[leetcode]944. Delete Columns to Make Sorted

• Easy
• Related Topic: Greedy

Problem:

We are given an array A of N lowercase letter strings, all of the same length.

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

For example, if we have an array A = [“abcdef”,“uvwxyz”] and deletion indices {0, 2, 3}, then the final array after deletions is [“bef”, “vyz”], and the remaining columns of A are [“b”,“v”], [“e”,“y”], and [“f”,“z”]. (Formally, the c-th column is [A[0][c], A[1][c], …, A[A.length-1][c]].)

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.

Return the minimum possible value of D.length.

Example 1:

Input: [“cba”,“daf”,“ghi”]
Output: 1
Explanation:
After choosing D = {1}, each column [“c”,“d”,“g”] and [“a”,“f”,“i”] are in non-decreasing sorted order.
If we chose D = {}, then a column [“b”,“a”,“h”] would not be in non-decreasing sorted order.
Example 2:

Input: [“a”,“b”]
Output: 0
Explanation: D = {}
Example 3:

Input: [“zyx”,“wvu”,“tsr”]
Output: 3
Explanation: D = {0, 1, 2}

Note:

1 <= A.length <= 100
1 <= A[i].length <= 1000

---

Solution

• O(NM)

class Solution:
    def minDeletionSize(self, A):
        """
        :type A: List[str]
        :rtype: int
        """
        if A == []:
            return 0
        lowest = []
        count = 0
        for c in A[0]:
            lowest.append(ord(c))
        for str in A[1:]:
            for i,c in enumerate(str):
                if lowest[i] == -1:
                    continue
                if ord(c)<lowest[i]:
                    count += 1
                    lowest[i] = -1
                    continue
                if ord(c)>lowest[i]:
                    lowest[i] = ord(c)
        return count

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其他人的解法
O(NlogNM), with N <= 100

    def minDeletionSize(self, A):
        return sum(list(col) != sorted(col) for col in zip(*A))

O(NM)

    def minDeletionSize(self, A):
        return sum(any(a > b for a, b in zip(col, col[1:])) for col in zip(*A))

其中，zip(*)，表示zip的逆操作，
例子

array=[[1,4,7],[2,5,8],[3,6,9]]
x,y,z=zip(*array)
print(x)
print(y)
print(z)

(1, 2, 3) (4, 5, 6) (7, 8, 9)
zip(*A))可以得到矩阵的转置 -> map(list, zip(*array)