minimum columns to be deleted in a matrix to make it row-wise lexicographically sorted

Question

I was trying to solve this hiring contest problem (now closed)

Lexicographic Rows

You are given a matrix of characters. In one operation you can remove a column of the matrix. You can perform as many operations as you want. Your task is to make the final matrix interesting i.e. the string formed by the characters of row is lexicographically smaller or equal to the string formed by the characters of the row. You need to use minimum number of operations possible. An empty matrix is always a interesting matrix.

Input

The first line contains two integers and . The next lines contain letters each.

Output

In the output, you need to print the minimum number of operations to make the matrix interesting.

Constraints

There are only lowercase English alphabets as characters in the input.

Sample Input

3 3

cfg

agk

dlm

Sample Output

1

Explanation

Delete the first column to make the matrix interesting.

I'm pretty convinced this is a DP problem. I was having difficulties finding the optimal subproblem though. I managed to pass only a couple of test cases

I defined dp[i][j] as the minimum number of the columns to be removed to have an interesting matrix.

And for every character input[i][j] there are two possibilities.

if the previous entry is lexicographically valid we can take dp[i][j - 1] and the current input isn't going to change anything.
else we check if the input[i -1][j] and input[i][j] if they are in the correct order we consider dp[i][j - 1] else this column is invalid too so we add 1 to dp[i][j-1]

My soln. code

int n, m;
cin >> n >> m;
vector<string> input(n);
for (int i = 0; i < n; ++i) {
    string temp = "";
    for (int j = 0; j < m; ++j) {
        char c;
        cin >> c;
        temp = temp + c;
    }
    input[i] = temp;
}

vector<vector<int> > dp(n, vector<int>(m, 0));

for (int i = 1; i < n; ++i) {
    for (int j = 1; j < m; ++j) {
        //Left is valid
        if (input[i - 1][j - 1] < input[i][j - 1]) {
            dp[i][j] = dp[i][j - 1];
        }
        else {
            //Current is valid
            if (input[i - 1][j] <= input[i][j]) {
                dp[i][j] = dp[i][j - 1];
            }
            else {
                dp[i][j] = dp[i][j - 1] + 1;
            }
        }
    }
}
cout << dp[n - 1][m - 1] << endl;

What's the link to test submissions? What are the constraints on the size of the input matrix? — גלעד ברקן
How did you solve the first questions (LCS Again)? Even though I tried with space optimized but I manage to pass only 3 test cases. — Mukesh Kumar Gupta

David Eisenstat David Eisenstat · Accepted Answer · 2019-03-25T00:00:41

We can iterate through the columns left to right, choosing the ones whose inclusion wouldn't make the current matrix uninteresting. Properly implemented, this will take time linear in the size of the input.

The key fact supporting this algorithm is that, given two interesting subsets of columns, we can add the first column missing from one to the other without making it uninteresting.

minimum columns to be deleted in a matrix to make it row-wise lexicographically sorted

1 Answers