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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: Dynamic Programming

We define $f[i]$ as the length of the longest non-decreasing subsequence ending at column $i$. Initially, $f[i] = 1$, and the final answer is $n - \max(f)$.

To compute $f[i]$, we iterate over all $j &lt; i$. If for all strings $s$, we have $s[j] \leq s[i]$, then we update $f[i]$ as follows: $$ f[i] = \max(f[i], f[j] + 1) $$

Finally, we return $n - \max(f)$.

The time complexity is $O(n^2 \times m)$, and the space complexity is $O(n)$, where $n$ is the length of each string in the array $\textit{strs}$, and $m$ is the number of strings in the array.

Python3

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

Java

class Solution {
    public int minDeletionSize(String[] strs) {
        int n = strs[0].length();
        int[] f = new int[n];
        Arrays.fill(f, 1);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                boolean ok = true;
                for (String s : strs) {
                    if (s.charAt(j) > s.charAt(i)) {
                        ok = false;
                        break;
                    }
                }
                if (ok) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
        }
        return n - Arrays.stream(f).max().getAsInt();
    }
}

C++

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

Go

func minDeletionSize(strs []string) int {
	n := len(strs[0])
	f := make([]int, n)
	for i := range f {
		f[i] = 1
	}
	for i := 1; i < n; i++ {
		for j := 0; j < i; j++ {
			ok := true
			for _, s := range strs {
				if s[j] > s[i] {
					ok = false
					break
				}
			}
			if ok {
				f[i] = max(f[i], f[j]+1)
			}
		}
	}
	return n - slices.Max(f)
}

TypeScript

function minDeletionSize(strs: string[]): number {
    const n = strs[0].length;
    const f: number[] = Array(n).fill(1);
    for (let i = 1; i < n; i++) {
        for (let j = 0; j < i; j++) {
            let ok = true;
            for (const s of strs) {
                if (s[j] > s[i]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                f[i] = Math.max(f[i], f[j] + 1);
            }
        }
    }
    return n - Math.max(...f);
}

Rust

impl Solution {
    pub fn min_deletion_size(strs: Vec<String>) -> i32 {
        let n = strs[0].len();
        let mut f = vec![1; n];

        for i in 1..n {
            for j in 0..i {
                if strs.iter().all(|s| s.as_bytes()[j] <= s.as_bytes()[i]) {
                    f[i] = f[i].max(f[j] + 1);
                }
            }
        }

        (n - *f.iter().max().unwrap()) as i32
    }
}