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.

&nbsp;
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.


&nbsp;
Constraints:


	n == strs.length
	1 &lt;= n &lt;= 100
	1 &lt;= strs[i].length &lt;= 1000
	strs[i] consists of lowercase English letters.


class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:

        cols={}
        l=len(strs)
        l_s = len(strs[0])
        delete = set()
        for i in range(l):
            for col in range(l_s):
                if col in cols:
                    if cols[col]>strs[i][col]:
                        delete.add(col)
                cols[col] = strs[i][col]
        return len(delete)
class Solution {
    public int minDeletionSize(String[] strs) {
        int count = 0;
        for(int i = 0; i < strs[0].length(); i++){ //strs[0].length() is used to find the length of the column
            for(int j = 0; j < strs.length-1; j++){
                if((int) strs[j].charAt(i) <= (int) strs[j+1].charAt(i)){
                    continue;
                }else{
                    count++;
                    break;
                }
            }
        }
        return count;
    }
}
class Solution {
public:
    int minDeletionSize(vector<string>& strs) {
        int col = strs[0].size();
        int row = strs.size();
        int count = 0;
        for(int c = 0 ; c < col ; c++){
            for(int r = 1 ; r < row ; r++){
                if(strs[r][c] - strs[r - 1][c] < 0){
                    count++;
                    break;
                }
            }
        }
        return count;
    }
};
/**
 * @param {string[]} strs
 * @return {number}
 */
var minDeletionSize = function(strs) {
    let count = 0;
    for(let i=0; i<strs[0].length; i++){
        for(let j=0; j<strs.length-1; j++){
            if(strs[j].charAt(i) > strs[j+1].charAt(i)){
                count++;
                break;
            }
        }
    }
    return count;
};
Delete Columns to Make Sorted