Introduction

In this blog post, we will explore how to solve the LeetCode problem "960" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.length 1 <= n <= 100 1 <= strs[i].length <= 100 strs[i] consists of lowercase English letters.

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• Key Idea: Iterate through the columns of the input strings. For each column, check if deleting that column would violate the lexicographical order of any of the strings. If it does, then that column must be deleted.
• Optimization: A column can be skipped if all the strings are already sorted up to the current column.

• Algorithm: Initialize a count of deletion indices to zero. Iterate through each column. For each column, check if deleting it would violate the lexicographical order of any row. If a violation is found, increment the deletion count.

• Complexity: Time: O(n*m), where n is the number of strings and m is the length of each string. Space: O(1).

Code

    def minDeletionSize(strs):
    """
    Calculates the minimum number of columns to delete to make all rows lexicographically sorted.

    Args:
        strs: A list of strings, all of the same length.

    Returns:
        The minimum number of columns to delete.
    """
    n = len(strs)
    m = len(strs[0])
    delete_count = 0

    for j in range(m):
        for i in range(1, n):
            if strs[i][j] < strs[i - 1][j]:
                delete_count += 1
                break  # Move to the next column if this column needs to be deleted
    return delete_count