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by Sumit Chourasia | Oct 05, 2020 | Category :coding | Tags : algorithm data-structure easy greedy leetcode

Delete Columns to Make Sorted - Greedy - Easy - LeetCode

Delete Columns to Make Sorted - Greedy - Easy - LeetCode

We are given an array A of N lowercase letter strings, all of the same length.

Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.

For example, if we have an array A = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef", "vyz"], and the remaining columns of A are ["b","v"], ["e","y"], and ["f","z"].  (Formally, the c-th column is [A[0][c], A[1][c], ..., A[A.length-1][c]]).

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.

Return the minimum possible value of D.length.


Example 1:

Input: A = ["cba","daf","ghi"]
Output: 1
After choosing D = {1}, each column ["c","d","g"] and ["a","f","i"] are in non-decreasing sorted order.
If we chose D = {}, then a column ["b","a","h"] would not be in non-decreasing sorted order.
Example 2:

Input: A = ["a","b"]
Output: 0
Explanation: D = {}
Example 3:

Input: A = ["zyx","wvu","tsr"]
Output: 3
Explanation: D = {0, 1, 2}


1 <= A.length <= 100
1 <= A[i].length <= 1000

public class Solution {
    public int MinDeletionSize(string[] A) {
        int count = 0;
        for(int c=0;c<A[0].Length;c++){
            for(int r=1;r<A.Length;r++){
        return count;

Time Complexity: O(m*n)

Space Complexity: O(1)

where m and n are the count of string and max length of string.

Contributed By: Sumit Chourasia
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