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944. Delete Columns to Make Sorted

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 follows:
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.

Solutions

  • class Solution {
    public int minDeletionSize(String[] strs) {
        int m = strs[0].length(), n = strs.length;
        int ans = 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)) {
                    ++ans;
                    break;
                }
            }
        }
        return ans;
    }
}

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


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


  • func minDeletionSize(strs []string) int {
	m, n := len(strs[0]), len(strs)
	ans := 0
	for j := 0; j < m; j++ {
		for i := 1; i < n; i++ {
			if strs[i][j] < strs[i-1][j] {
				ans++
				break
			}
		}
	}
	return ans
}

  • impl Solution {
    pub fn min_deletion_size(strs: Vec<String>) -> i32 {
        let n = strs.len();
        let m = strs[0].len();
        let mut res = 0;
        for i in 0..m {
            for j in 1..n {
                if strs[j - 1].as_bytes()[i] > strs[j].as_bytes()[i] {
                    res += 1;
                    break;
                }
            }
        }
        res
    }
}


  • function minDeletionSize(strs: string[]): number {
    const [m, n] = [strs[0].length, strs.length];
    let ans = 0;
    for (let j = 0; j < m; ++j) {
        for (let i = 1; i < n; ++i) {
            if (strs[i][j] < strs[i - 1][j]) {
                ++ans;
                break;
            }
        }
    }
    return ans;
}

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