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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

• Java
• C++
• Python
• Go
• RenderScript

• 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
      }
  }
  
  

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