10.leetcode Delete Columns to Make Sorted

题目

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, A1, ..., AA.length-1].)

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.
就是给一个数组， 把每一项的对应的index组合成一个新的数组，再算出那些不是递增的个数。
Return the minimum possible value of D.length.

例子

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.

Input: ["a","b"]
Output: 0
Explanation: D = {}

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

个人算法

var minDeletionSize = function(A) {
    const b = A.map(v => v.split(""))
    let len = A.length
    let len2 = b[0].length
    let n = 0
    for (var i=0;i<len2;i++){
        for (var j=0; j < len-1; j++){
            if( b[j][i] > b[j+1][i]){
                n++
                break
            }
        }
    }
    return n
};

Runtime: 88 ms, faster than 64.27% of JavaScript online submissions for Delete Columns to Make Sorted.
Memory Usage: 43.9 MB, less than 7.69% of JavaScript online submissions for Delete Columns to Make Sorted.

其余算法

class Solution:
    def minDeletionSize(self, A):
        return sum(any(a[j] > b[j] for a, b in zip(A, A[1:])) for j in range(len(A[0])))

思路跟个人差很少，只不过用了python的zip api