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 = ["babca","bbazb"] and deletion indices {0, 1, 4}, then the final array after deletions is ["bc","az"].
Suppose we chose a set of deletion indices D such that after deletions, the final array has every element (row) in lexicographic order.
For clarity, A[0] is in lexicographic order (ie. A[0][0] <= A[0][1] <= ... <= A[0][A[0].length - 1]), A[1] is in lexicographic order (ie. A[1][0] <= A[1][1] <= ... <= A[1][A[1].length - 1]), and so on.
Return the minimum possible value of D.length.
 

Example 1:
Input: ["babca","bbazb"]
Output: 3
Explanation: After deleting columns 0, 1, and 4, the final array is A = ["bc", "az"].
Both these rows are individually in lexicographic order (ie. A[0][0] <= A[0][1] and A[1][0] <= A[1][1]).
Note that A[0] > A[1] - the array A isn't necessarily in lexicographic order.


Example 2:
Input: ["edcba"]
Output: 4
Explanation: If we delete less than 4 columns, the only row won't be lexicographically sorted.


Example 3:
Input: ["ghi","def","abc"]
Output: 0
Explanation: All rows are already lexicographically sorted.

 



Note:

1 <= A.length <= 100
1 <= A[i].length <= 100

Source: leetcode | Difficulty: MEDIUM
Tags: String algorithms

Dynamic programming

Data structures

Starter code: class Solution:
    def minDeletionSize(self, A: List[str]) -> int:
        

Solutions:
class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		dp = [(1, 1)] * len(A[0])
		for i in range(len(dp)):
			if i > 0:
				max_pre = None
				for pre in range(i - 1, -1, -1):
					for word in A:
						if word[pre] > word[i]:
							pre -= 1
							break
					else:
						if max_pre is None or dp[pre][1] > dp[max_pre][1]:
							max_pre = pre
				max_len = 1 if max_pre is None else max(1, dp[max_pre][1] + 1)
				overall = max(dp[i - 1][0], max_len)
				dp[i] = (overall, max_len)
		return len(dp) - dp[-1][0]


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		(H, W) = (len(A), len(A[0]))

		def is_valid(j1: int, j2: int) -> bool:
			for i in range(H):
				if A[i][j1] > A[i][j2]:
					return False
			return True
		f = [1 for _ in range(W)]
		max_v = 1
		for i in reversed(range(W - 1)):
			for j in range(i + 1, W):
				if is_valid(i, j):
					f[i] = max(f[i], f[j] + 1)
			max_v = max(max_v, f[i])
		print(f)
		return W - max_v


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		W = len(A[0])
		dp = [1] * W
		for i in range(W - 2, -1, -1):
			for j in range(i + 1, W):
				if all((row[i] <= row[j] for row in A)):
					dp[i] = max(dp[i], 1 + dp[j])
		return W - max(dp)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		n = len(A[0])
		dp = [1 for _ in range(n)]
		for i in range(1, n):
			for j in range(i):
				if all((a[j] <= a[i] for a in A)):
					dp[i] = max(dp[i], dp[j] + 1)
		return n - max(dp)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		m = len(A)
		n = len(A[0])
		dp = [1] * n
		res = n - 1
		for j in range(1, n):
			for i in range(j):
				k = 0
				while k < m and A[k][i] <= A[k][j]:
					k += 1
				if k == m and 1 + dp[i] > dp[j]:
					dp[j] = 1 + dp[i]
			res = min(res, n - dp[j])
		return res


class Solution:

	def minDeletionSize(self, A):
		W = len(A[0])
		dp = [1] * W
		for j in range(1, W):
			for i in range(j):
				if all((row[i] <= row[j] for row in A)):
					dp[j] = max(dp[i] + 1, dp[j])
		return W - max(dp)


class Solution:

	def minDeletionSize(self, A):
		(m, n) = (len(A), len(A[0]))
		dp = [1] * n
		for j in range(1, n):
			for k in range(j):
				if all((A[i][k] <= A[i][j] for i in range(m))):
					dp[j] = max(dp[j], dp[k] + 1)
		return n - max(dp)


class Solution:

	def minDeletionSize(self, a: List[str]) -> int:
		(n, m) = (len(a), len(a[0]))
		dp = [1 for i in range(m)]
		maxx = 1
		for i in range(1, m):
			for j in range(i):
				if all((a[k][i] >= a[k][j] for k in range(n))):
					dp[i] = max(dp[i], dp[j] + 1)
					maxx = max(maxx, dp[i])
		return m - maxx


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		memo = [1 for _ in range(len(A[0]))]
		for i in range(len(A[0])):
			for j in range(i):
				if all((A[k][j] <= A[k][i] for k in range(len(A)))):
					memo[i] = max(memo[i], memo[j] + 1)
		return len(A[0]) - max(memo)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		size = len(A[0])
		best = [0] * size
		best[0] = 1
		for i in range(1, size):
			best[i] = 1
			for j in range(0, i):
				if best[j] >= best[i] and self.beats(A, i, j):
					best[i] = best[j] + 1
		return size - max(best)

	def beats(self, A, a, b):
		for s in A:
			if s[b] > s[a]:
				return False
		return True


class Solution:

	def less_than(self, A, i, j):
		for r in range(len(A)):
			if ord(A[r][i]) > ord(A[r][j]):
				return False
		return True

	def minDeletionSize(self, A: List[str]) -> int:
		n = len(A[0])
		dp = [1 for i in range(n)]
		for i in range(1, n):
			j = i - 1
			while j >= 0:
				if self.less_than(A, j, i):
					dp[i] = max(dp[i], dp[j] + 1)
				j -= 1
		return n - max(dp)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		N = len(A)
		wordLen = len(A[0])
		table = [1 for _ in range(wordLen)]
		for i in range(wordLen - 2, -1, -1):
			subAnsSet = set()
			for k in range(i + 1, wordLen):
				if A[0][i] <= A[0][k]:
					subAnsSet.add(k)
			for j in range(1, N):
				nextSubAnsSet = set()
				for k in subAnsSet:
					if A[j][i] <= A[j][k]:
						nextSubAnsSet.add(k)
				subAnsSet &= nextSubAnsSet
			table[i] = max([table[k] + 1 for k in subAnsSet] + [1])
		return wordLen - max(table)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		(m, n) = (len(A), len(A[0]))
		dp = [0] * (n + 1)
		for i in range(n - 1, -1, -1):
			for j in range(i + 1):
				a = 1 + dp[j]
				b = 10 ** 9
				if j == 0 or all((A[k][i] >= A[k][j - 1] for k in range(m))):
					b = dp[i + 1]
				dp[j] = min(a, b)
		return dp[0]


from functools import lru_cache

class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		(R, C) = (len(A), len(A[0]))

		def col_lower(j1, j2):
			return all((A[i][j1] <= A[i][j2] for i in range(R)))

		@lru_cache(None)
		def LIS(i):
			if i == 0:
				return 1
			return max((LIS(j) for j in range(i) if col_lower(j, i)), default=0) + 1
		return C - max((LIS(i) for i in range(C)))


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		dp = {-1: 0}
		for j in range(len(A[0])):
			new = defaultdict(lambda : 0)
			for (k, v) in list(dp.items()):
				if k == -1:
					new[-1] = 0
					new[j] = 1
				else:
					if all((ord(A[i][k]) <= ord(A[i][j]) for i in range(len(A)))):
						new[j] = max(new[j], v + 1)
					new[k] = max(new[k], v)
			dp = new
		return len(A[0]) - max(dp.values())


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		(l, m) = (len(A), len(A[0]))
		f = [i for i in range(m)]
		f[0] = 0
		for i in range(1, m):
			for j in range(0, i):
				flag = True
				for k in range(l):
					if A[k][j] > A[k][i]:
						flag = False
						break
				if flag:
					f[i] = min(f[i], f[j] + i - j - 1)
		return min([f[i] + m - 1 - i for i in range(m)])


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		word_length = len(A[0])
		dp = [1] * word_length
		for index in range(word_length):
			for curr in range(index - 1, -1, -1):
				poss = True
				for k in range(len(A)):
					if A[k][index] < A[k][curr]:
						poss = False
						break
				if poss:
					dp[index] = max(dp[index], 1 + dp[curr])
		return word_length - max(dp)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		(m, n) = (len(A), len(A[0]))
		memo = [1] * n
		for i in range(n):
			for j in range(i):
				for k in range(m):
					if A[k][j] > A[k][i]:
						break
				else:
					memo[i] = max(memo[i], memo[j] + 1)
		return n - max(memo)


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		m = len(A)
		n = len(A[0])
		dp = [1 for i in range(n)]
		maxVal = 1

		def check(i, j):
			for k in range(m):
				if A[k][i] < A[k][j]:
					return False
			return True
		for i in range(1, n):
			for j in range(i):
				if check(i, j):
					dp[i] = max(dp[i], dp[j] + 1)
			maxVal = max(maxVal, dp[i])
		return n - maxVal


class Solution:

	def minDeletionSize(self, A: List[str]) -> int:
		dp = [1] * len(A[0])
		for j in range(1, len(A[0])):
			for k in range(j):
				isAfter = True
				for i in range(len(A)):
					if A[i][k] > A[i][j]:
						isAfter = False
						break
				if isAfter:
					dp[j] = max(dp[j], dp[k] + 1)
		return len(A[0]) - max(dp)