Approach

# for

Time Complexity

n = row
m = column
O(n * m)

Space Complexity

n = row
m = column
O(n * m)

Code

class Solution {
public:
    int minDeletionSize(vector<string>& strs) {
        int cnt = 0;

        for (int i = 0 ; i < strs[0].size(); i++) {
            bool check = true;

            for (int j = 1 ; j < strs.size() ; j++) {
                if (strs[j][i] - strs[j - 1][i] < 0) {
                    check = false;
                }
            }

            if (!check) {
                cnt++;
            }
        }

        return cnt;
    }
};