Codeforces Round #116 (Div. 2, ACM-ICPC Rules), E. Cubes

Codeforces Round #116 (Div. 2, ACM-ICPC Rules)
Problem E. Cubes

Problem statement:
http://codeforces.com/problemset/problem/180/E

First, we define an array of lists where each element contains the list of the cube indexes, for example:

    List<int>[] colorIndex = new List<int>[m + 1];

Since the number of colors is m <= 10^5, and the total number of cubes is n <= 2.10^5, we could make a fixed size array of (m+1). The space needed will be O(n+m).

Let's see for this input:
10 3 2
1 2 1 1 3 2 1 1 2 2

n=10, m=3, k=2

We could see it as a table of:
Index: 0 1 2 3 4 5 6 7 8 9
Color: 1 2 1 1 3 2 1 1 2 2

Then contents of the array colorIndex will be:
    colorIndex[0]: null

    colorIndex[1]: {0,2,3,6,7}

    colorIndex[2]: {1,5,8,9}

    colorIndex[3]: {4}

The codes will be something like:
            for ( int i = 0; i < n; i++ )
            {
                int clr = c[i];
                if ( colidx[clr] == null )
                    colidx[clr] = new List<int> ( );
                colidx[clr].Add ( i );

            }

Then we could calculate the score for each color by using 2 pointers: the first one points to the start of a sequence, and the second one points to the end of the sequence. Let's call them iStart and iEnd.

For example, if we look at color #1, where the indexes are {0,2,3,6,7}.
First we will set iStart and iEnd to 0.
Then we shift iEnd to the right repeatedly until the number of cubes needed to remove is greater than k, or until the end of the list.
In each iteration we could calculate the score of color #1 (the number of cubes with color #1) as:

    score = iEnd - iStart + 1;

The total number of cubes (all colors included) between iStart and iEnd, inclusive, is:

    numAllCubes = colidx[col][iend] - colidx[col][istart] + 1;

Then we calculate the number of cubes with different color that we need to remove as:

    numCubesToDelete = numAllCubes - score;

As long as numCubesToDelete <= k, the score is valid and we could update the maximum score with the current score.
Once numCubesToDelete > k, the score is not valid anymore and we need to shift the left pointer, iStart, to the right, by increasing its value by 1.
Then we calculate again the value of score, numAllCubes, and numCubesToDelete, and update the maximum score as needed.