Let's dry-run our algorithm to see how the solution works. Check if we should turn our direction or not.Source: LeetCode Input: n = 3 Output:, , ]Ĭonstraints: - 1 > /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ - loop while k = l i- set result = value - increment value++ - m- loop for i = m - 1 i >= k i- set result = value - increment value++ - l++ - return result C++ solution class Solution
Initial position is at top left corner with direction towards right. Condition to turn when current position in other parts.īool should_turn(int cur_row, int cur_col, int rows, int cols)
Condition to turn when current position is in top-left part. < cols / 2 will give priority to right part when current position is vertically centered. < (rows + 1) / 2 will give priority to top part when current position is horizontally centered.
SPIRAL ORDER MATRIX IBOOKMARK SUGGEST EDIT HOW TO
Now you know where to put the signs! How to check if point is in top-left or other parts?Ĭonsider these grids to understand what the below code does. Now for the top-left part we need to check: Min(cur_row, rows - 1 - cur_row) = min(cur_col, cols - 1 - cur_col). Once you've got all four functions working properly to draw the outer ring, you can call the functions a second time, and see how that goes. When that works properly, you can write and test the next function. Then print the matrix to see the partial answer. We can write conditions separately or combine them as: You should start with one function, that you call from main once. If matrix size is rows * cols then for any point (at position cur_row and cur_col) if we want to check if there is a sign or not simply check:ġ) top-right: cur_row = cols - 1 - cur_colĢ) bottom-right: rows - 1 - cur_row = cols - 1 - cur_colģ) bottom-left: rows - 1 - cur_row = cur_col In simple words, what the program does is ask for the number of elements the user wants, and creates a square matrix ( note: the matrix created needs to be a square). Vertically centered points: Consider them in right parts.įor top-right, bottom-right and bottom-left pattern is same! Hello everyone, in this tutorial, I will guide you with the basic of filling up a Double-Dimension matrix, in spiral order in C or C++. Horizontally centered points: Consider them in top parts. Now for most of the points we can easily decide in which part they will fall, except points which are horizontally centered or vertically centered. We can divide the grid in 4 parts and then follow some patterns. Then we will visit the matrix in spiral order. Suppose we start from the top-left corner (0, 0) and turn right at certain locations (indicated by # signs).
Here we will provide one interesting solution that uses only one for loop. There are many solutions possible for this problem. That will also help your interviewer follow along more easily. Instead of i,j,k,l etc, try naming them like row, col, start, end etc. To rotate a matrix in spiral order, we need to do following until all the inner matrix and the outer matrix are covered. Hint - It may be faster to write this, if you name your variables clearly. How to print a matrix of size nn in spiral order using C Csharp Server Side Programming Programming. This problem is less about logic, but more about careful index manipulation.