Styjun

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.

1. if the previous entry is lexicographically valid we can take dp[i][j - 1] and the current input isn't going to change anything.


2. 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;

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.