[LeetCode] 944. Delete Columns to Make Sorted

944. Delete Columns to Make Sorted

• Hardness: \(\color{green}\textsf{Easy}\)
• Ralated Topics: Array、String

一、題目

You are given an array of n string strs, all of the same length.
The string s can be arranged such that there is one on each line, making a grid. For example, strs = ["abc", "bce", "cae" can be arranged as :

abc
bce
cae

You want to delete the columns that are not sorted lexicographically. In the aove 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 3.

Constraints:

• n == strs.length
• 1 <= n <= 100
• 1 <= strs[i].length <= 1000

二、分析

• 根據題意，逐個檢查是否字元有隨著列增加而呈 lexicographically sorted。
• 注意迴圈的走法，若是外圈為欄，內圈為列的話，在發現沒有排序時，可以提早跳出，加快速度。

三、解題

1. String

• Time complexity: \(O(m\times n)\)
• Space complexity: \(O(1)\)

int minDeletionSize(vector<string>& strs) {
    int m = strs.size(), n = strs[0].size();
    int cnt = 0;
    for (int col = 0; col < n; ++col) {
        for (int row = 1; row < m; ++row) {
            if (strs[row-1][col] > strs[row][col]) {
                cnt++;
                break;
            }
        }
    }
    return cnt;
}

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