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

问题描述

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;

标签： algorithmdynamic-programming

解决方案

我们可以从左到右遍历列，选择包含不会使当前矩阵无趣的列。如果实施得当，这将花费与输入大小成线性关系的时间。

支持该算法的关键事实是，给定两个有趣的列子集，我们可以将缺少的第一列从一个列添加到另一个列，而不会使其无趣。

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