Leetcode #960: Delete Columns to Make Sorted III

In this guide, we solve Leetcode #960 Delete Columns to Make Sorted III in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

Difficulty: Hard
Premium: No
Tags: Array, String, Dynamic Programming

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

Choose a DP state definition.
Write the recurrence and base cases.
Compute states in the correct order.

Example

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.

Python Solution

class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:
        n = len(strs[0])
        f = [1] * n
        for i in range(n):
            for j in range(i):
                if all(s[j] <= s[i] for s in strs):
                    f[i] = max(f[i], f[j] + 1)
        return n - max(f)

Complexity

The time complexity is $O(n^2 \times m)$ , and the space complexity is $O(n)$ , where $n$ is the length of each string in the array $\textit{strs}$ , and $m$ is the number of strings in the array. The space complexity is $O(n)$ , where $n$ is the length of each string in the array $\textit{strs}$ , and $m$ is the number of strings in the array.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Example

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.

Python Solution

class Solution:
    def minDeletionSize(self, strs: List[str]) -> int:
        n = len(strs[0])
        f = [1] * n
        for i in range(n):
            for j in range(i):
                if all(s[j] <= s[i] for s in strs):
                    f[i] = max(f[i], f[j] + 1)
        return n - max(f)

Complexity

The time complexity is

O(n^2 \times m)

, and the space complexity is

O(n)

, where

n

is the length of each string in the array

\textit{strs}

, and

m

is the number of strings in the array. The space complexity is

O(n)

, where

n

is the length of each string in the array

\textit{strs}

, and

m

is the number of strings in the array.

Leetcode #960: Delete Columns to Make Sorted III

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #960: Delete Columns to Make Sorted III

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview