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.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
You are given an array of n strings strs, all of the same length.
The strings can be arranged such that there is one on each line, making a grid.
strs = ["abc", "bce", "cae"] can be arranged as follows:abc bce cae
You want to delete the columns that are not sorted lexicographically. In the above example (0-indexed), columns 0 ('a', 'b', 'c') and 2 ('c', 'e', 'e') are sorted, while column 1 ('b', 'c', 'a') is not, so you would delete column 1.
Return the number of columns that you will delete.
Example 1:
Input: strs = ["cba","daf","ghi"] Output: 1 Explanation: The grid looks as follows: cba daf ghi Columns 0 and 2 are sorted, but column 1 is not, so you only need to delete 1 column.
Example 2:
Input: strs = ["a","b"] Output: 0 Explanation: The grid looks as follows: a b Column 0 is the only column and is sorted, so you will not delete any columns.
Example 3:
Input: strs = ["zyx","wvu","tsr"] Output: 3 Explanation: The grid looks as follows: zyx wvu tsr All 3 columns are not sorted, so you will delete all 3.
Constraints:
n == strs.length1 <= n <= 1001 <= strs[i].length <= 1000strs[i] consists of lowercase English letters.Problem summary: You are given an array of n strings strs, all of the same length. The strings can be arranged such that there is one on each line, making a grid. For example, strs = ["abc", "bce", "cae"] can be arranged as follows: abc bce cae You want to delete the columns that are not sorted lexicographically. In the above example (0-indexed), columns 0 ('a', 'b', 'c') and 2 ('c', 'e', 'e') are sorted, while column 1 ('b', 'c', 'a') is not, so you would delete column 1. Return the number of columns that you will delete.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
["cba","daf","ghi"]
["a","b"]
["zyx","wvu","tsr"]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #944: Delete Columns to Make Sorted
class Solution {
public int minDeletionSize(String[] strs) {
int m = strs[0].length(), n = strs.length;
int ans = 0;
for (int j = 0; j < m; ++j) {
for (int i = 1; i < n; ++i) {
if (strs[i].charAt(j) < strs[i - 1].charAt(j)) {
++ans;
break;
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #944: Delete Columns to Make Sorted
func minDeletionSize(strs []string) (ans int) {
m, n := len(strs[0]), len(strs)
for j := 0; j < m; j++ {
for i := 1; i < n; i++ {
if strs[i][j] < strs[i-1][j] {
ans++
break
}
}
}
return
}
# Accepted solution for LeetCode #944: Delete Columns to Make Sorted
class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
m, n = len(strs[0]), len(strs)
ans = 0
for j in range(m):
for i in range(1, n):
if strs[i][j] < strs[i - 1][j]:
ans += 1
break
return ans
// Accepted solution for LeetCode #944: Delete Columns to Make Sorted
impl Solution {
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
let n = strs.len();
let m = strs[0].len();
let mut ans = 0;
for j in 0..m {
for i in 1..n {
if strs[i].as_bytes()[j] < strs[i - 1].as_bytes()[j] {
ans += 1;
break;
}
}
}
ans
}
}
// Accepted solution for LeetCode #944: Delete Columns to Make Sorted
function minDeletionSize(strs: string[]): number {
const [m, n] = [strs[0].length, strs.length];
let ans = 0;
for (let j = 0; j < m; ++j) {
for (let i = 1; i < n; ++i) {
if (strs[i][j] < strs[i - 1][j]) {
++ans;
break;
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
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.