Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming
["babca","bbazb"]
["edcba"]
["ghi","def","abc"]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #960: Delete Columns to Make Sorted III
class Solution {
public int minDeletionSize(String[] strs) {
int n = strs[0].length();
int[] f = new int[n];
Arrays.fill(f, 1);
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
boolean ok = true;
for (String s : strs) {
if (s.charAt(j) > s.charAt(i)) {
ok = false;
break;
}
}
if (ok) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
}
return n - Arrays.stream(f).max().getAsInt();
}
}
// Accepted solution for LeetCode #960: Delete Columns to Make Sorted III
func minDeletionSize(strs []string) int {
n := len(strs[0])
f := make([]int, n)
for i := range f {
f[i] = 1
}
for i := 1; i < n; i++ {
for j := 0; j < i; j++ {
ok := true
for _, s := range strs {
if s[j] > s[i] {
ok = false
break
}
}
if ok {
f[i] = max(f[i], f[j]+1)
}
}
}
return n - slices.Max(f)
}
# Accepted solution for LeetCode #960: Delete Columns to Make Sorted III
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)
// Accepted solution for LeetCode #960: Delete Columns to Make Sorted III
impl Solution {
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
let n = strs[0].len();
let mut f = vec![1; n];
for i in 1..n {
for j in 0..i {
if strs.iter().all(|s| s.as_bytes()[j] <= s.as_bytes()[i]) {
f[i] = f[i].max(f[j] + 1);
}
}
}
(n - *f.iter().max().unwrap()) as i32
}
}
// Accepted solution for LeetCode #960: Delete Columns to Make Sorted III
function minDeletionSize(strs: string[]): number {
const n = strs[0].length;
const f: number[] = Array(n).fill(1);
for (let i = 1; i < n; i++) {
for (let j = 0; j < i; j++) {
let ok = true;
for (const s of strs) {
if (s[j] > s[i]) {
ok = false;
break;
}
}
if (ok) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
}
return n - Math.max(...f);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.