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

Ways to arrange Balls such that adjacent balls are of different types - GeeksforGeeks
There are 'a' balls of type A, 'b' balls of type B and 'c' balls of type C. Using the balls we want to create a straight colorful line such that no two balls of same type are adjacent.

Input  : a = 1, b = 1, c = 0
Output : 2
There are only two arrangements AB and BA

Input  : a = 1, b = 1, c = 1
Output : 6
There are only six arrangements ABC, BAC, 
BCA, CBA, ACB and CAB

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.

Time complexity : O(a*b*c)
Auxiliary Space : O(a*b*c*3)

// table to store to store results of subproblems

int dp[MAX][MAX][MAX][3];

 

// Returns count of arrangements where last placed ball is

// 'last'.  'last' is 0 for 'a', 1 for 'b' and 2 for 'c'

int countWays(int a, int b, int c, int last)

{

    // if number of balls of any color becomes less

    // than 0 the number of ways arrangements is 0.

    if (a<0 || b<0 || c<0)

        return 0;

 

    // If last ball required is of type A and the number

    // of balls of A type is 1 while number of balls of

    // other color is 0 the number of ways is 1.

    if (a==1 && b==0 && c==0 && last==0)

        return 1;

 

    // Same case as above for 'b' and 'c'

    if (a==0 && b==1 && c==0 && last==1)

        return 1;

    if (a==0 && b==0 && c==1 && last==2)

        return 1;

 

    // If this subproblem is already evaluated

    if (dp[a][b][last] != -1)

        return dp[a][b][last];

 

    // if last ball required is A and the number of ways is

    // the sum of number of ways to form sequence with 'a-1' A

    // balls, b B Balls and c C balls ending with B and C.

    if (last==0)

       dp[a][b]1[last] = countWays(a-1,b,c,1) + countWays(a-1,b,c,2);

 

    // Same as above case for 'b' and 'c'

    else if (last==1)

       dp[a][b]1[last] = countWays(a,b-1,c,0) + countWays(a,b-1,c,2);

    else //(last==2)

       dp[a][b]1[last] =  countWays(a,b,c-1,0) + countWays(a,b,c-1,1);

 

    return dp[a][b]1[last];

}

// Returns count of required arrangements

int countUtil(int a, int b, int c)

{

   // Initialize 'dp' array

   memset(dp, -1, sizeof(dp));

 

   // Three cases arise:

   return countWays(a, b, c, 0) +  // Last required balls is type A

          countWays(a, b, c, 1) +  // Last required balls is type B

          countWays(a, b, c, 2); // Last required balls is type C

}

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 A
2) Last ball to be placed is of type B
3) Last ball to be placed is of type C

// Returns count of arrangements where last placed ball is

// 'last'.  'last' is 0 for 'a', 1 for 'b' and 2 for 'c'

int countWays(int a, int b, int c, int last)

{

    // if number of balls of any color becomes less

    // than 0 the number of ways arrangements is 0.

    if (a<0 || b<0 || c<0)

        return 0;

 

    // If last ball required is of type A and the number

    // of balls of A type is 1 while number of balls of

    // other color is 0 the number of ways is 1.

    if (a==1 && b==0 && c==0 && last==0)

        return 1;

 

    // Same case as above for 'b' and 'c'

    if (a==0 && b==1 && c==0 && last==1)

        return 1;

    if (a==0 && b==0 && c==1 && last==2)

        return 1;

 

    // if last ball required is A and the number of ways is

    // the sum of number of ways to form sequence with 'a-1' A

    // balls, b B Balls and c C balls ending with B and C.

    if (last==0)

        return countWays(a-1,b,c,1) + countWays(a-1,b,c,2);

 

    // Same as above case for 'b' and 'c'

    if (last==1)

        return countWays(a,b-1,c,0) + countWays(a,b-1,c,2);

    if (last==2)

        return countWays(a,b,c-1,0) + countWays(a,b,c-1,1);

}

 

// Returns count of required arrangements

int countUtil(int a, int b, int c)

{

   // Three cases arise:

   return countWays(a, b, c, 0) +  // Last required balls is type A

          countWays(a, b, c, 1) +  // Last required balls is type B

          countWays(a, b, c, 2); // Last required balls is type C

}

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