955. Delete Columns to Make Sorted II

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 its elements in lexicographic order (i.e., strs[0] <= strs[1] <= strs[2] <= … <= strs[n - 1]). Return the minimum possible value of answer.length.
 

Example 1:

Input: strs = [“ca”,“bb”,“ac”]
Output: 1
**Explanation: **
After deleting the first column, strs = [“a”, “b”, “c”].
Now strs is in lexicographic order (ie. strs[0] <= strs[1] <= strs[2]).
We require at least 1 deletion since initially strs was not in lexicographic order, so the answer is 1.

Example 2:

Input: strs = [“xc”,“yb”,“za”]
Output: 0
**Explanation: **
strs is already in lexicographic order, so we do not need to delete anything.
Note that the rows of strs are not necessarily in lexicographic order:
i.e., it is NOT necessarily true that (strs[0][0] <= strs[0][1] <= …)

Example 3:

Input: strs = [“zyx”,“wvu”,“tsr”]
Output: 3
Explanation: We have to delete every column.

Constraints:

• n == strs.length
• 1 <= n <= 100
• 1 <= strs[i].length <= 100
• strs[i] consists of lowercase English letters.

From: LeetCode
Link: 955. Delete Columns to Make Sorted II

Solution:

Ideas:

• We scan columns from left to right and decide for each column whether we must delete it.

• Maintain an array sorted[i] indicating whether the pair (strs[i], strs[i+1]) is already confirmed to be in correct order based on previous kept columns.

• For each column:

  • First check if any unsorted pair would become invalid (strs[i][col] > strs[i+1][col]) → must delete this column.
  • If not deleted, then for any unsorted pair where strs[i][col] < strs[i+1][col], mark that pair as sorted going forward.

Code:

int minDeletionSize(char** strs, int strsSize) {
    if (strsSize <= 1) return 0;

    int n = strlen(strs[0]);      // number of columns
    bool sorted[100];             // up to 100 strings, so at most 99 pairs
    for (int i = 0; i < strsSize - 1; ++i) {
        sorted[i] = false;
    }

    int deletions = 0;

    for (int col = 0; col < n; ++col) {
        bool needDelete = false;

        // Check if keeping this column would break lexicographic order
        for (int i = 0; i < strsSize - 1; ++i) {
            if (!sorted[i] && strs[i][col] > strs[i + 1][col]) {
                needDelete = true;
                break;
            }
        }

        if (needDelete) {
            // Delete this column
            deletions++;
            continue;
        }

        // Update which adjacent pairs are now confirmed sorted
        for (int i = 0; i < strsSize - 1; ++i) {
            if (!sorted[i] && strs[i][col] < strs[i + 1][col]) {
                sorted[i] = true;
            }
        }
    }

    return deletions;
}