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
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}
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.
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