Skip to content

Latest commit

 

History

History
186 lines (150 loc) · 4.85 KB

README_EN.md

File metadata and controls

186 lines (150 loc) · 4.85 KB
comments difficulty edit_url tags
true
Hard
Array
String
Dynamic Programming

960. Delete Columns to Make Sorted III

中文文档

Description

You are given an array of n strings strs, all of the same length.

We may choose any deletion indices, and we delete all the characters in those indices for each string.

For example, if we have strs = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef", "vyz"].

Suppose we chose a set of deletion indices answer such that after deletions, the final array has every string (row) in lexicographic order. (i.e., (strs[0][0] <= strs[0][1] <= ... <= strs[0][strs[0].length - 1]), and (strs[1][0] <= strs[1][1] <= ... <= strs[1][strs[1].length - 1]), and so on). Return the minimum possible value of answer.length.

 

Example 1:

Input: strs = ["babca","bbazb"]
Output: 3
Explanation: After deleting columns 0, 1, and 4, the final array is strs = ["bc", "az"].
Both these rows are individually in lexicographic order (ie. strs[0][0] <= strs[0][1] and strs[1][0] <= strs[1][1]).
Note that strs[0] > strs[1] - the array strs is not necessarily in lexicographic order.

Example 2:

Input: strs = ["edcba"]
Output: 4
Explanation: If we delete less than 4 columns, the only row will not be lexicographically sorted.

Example 3:

Input: strs = ["ghi","def","abc"]
Output: 0
Explanation: All rows are already lexicographically sorted.

 

Constraints:

  • n == strs.length
  • 1 <= n <= 100
  • 1 <= strs[i].length <= 100
  • strs[i] consists of lowercase English letters.
  •  

Solutions

Solution 1

Python3

class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:
        n = len(strs[0])
        dp = [1] * n
        for i in range(1, n):
            for j in range(i):
                if all(s[j] <= s[i] for s in strs):
                    dp[i] = max(dp[i], dp[j] + 1)
        return n - max(dp)

Java

class Solution {
    public int minDeletionSize(String[] strs) {
        int n = strs[0].length();
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        int mx = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (check(i, j, strs)) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            mx = Math.max(mx, dp[i]);
        }
        return n - mx;
    }

    private boolean check(int i, int j, String[] strs) {
        for (String s : strs) {
            if (s.charAt(i) < s.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    int minDeletionSize(vector<string>& strs) {
        int n = strs[0].size();
        vector<int> dp(n, 1);
        int mx = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (check(i, j, strs)) {
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
            mx = max(mx, dp[i]);
        }
        return n - mx;
    }

    bool check(int i, int j, vector<string>& strs) {
        for (string& s : strs)
            if (s[i] < s[j])
                return false;
        return true;
    }
};

Go

func minDeletionSize(strs []string) int {
	n := len(strs[0])
	dp := make([]int, n)
	mx := 1
	dp[0] = 1
	check := func(i, j int) bool {
		for _, s := range strs {
			if s[i] < s[j] {
				return false
			}
		}
		return true
	}
	for i := 1; i < n; i++ {
		dp[i] = 1
		for j := 0; j < i; j++ {
			if check(i, j) {
				dp[i] = max(dp[i], dp[j]+1)
			}
		}
		mx = max(mx, dp[i])
	}
	return n - mx
}