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This approach utilizes dynamic programming to identify the minimum columns to delete to make each row lexicographically sorted. The key idea is to use a dp array where dp[j] represents the length of the longest increasing subsequence of columns ending at index j.
We'll iterate over each column and compare with the previous columns to update the dp array by checking each row for their lexicographic order. We finally return the total number of columns minus the maximum value in the dp array, which gives the least number of deletions needed.
Time Complexity: O(n^2 * m), where n is the length of each string and m is the number of strings.
Space Complexity: O(n) for the dp array.
1#include <stdio.h>
2#include <string.h>
3
4int minDeletionSize(char** strs, int strsSize) {
5 int n
The C code implements a dynamic programming solution. It initializes a dp array to track the longest increasing subsequence of column indices which maintain the row order. For each pair of columns, it checks if swapping the columns maintains the lexicographic order in each row. The solution updates the dp array accordingly and returns the total columns minus the maximum value in this array.
This approach involves a greedy strategy where we incrementally build the sorted subset of columns. By comparing columns, we find those that necessarily break the order when removed and only retain these to maintain row order.
At each step, the algorithm compares characters in successive columns, choosing deletions based on maintaining an overall least sequence of lexicographically ordered characters per row.
Time Complexity: O(m * n), where m is the number of strings and n is the length of each string.
Space Complexity: O(1), as only constant space is used.
Traversing column by column, JavaScript finds necessary deletions to maintain order in a greedy manner. A column is marked if any row fails in sequence compared to the one before.