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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.
1var minDeletionSize = function(strs) {
2 const n = strs[0].length;
3 const dp = Array(n).fill(1);
4
5 for (let j = 1; j < n; ++j) {
6 for (let i = 0; i < j; ++i) {
7 if (strs.every(s => s[i] <= s[j])) {
8 dp[j] = Math.max(dp[j], dp[i] + 1);
9 }
10 }
11 }
12
13 return n - Math.max(...dp);
14};
15The JavaScript solution also capitalizes on dynamic programming by maintaining an array for valid column subsequences. As rows are processed, if maintaining order, the array is updated to reflect subsequences' lengths. The outermost step reflects total column counts against this subsequence maximum for minimal deletions.
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.
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.