> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Delete Columns to Make Sorted II

> Tested Python solution for LeetCode 955 with 24 pytest cases. Generate a practice environment with lcpy.

LeetCode 955, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [String](/catalog/topics/string), [Greedy](/catalog/topics/greedy). [View on LeetCode](https://leetcode.com/problems/delete-columns-to-make-sorted-ii/description/).

Generate this problem as a practice environment: tested reference solution, 24 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 955   # by problem number
lcpy gen -s delete_columns_to_make_sorted_ii   # by problem name
```

## Problem

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`.

### Examples

```
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.
```

```
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] <= ...)
```

```
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.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/delete_columns_to_make_sorted_ii/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/delete_columns_to_make_sorted_ii/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n * w^2)
    # Space: O(n * w)
    def min_deletion_size(self, strs: list[str]) -> int:
        keep = [""] * len(strs)
        deleted = 0
        for j in range(len(strs[0])):
            candidate = [row + s[j] for row, s in zip(keep, strs, strict=True)]
            if all(candidate[i] <= candidate[i + 1] for i in range(len(candidate) - 1)):
                keep = candidate
            else:
                deleted += 1
        return deleted
```

## Complexity

| Time | Space |
| - | - |
| O(n \* w^2) | O(n \* w) |

## Tags


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