944. Delete Columns to Make Sorted

Problem Description

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 like this:
  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 like this:
  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 like this:
  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.

Python Solution

class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:
        n, m = len(strs), len(strs[0])
        count = 0
        
        for j in range(m):
            for i in range(1, n):
                if strs[i][j] < strs[i-1][j]:
                    count += 1
                    break
                    
        return count

Java Solution

class Solution {
    public int minDeletionSize(String[] strs) {
        int n = strs.length, m = strs[0].length();
        int count = 0;
        
        for (int j = 0; j < m; j++) {
            for (int i = 1; i < n; i++) {
                if (strs[i].charAt(j) < strs[i-1].charAt(j)) {
                    count++;
                    break;
                }
            }
        }
        
        return count;
    }
}

C++ Solution

class Solution {
public:
    int minDeletionSize(vector& strs) {
        int n = strs.size(), m = strs[0].length();
        int count = 0;
        
        for (int j = 0; j < m; j++) {
            for (int i = 1; i < n; i++) {
                if (strs[i][j] < strs[i-1][j]) {
                    count++;
                    break;
                }
            }
        }
        
        return count;
    }
};

JavaScript Solution

/**
 * @param {string[]} strs
 * @return {number}
 */
var minDeletionSize = function(strs) {
    const n = strs.length, m = strs[0].length;
    let count = 0;
    
    for (let j = 0; j < m; j++) {
        for (let i = 1; i < n; i++) {
            if (strs[i][j] < strs[i-1][j]) {
                count++;
                break;
            }
        }
    }
    
    return count;
};

C# Solution

public class Solution {
    public int MinDeletionSize(string[] strs) {
        int n = strs.Length, m = strs[0].Length;
        int count = 0;
        
        for (int j = 0; j < m; j++) {
            for (int i = 1; i < n; i++) {
                if (strs[i][j] < strs[i-1][j]) {
                    count++;
                    break;
                }
            }
        }
        
        return count;
    }
}