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# Find minimum elements after considering all possible transformations

Given an array of three colors. The array elements have a special property. Whenever two elements of different colors become adjacent to each other, they merge into an element of the third color. How many minimum numbers of elements can be there in the array after considering all possible transformations.
Example:

```Input : arr[] = {R, G}
Output : 1
G B -> {G B} R -> R

Input : arr[] = {R, G, B}
Output : 2
Explanation :
R G B -> [R G] B ->  B B
OR
R G B -> R {G B} ->  R R ```

Recommended Practice

Scenarios

Before you rush into solution, we would suggest you try out different examples and see if you can find any pattern.

```Let us see few more scenarios:
1. R R R, Output 3
2. R R G B -> R [R G] B -> [R B] B -> [G B] -> R, Output 1
3. R G R G -> [R G] R G -> [B R] G ->G G, Output 2
4. R G B B G R -> R [G B] B G R ->R [R B] G R ->[R G] G R
-> [B G] R ->R R, Output 2
5. R R B B G -> R [R B] [B G] -> R [G R] -> [R B] -> G,
Output 1```

Did you find any pattern in output?

Possible Patterns

Let n be number of elements in array. No matter what the input is, we always end up in three types of outputs:

• n: When no transformation can take place at all
• 2: When number of elements of each color are all odd or all even
• 1: When number of elements of each color are mix of odd and even

Steps:

```1) Count the number of elements of each color.
2) Then only one out of the following four cases can happen:
......1) There are elements of only one color, return n.
......2) There are even number of elements of each color, return 2.
......3) There are odd number of elements of each color, return 2.
......4) In every other case, return 1.
(The elements will combine with each other repeatedly until
only one of them is left)```

Below is the implementation of the above algorithm.

## C++

 `// C++ program to find count of minimum elements``// after considering all possible transformations.``#include ``using` `namespace` `std;` `// Returns minimum possible elements after considering``// all possible transformations.``int` `findMin(``char` `arr[], ``int` `n)``{``    ``// Initialize counts of all colors as 0``    ``int` `b_count = 0, g_count = 0, r_count = 0;` `    ``// Count number of elements of different colors``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(arr[i] == ``'B'``) b_count++;``        ``if` `(arr[i] == ``'G'``) g_count++;``        ``if` `(arr[i] == ``'R'``) r_count++;``    ``}` `    ``// Check if elements are of same color``    ``if` `(b_count==n || g_count==n || r_count==n)``        ``return` `n;` `    ``// If all are odd, return 2``    ``if` `((r_count&1 && g_count&1 && b_count&1) )``        ``return` `2;` `    ``// If all are even, then also return 2``    ``if` `(!(r_count & 1) && !(g_count & 1) &&``        ``!(b_count & 1) )``        ``return` `2;` `    ``// If none above the cases are true, return 1``    ``return` `1;``}` `// Driver code``int` `main()``{``    ``char` `arr[] = {``'R'``, ``'G'``, ``'B'``, ``'R'``};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr);``    ``cout << findMin(arr, n);``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `// Java program to find count of minimum elements``// after considering all possible transformations.``class` `solution``{` `// Returns minimum possible elements after considering``// all possible transformations.``static` `int` `findMin(``char` `arr[], ``int` `n)``{``    ``// Initialize counts of all colors as 0``    ``int` `b_count = ``0``, g_count = ``0``, r_count = ``0``;` `    ``// Count number of elements of different colors``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``if` `(arr[i] == ``'B'``) b_count++;``        ``if` `(arr[i] == ``'G'``) g_count++;``        ``if` `(arr[i] == ``'R'``) r_count++;``    ``}` `    ``// Check if elements are of same color``    ``if` `(b_count==n || g_count==n || r_count==n)``        ``return` `n;` `    ``// If all are odd, return 2``    ``if``((r_count&``1``) == ``1``)    {``     ``if``((g_count&``1``) == ``1``)     {``      ``if``((b_count&``1``) == ``1``)``        ``return` `2``;``     ``}``    ``}` `    ``// If all are even, then also return 2``    ``if``((r_count & ``1``) == ``0``)``    ``{``      ``if` `((g_count & ``1``) == ``0``)``      ``{``          ``if` `((b_count & ``1``) == ``0``)``                ``return` `2``;``      ``}``    ``}` `    ``// If none above the cases are true, return 1``    ``return` `1``;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``char` `arr[] = {``'R'``, ``'G'``, ``'B'``, ``'R'``};``    ``int` `n = arr.length;``    ``System.out.println(findMin(arr, n));``    ` `}``}``// This code is contributed byte``// Surendra_Gangwar`

## Python 3

 `# Python 3 program to find count of minimum elements``# after considering all possible transformations.` `# Returns minimum possible elements after``# considering all possible transformations.``def` `findMin(arr, n):` `    ``# Initialize counts of all``    ``# colors as 0``    ``b_count ``=` `0``    ``g_count ``=` `0``    ``r_count ``=` `0` `    ``# Count number of elements of``    ``# different colors``    ``for` `i ``in` `range``(n):``        ``if` `(arr[i] ``=``=` `'B'``):``            ``b_count ``+``=` `1``        ``if` `(arr[i] ``=``=` `'G'``):``            ``g_count ``+``=` `1``        ``if` `(arr[i] ``=``=` `'R'``):``            ``r_count ``+``=` `1``            ` `    ``# Check if elements are of same color``    ``if` `(b_count ``=``=` `n ``or``        ``g_count ``=``=` `n ``or` `r_count ``=``=` `n):``        ``return` `n` `    ``# If all are odd, return 2``    ``if` `((r_count&``1` `and``         ``g_count&``1` `and` `b_count&``1``)):``        ``return` `2` `    ``# If all are even, then also return 2``    ``if` `(``not` `(r_count & ``1``) ``and` `not``            ``(g_count & ``1``) ``and` `not` `(b_count & ``1``)):``        ``return` `2` `    ``# If none above the cases``    ``# are true, return 1``    ``return` `1` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[``'R'``, ``'G'``, ``'B'``, ``'R'``]``    ``n ``=` `len``(arr)``    ``print``(findMin(arr, n))` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# program to find count of minimum elements``// after considering all possible transformations.``using` `System;` `class` `GFG``{` `// Returns minimum possible elements after``// considering all possible transformations.``static` `int` `findMin(``char` `[]arr, ``int` `n)``{``    ``// Initialize counts of all colors as 0``    ``int` `b_count = 0, g_count = 0, r_count = 0;` `    ``// Count number of elements of different colors``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(arr[i] == ``'B'``) b_count++;``        ``if` `(arr[i] == ``'G'``) g_count++;``        ``if` `(arr[i] == ``'R'``) r_count++;``    ``}` `    ``// Check if elements are of same color``    ``if` `(b_count == n || g_count == n || r_count == n)``        ``return` `n;` `    ``// If all are odd, return 2``    ``if``((r_count&1) == 1)``    ``{``        ``if``((g_count&1) == 1)   ``        ``{``            ``if``((b_count&1) == 1)``                ``return` `2;``        ``}``    ``}` `    ``// If all are even, then also return 2``    ``if``((r_count & 1) == 0)``    ``{``        ``if` `((g_count & 1) == 0)``        ``{``            ``if` `((b_count & 1) == 0)``                    ``return` `2;``        ``}``    ``}` `    ``// If none above the cases are true,``    ``// return 1``    ``return` `1;``}` `// Driver code``public` `static` `void` `Main()``{``    ``char` `[]arr = {``'R'``, ``'G'``, ``'B'``, ``'R'``};``    ``int` `n = arr.Length;``    ``Console.WriteLine(findMin(arr, n));``}``}` `// This code is contributed byte``// nitin mittal`

## PHP

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## Javascript

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Output :

`1`

Time complexity: O(n)
Auxiliary Space: O(1)
Exercise:

1. How many transformations are needed in above problem?
2. Is it possible to print the sequence in which elements transform? If yes, what will be the approach? Discuss time and space complexity

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