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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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