Algorithm
Problem Name: 960. Delete Columns to Make Sorted III
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 every string (row) in lexicographic order. (i.e., (strs[0][0] <= strs[0][1] <= ... <= strs[0][strs[0].length - 1]), and (strs[1][0] <= strs[1][1] <= ... <= strs[1][strs[1].length - 1]), and so on). Return the minimum possible value of answer.length.
Example 1:
Input: strs = ["babca","bbazb"] Output: 3 Explanation: After deleting columns 0, 1, and 4, the final array is strs = ["bc", "az"]. Both these rows are individually in lexicographic order (ie. strs[0][0] <= strs[0][1] and strs[1][0] <= strs[1][1]). Note that strs[0] > strs[1] - the array strs is not necessarily in lexicographic order.
Example 2:
Input: strs = ["edcba"] Output: 4 Explanation: If we delete less than 4 columns, the only row will not be lexicographically sorted.
Example 3:
Input: strs = ["ghi","def","abc"] Output: 0 Explanation: All rows are already lexicographically sorted.
Constraints:
n == strs.length1 <= n <= 1001 <= strs[i].length <= 100strs[i]consists of lowercase English letters.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const minDeletionSize = function(A) {
const dp = new Array(A[0].length).fill(1)
for (let i = 0; i < A[0].length; i++) {
for (let j = 0; j < i; j++) {
for (let k = 0; k < = A.length; k++) {
if (k === A.length) dp[i] = Math.max(dp[i], dp[j] + 1)
else if (A[k][j] > A[k][i]) break
}
}
}
return A[0].length - Math.max(...dp)
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def minDeletionSize(self, A):
m, n = len(A), len(A[0])
dp = [1] * n
for j in range(1, n):
for i in range(j):
if all(A[k][i] <= A[k][j] for k in range(m)):
dp[j] = max(dp[j], dp[i] + 1)
return n - max(dp)
Input
Output