWays(p, q, r, 1) +  // Last required balls is type Q           countWays(p, q, r, 2); // Last required balls is type R }    // Driver code to test above int main() {     int p = 1, q = 1, r = 1;     printf("%d", countUtil(p, q, r));     return 0; }

Java

 // Java program to count number // of ways to arrange three // types of balls such that no // two balls of same color // are adjacent to each other import java.util.Arrays;    class GFG  {        static final int MAX = 100;            // table to store to store results of subproblems     static int dp[][][][] = new int[MAX][MAX][MAX];            // Returns count of arrangements     // where last placed ball is     // 'last'. 'last' is 0 for 'p',     // 1 for 'q' and 2 for 'r'     static int countWays(int p, int q, int r, int last)     {         // if number of balls of any          // color becomes less than 0         // the number of ways arrangements is 0.         if (p < 0 || q < 0 || r < 0)         return 0;                // If last ball required is          // of type P and the number         // of balls of P type is 1          // while number of balls of         // other color is 0 the number          // of ways is 1.         if (p == 1 && q == 0 && r == 0 && last == 0)             return 1;                // Same case as above for 'q' and 'r'         if (p == 0 && q == 1 && r == 0 && last == 1)             return 1;                    if (p == 0 && q == 0 && r == 1 && last == 2)             return 1;                // If this subproblem is already evaluated         if (dp[p][q][r][last] != -1)             return dp[p][q][r][last];                // if last ball required is P and          // the number of ways is the sum         // of number of ways to form sequence         // with 'p-1' P balls, q Q balss and         // r R balls ending with Q and R.         if (last == 0)         dp[p][q][r][last] = countWays(p - 1, q, r, 1) +                              countWays(p - 1, q, r, 2);                // Same as above case for 'q' and 'r'         else if (last == 1)         dp[p][q][r][last] = countWays(p, q - 1, r, 0) +                              countWays(p, q - 1, r, 2);         //(last==2)         else          dp[p][q][r][last] = countWays(p, q, r - 1, 0) +                              countWays(p, q, r - 1, 1);                return dp[p][q][r][last];     }            // Returns count of required arrangements     static int countUtil(int p, int q, int r)     {         // Initialize 'dp' array         for (int[][][] row : dp)         {             for (int[][] innerRow : row)              {                 for (int[] innerInnerRow : innerRow)                 {                     Arrays.fill(innerInnerRow, -1);                 }             }         };                // Three cases arise:         return countWays(p, q, r, 0) + // Last required balls is type P             countWays(p, q, r, 1) +    // Last required balls is type Q             countWays(p, q, r, 2);       // Last required balls is type R     }        // Driver code     public static void main(String[] args)     {         int p = 1, q = 1, r = 1;         System.out.print(countUtil(p, q, r));     } }    // This code is contributed by Anant Agarwal.

C#

 // C# program to count number  // of ways to arrange three  // types of balls such that no  // two balls of same color  // are adjacent to each other  using System;    class GFG  {         static int MAX = 101;             // table to store to store results of subproblems      static int[,,,] dp = new int[MAX, MAX, MAX, 4];             // Returns count of arrangements      // where last placed ball is      // 'last'. 'last' is 0 for 'p',      // 1 for 'q' and 2 for 'r'      static int countWays(int p, int q, int r, int last)      {          // if number of balls of any          // color becomes less than 0          // the number of ways arrangements is 0.          if (p < 0 || q < 0 || r < 0)          return 0;                 // If last ball required is          // of type P and the number          // of balls of P type is 1          // while number of balls of          // other color is 0 the number          // of ways is 1.          if (p == 1 && q == 0 && r == 0 && last == 0)              return 1;                 // Same case as above for 'q' and 'r'          if (p == 0 && q == 1 && r == 0 && last == 1)              return 1;                     if (p == 0 && q == 0 && r == 1 && last == 2)              return 1;                 // If this subproblem is already evaluated          if (dp[p, q, r, last] != -1)              return dp[p, q, r, last];                 // if last ball required is P and          // the number of ways is the sum          // of number of ways to form sequence          // with 'p-1' P balls, q Q balss and          // r R balls ending with Q and R.          if (last == 0)          dp[p, q, r, last] = countWays(p - 1, q, r, 1) +                              countWays(p - 1, q, r, 2);                 // Same as above case for 'q' and 'r'          else if (last == 1)          dp[p, q, r, last] = countWays(p, q - 1, r, 0) +                              countWays(p, q - 1, r, 2);          //(last==2)          else         dp[p, q, r, last] = countWays(p, q, r - 1, 0) +                              countWays(p, q, r - 1, 1);                 return dp[p, q, r, last];      }             // Returns count of required arrangements      static int countUtil(int p, int q, int r)      {          // Initialize 'dp' array          for(int i = 0; i < MAX; i++)         for(int j = 0; j < MAX; j++)         for(int k = 0; k < MAX; k++)         for(int l = 0; l < 4; l++)         dp[i, j, k, l] = -1;                // Three cases arise:          return countWays(p, q, r, 0) + // Last required balls is type P              countWays(p, q, r, 1) + // Last required balls is type Q              countWays(p, q, r, 2); // Last required balls is type R      }         // Driver code      static void Main()      {          int p = 1, q = 1, r = 1;          Console.WriteLine(countUtil(p, q, r));      }  }     // This code is contributed by mits.

Python3

 # Python3 program to count number of ways to  # arrange three types of balls such that no  # two balls of same color are adjacent to each other MAX = 100;     # table to store to store results of subproblems  dp = [[[[-1] * 4 for i in range(MAX)]                   for j in range(MAX)]                   for k in range(MAX)];     # Returns count of arrangements where last  # placed ball is 'last'. 'last' is 0 for 'p',  # 1 for 'q' and 2 for 'r'  def countWays(p, q, r, last):        # if number of balls of any color becomes less      # than 0 the number of ways arrangements is 0.      if (p < 0 or q < 0 or r < 0):          return 0;         # If last ball required is of type P and the      # number of balls of P type is 1 while number      # of balls of other color is 0 the number of     # ways is 1.      if (p == 1 and q == 0 and          r == 0 and last == 0):          return 1;         # Same case as above for 'q' and 'r'      if (p == 0 and q == 1 and         r == 0 and last == 1):          return 1;      if (p == 0 and q == 0 and          r == 1 and last == 2):          return 1;         # If this subproblem is already evaluated      if (dp[p][q][r][last] != -1):          return dp[p][q][r][last];         # if last ball required is P and the number      # of ways is the sum of number of ways to       # form sequence with 'p-1' P balls, q Q Balls     # and r R balls ending with Q and R.      if (last == 0):         dp[p][q][r][last] = (countWays(p - 1, q, r, 1) +                               countWays(p - 1, q, r, 2));         # Same as above case for 'q' and 'r'      elif (last == 1):         dp[p][q][r][last] = (countWays(p, q - 1, r, 0) +                               countWays(p, q - 1, r, 2));      else:                    #(last==2)         dp[p][q][r][last] = (countWays(p, q, r - 1, 0) +                               countWays(p, q, r - 1, 1));         return dp[p][q][r][last];     # Returns count of required arrangements  def countUtil(p, q, r):            # Three cases arise:     # Last required balls is type P     # Last required balls is type Q     # Last required balls is type R     return (countWays(p, q, r, 0) +              countWays(p, q, r, 1) +              countWays(p, q, r, 2));    # Driver Code p, q, r = 1, 1, 1;  print(countUtil(p, q, r));     # This code is contributed by mits

PHP



Output:

6

Time complexity : O(p*q*r)
Auxiliary Space : O(p*q*r*3)

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Improved By : nitin mittal, Mithun Kumar