# 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 number of elements can be there in 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: Please solve it on “PRACTICE” first, before moving on to the solution.

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 folowing 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 C++ implementation of above algorithm.

```// C++ program to find count of minimum elements
// after considering all possible transformations.
#include <iostream>
using namespace std;

// Returns minium 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[0]);
cout << findMin(arr, n);
return 0;
}
```

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

This article is contributed by Aashish Barnwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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