10.leetcode Delete Columns to Make Sorted

題目

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

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

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].)api

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.
就是給一個數組, 把每一項的對應的index組合成一個新的數組,再算出那些不是遞增的個數。
Return the minimum possible value of D.length.數組

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}

個人算法

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.

其餘算法

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])))

思路跟個人差很少,只不過用了python的zip apiless

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