Delete Columns to Make Sorted III - Solution & Explanation

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.

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.

In this greedy C solution, we traverse each column, comparing successive characters for lexicographic order. If for any column, the order breaks, the column is marked for deletion, thereby incrementing the answer tally, which is returned as the result.