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0944 - Delete Columns to Make Sorted (Easy)

https://leetcode.com/problems/delete-columns-to-make-sorted/

Problem Statement​

You are given an array of n strings strs, all of the same length.

The strings can be arranged such that there is one on each line, making a grid. For example, strs = ["abc", "bce", "cae"] can be arranged as:

abc
bce
cae

You want to delete the columns that are not sorted lexicographically. In the above example (0-indexed), columns 0 ('a', 'b', 'c') and 2 ('c', 'e', 'e') are sorted while column 1 ('b', 'c', 'a') is not, so you would delete column 1.

Return the number of columns that you will delete.

Example 1:

Input: strs = ["cba","daf","ghi"]
Output: 1
Explanation: The grid looks as follows:
  cba
  daf
  ghi
Columns 0 and 2 are sorted, but column 1 is not, so you only need to delete 1 column.

Example 2:

Input: strs = ["a","b"]
Output: 0
Explanation: The grid looks as follows:
  a
  b
Column 0 is the only column and is sorted, so you will not delete any columns.

Example 3:

Input: strs = ["zyx","wvu","tsr"]
Output: 3
Explanation: The grid looks as follows:
  zyx
  wvu
  tsr
All 3 columns are not sorted, so you will delete all 3.

Constraints:

  • n == strs.length
  • 1 <= n <= 100
  • 1 <= strs[i].length <= 1000
  • strs[i] consists of lowercase English letters.

Approach 1: Matrix Traversing​

Written by @wingkwong
class Solution {
public:
    int minDeletionSize(vector<string>& s) {
        int ans = 0;
        // iterate cols
        for (int col = 0; col < s[0].size(); col++) {
            // iterate rows
            for (int row = 1; row < s.size(); row++) {
                // if the character in the previous row is greater than the character in the current row
                // then we can delete this column
                if (s[row - 1][col] > s[row][col]) {
                    ans += 1;
                    break;
                }
            }
        }
        return ans;
    }
};