Medium
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 its elements in lexicographic order (i.e., strs[0] <= strs[1] <= strs[2] <= ... <= strs[n - 1]
). Return the minimum possible value of answer.length
.
Example 1:
Input: strs = [“ca”,”bb”,”ac”]
Output: 1
Explanation: After deleting the first column, strs = [“a”, “b”, “c”]. Now strs is in lexicographic order (ie. strs[0] <= strs[1] <= strs[2]). We require at least 1 deletion since initially strs was not in lexicographic order, so the answer is 1.
Example 2:
Input: strs = [“xc”,”yb”,”za”]
Output: 0
Explanation: strs is already in lexicographic order, so we do not need to delete anything. Note that the rows of strs are not necessarily in lexicographic order: i.e., it is NOT necessarily true that (strs[0][0] <= strs[0][1] <= …)
Example 3:
Input: strs = [“zyx”,”wvu”,”tsr”]
Output: 3
Explanation: We have to delete every column.
Constraints:
n == strs.length
1 <= n <= 100
1 <= strs[i].length <= 100
strs[i]
consists of lowercase English letters.class Solution {
fun minDeletionSize(strs: Array<String>): Int {
val sorted = BooleanArray(strs.size)
var res = 0
for (i in 0 until strs[0].length) {
var j = 0
while (j < strs.size - 1) {
if (!sorted[j] && strs[j][i] > strs[j + 1][i]) {
res++
break
}
j++
}
if (j < strs.size - 1) {
continue
}
j = 0
while (j < strs.size - 1) {
if (strs[j][i] < strs[j + 1][i]) {
sorted[j] = true
}
j++
}
}
return res
}
}