## From an interview: Removing rows and columns in an n×n matrix to maximize the sum of remaining values

unit area of largest region of 1's

maximal rectangle python

Given an n×n matrix of real numbers. You are allowed to erase any number (from 0 to n) of rows and any number (from 0 to n) of columns, and after that the sum of the remaining entries is computed. Come up with an algorithm which finds out which rows and columns to erase in order to maximize that sum.

**Remove any corner X rows and columns from a matrix ,** Simple Approach: Skip the first x rows and columns and print the remaining elements in the matrix. Start from the xth row and print till the n-� To do this, select the row or column and then press the Delete key. Right-click in a table cell, row, or column you want to delete. On the menu, click Delete Cells. To delete one cell, choose Shift cells left or Shift cells up. To delete the row, click Delete entire row.

**Remove first X rows and columns from a matrix,** Approach: Print the elements of the matrix arr[i][j] for all i, j ∈ [X, N – 1] where N is the order of the given square matrix. Below is the� Use the mini toolbar to add rows and columns. To open the mini toolbar, right-click in a table cell or tap in a selected table cell next to where you want to add a row or column. On the mini toolbar, click Insert and choose where you’d like to insert the row or column. Use Insert Controls to add rows and columns. Insert Controls work only

We can improve on Cletus's generalized brute-force solution by modelling this as a directed graph. The initial matrix is the start node of the graph; its leaves are all the matrices missing one row or column, and so forth. It's a graph rather than a tree, because the node for the matrix without both the first column and row will have two parents - the nodes with just the first column or row missing.

We can optimize our solution by turning the graph into a tree: There's never any point exploring a submatrix with a column or row deleted that comes before the one we deleted to get to the current node, as that submatrix will be arrived at anyway.

This is still a brute-force search, of course - but we've eliminated the duplicate cases where we remove the same rows in different orders.

Here's an example implementation in Python:

def maximize_sum(m): frontier = [(m, 0, False)] best = None best_score = 0 while frontier: current, startidx, cols_done = frontier.pop() score = matrix_sum(current) if score > best_score or not best: best = current best_score = score w, h = matrix_size(current) if not cols_done: for x in range(startidx, w): frontier.append((delete_column(current, x), x, False)) startidx = 0 for y in range(startidx, h): frontier.append((delete_row(current, y), y, True)) return best_score, best

And here's the output on 280Z28's example matrix:

>>> m = ((1, 1, 3), (1, -89, 101), (1, 102, -99)) >>> maximize_sum(m) (106, [(1, 3), (1, 101)])

**Google Interview Question: Given a sorted matrix where t,** Interview question for Software Engineer in Mountain View, CA. if the number exists in the matrix, will be in this row. or log(sqrt(n^2+ m^2)) more precisely First, the binary search runs on the items in the first column to identify the column and then run a binary search One or more comments have been removed. The rows and column values may be scalar values, lists, slice objects or boolean. Select all the rows, and 4th, 5th and 7th column: To replicate the above DataFrame, pass the column names as a list to the .loc indexer: Selecting disjointed rows and columns To select a particular number of rows and columns, you can do the following using .iloc

Since nobody asked for an efficient algorithm, use brute force: generate every possible matrix that can be created by removing rows and/or columns from the original matrix, choose the best one. A slightly more efficent version, which most likely can be proved to still be correct, is to generate only those variants where the removed rows and columns contain at least one negative value.

**Delete elements, rows or columns from a Numpy Array by index ,** Let's see how to use np.delete() to remove elements by index positions from 1D & 2D numpy arrays and also how to delete rows & columns� The DML statements are used to add new rows to a table, update or modify data in existing rows, or remove existing rows from a table. What are the various multiple row comparison operators in SQL? IN, ANY, ALL.

To try it in a simple way:

We need the *valid* subset of the set of entries {A00, A01, A02, ..., A0n, A10, ...,Ann} which max. sum.

First compute *all* subsets (the power set).

A *valid* subset is a member of the power set that for each two contained entries Aij and A(i+x)(j+y), contains also the elements A(i+x)j and Ai(j+y) (which are the remaining corners of the rectangle spanned by Aij and A(i+x)(j+y)).

Aij ... . . . . ... A(i+x)(j+y)

By that you can eliminate the invalid ones from the power set and find the one with the biggest sum in the remaining.

I'm sure it can be improved by improving an algorithm for power set generation in order to generate only valid subsets and by that avoiding step 2 (adjusting the power set).

**Removing Rows or Columns from a Matrix,** The easiest way to remove a row or column from a matrix is to set that row or column equal to a pair of empty square brackets [] . For example, create a 4-by-4 � The main difference between rows and columns are discussed in this article. Row is an order in which people, objects or figures are placed alongside or in a straight line. A vertical division of facts, figures or any other details based on category, is called column.

**Solved: Remove rows with one or more empty values using JSL ,** Here is an example deleting all rows with missing cells in columns 2, 3 & 4: all null rowlist = {}; For(i= N Rows(m), i>=1,i--, If(V Sum(Transpose(m[i,0]))[1] == 0,� Once we have this info, we can use first row and first column as auxiliary arrays and apply method 1 for submatrix (matrix excluding first row and first column) of size (M-1)*(N-1). 1) Scan the first row and set a variable rowFlag to indicate whether we need to set all 1s in first row or not.

**Drop rows with empty values pandas,** Toggle navigation Data Interview Qs. Drop the duplicate by retaining last Output: Example 2: Using parameter n, which selects n numbers of rows randomly. Remove rows or columns by specifying label names and corresponding axis,� A relative weight used to distribute additional space between columns. A column with the weight 2 will grow twice as fast as a column with weight 1. The default is 0, which means that the column will not grow at all. grid_configure(**options) Same as grid. grid_forget() Remove this widget from the grid manager.

**Pandas Drop Duplicate Rows,** Identify Duplicate Rows based on Specific Columns; 2.5 5. If False, all the duplicate rows are deleted. inplace: if True, the source DataFrame is row result_df = source_df.drop_duplicates() print('Result DataFrame:\n', result_df) Java Tutorial � Python Tutorials � Java Interview Questions � Core Java Interview Questions� to remove just the a column you could do. Data <- subset( Data, select = -a ) and to remove the b and d columns you could do. Data <- subset( Data, select = -c(d, b ) ) You can remove all columns between d and b with: Data <- subset( Data, select = -c( d : b ) As I said above, this syntax works only when the column names are known.

##### Comments

- Does the rows includes negative numbers ?
- Dani: if there are no negative numbers, the answer is very simple: Don't remove anything ;-)
- Helluva interview question. What kind of position were you applying for to get a question like this?
- Sounds smart, but it's not an answer... In fact it has not so much to do with the question. What you do is constructing one special matrix of the class mentioned in the initial problem that represent a graph. The positive entries of this matrix represent the
*edges*of the graph (not vertexes with special property!). If you have a look at the matrix constructed by a simple square (corners = vertexes) you'll see there is not much connexion between the solution matrix for our initial problem (there are two) and the maximal clique (there are four). - With your a edits I can now suggest an analogy. However your values seem a bit arbitrary to me. Why not using 1 for 'being connected' (or 'edge') (and assuming every vertex is connected to itself) and -1 for 'not connected' (or 'no edge')?
- @280Z28: That's not relevant: it's the
*decision version*of the clique problem ("is there a clique of size of size ≥k?") that is NP-complete, so the equivalent here would be "does there exist a way of removing some rows and columns to get a sum ≥ s?" And given a certificate (which rows and columns to remove), this is trivial to check in polynomial time, so the problem is indeed in NP (and NP-complete). - @David Thornley, solving NP-complete problems with heuristics (and the aptitude to pick one that's good enough and prove its efficiency) is a valuable skill even in commercial development.
- @Pavel: Coming up with good approximations to solutions to NP-hard problems by whatever means is a valuable skill, and is often doable. Coming up with actual solutions to NP-hard problems would be an even more valuable skill, but it isn't doable unless P=NP. The problem statement was to find an algorithm that would maximize the sum, not a heuristic to find a large sum.
- I wonder if there is any smarter way to do it without brute forcing it... can't think of anything yet.
- Your answer is correct, btw ;-) See stackoverflow.com/questions/1720737/…
- from the question I conclude that if all numbers are negative, removing all rows and columns yields a maximum value of 0 which is better than a 1x1 matrix with a negatif number :-)