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+# Spiral Matrix
+
+Goal: Traverse a 2D array in spiral order.
+
+## Overview
+Given a 2D array, which can be a square array with `n == m`, we want to return a one-dimensional array of all the elements of the 2D array in spiral order.
+
+Spiral order will start at the top left corner of the 2D array move all the way to the right-most column on that row, then continue in a spiral pattern all the way until every element has been visited.
+
+
+
+## Example
+Given the 2D array
+
+```swift
+array = [
+ [1, 2, 3, 4],
+ [12, 13, 14, 5],
+ [11, 16, 15, 6],
+ [10, 9, 8, 7]
+]
+```
+
+we want to traverse this in spiral order and return the following one-dimensional array
+
+```swift
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
+```
+
+## Iterative approach
+We want to traverse this in clockwise spiral order. We go from the outer-most ‘layer’ to the inner-most ‘ layer’.
+
+Let’s hold the indices of
+
+```swift
+var startRow = 0
+var startCol = 0
+var endRow = array.count - 1
+var endCol = array[0].count - 1
+```
+
+The order in which we will move through the first layer of the spiral using these indices is
+
+### Steps
+
+1. Loop while `startRow <= endRow, startCol <= endCol`
+2. Horizontally traverse towards the right from `startCol` to (inclusive) `endCol` by `1`.
+3. Vertically traverse down from `startRow + 1` to (inclusive) `endRow` by `1` .
+4. Horizontally traverse towards the left from `endCol - 1` to (inclusive) `startCol` by `-1`.
+5. Vertically traverse up from `endRow - 1` to (inclusive) `startCol + 1` by `-1`.
+6. Increment indices `startRow += 1` `startCol += 1`
+7. Decrement indices `endRow -= 1` `endCol -= 1`
+
+### Edge cases
+The matrix may have single row or a single columns in the middle. In this case, we don’t want to double-count values during step `4` and `5`, which we have already counted in step `2` and `3`. We will do an index check in both cases and break if it is true.
+
+### Code
+
+```swift
+func spiralTraverse(array: [[Int]]) -> [Int] {
+ guard array.count != 0 else { return [] }
+
+ var result = [Int]()
+ var startRow = 0
+ var endRow = array.count - 1
+ var startCol = 0
+ var endCol = array[0].count - 1
+
+ while startRow <= endRow, startCol <= endCol {
+ for col in stride(from: startCol, through: endCol, by: 1) {
+ result.append(array[startRow][col])
+ }
+
+ for row in stride(from: startRow + 1, through: endRow, by: 1) {
+ result.append(array[row][endCol])
+ }
+
+ for col in stride(from: endCol - 1, through: startCol, by: -1) {
+ if startRow == endRow {
+ break
+ }
+ result.append(array[endRow][col])
+ }
+
+ for row in stride(from: endRow - 1, through: startRow + 1, by: -1) {
+ if startCol == endCol {
+ break
+ }
+ result.append(array[row][startCol])
+ }
+
+ startRow += 1
+ endRow -= 1
+ startCol += 1
+ endCol -= 1
+ }
+ return result
+}
+```
+
+## Recursive approach
+Now let’s check out the recursive implementation of the algorithm. In this version we don’t need to maintain and update indices since we will be calling the function recursively based on different values.
+
+### Code
+```swift
+func spiralTraverse(array: [[Int]]) -> [Int] {
+ var result = [Int]()
+ spiralHelper(array, 0, array.count - 1, 0, array[0].count - 1, &result)
+ return result
+}
+
+func spiralHelper(_ array: [[Int]], _ startRow: Int, _ endRow: Int, _ startCol: Int, _ endCol: Int, _ result: inout [Int]) {
+ if startRow > endRow || startCol > endCol {
+ return
+ }
+
+ for col in stride(from: startCol, through: endCol, by: 1) {
+ result.append(array[startRow][col])
+ }
+
+ for row in stride(from: startRow + 1, through: endRow, by: 1) {
+ result.append(array[row][endCol])
+ }
+
+ for col in stride(from: endCol - 1, through: startCol, by: -1) {
+ if startRow == endRow { break }
+ result.append(array[endRow][col])
+ }
+
+ for row in stride(from: endRow - 1, through: startRow + 1, by: -1) {
+ if startCol == endCol { break }
+ result.append(array[row][startCol])
+ }
+
+ spiralHelper(array, startRow + 1, endRow - 1, startCol + 1, endCol - 1, &result)
+}
+```
+
+## Test
+
+```swift
+let array = [
+ [1,2,3,4],
+ [12,13,14,5],
+ [11,16,15,6],
+ [10,9,8,7]
+]
+spiralTraverse(array: array)
+```
+
+## Complexity
+
+* Time: `O(n)` - where `n` is the total number of elements in the 2D array
+* Space: `O(n)` - since we are storing every element of the 2D array into a one-dimensional array
+---
+Written for Swift Algorithm Club by Azhar Anwar
diff --git a/Spiral Matrix/spiral-matrix.playground/Contents.swift b/Spiral Matrix/spiral-matrix.playground/Contents.swift
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+import Foundation
+
+/**
+ Complexity:
+ Time: O(n) - where n is the total number of elements in the 2D array
+ Space: O(n)
+ */
+
+/// Iterative Spiral Matrix traversal
+/// - Parameter array: 2D input array
+/// - Returns: One-dimensional result array
+func spiralTraverseIterative(array: [[Int]]) -> [Int] {
+ if array.count == 0 { return [] }
+
+ var result = [Int]()
+
+ var startRow = 0
+ var startCol = 0
+ var endRow = array.count - 1
+ var endCol = array[0].count - 1
+
+ while startRow <= endRow, startCol <= endCol {
+ for col in stride(from: startCol, through: endCol, by: 1) {
+ result.append(array[startRow][col])
+ }
+
+ for row in stride(from: startRow + 1, through: endRow, by: 1) {
+ result.append(array[row][endCol])
+ }
+
+ for col in stride(from: endCol - 1, through: startCol, by: -1) {
+ if startRow == endRow { break }
+ result.append(array[endRow][col])
+ }
+
+ for row in stride(from: endRow - 1, through: startCol + 1, by: -1) {
+ if startCol == endCol { break }
+ result.append(array[row][startCol])
+ }
+
+ startRow += 1
+ endRow -= 1
+ startCol += 1
+ endCol -= 1
+ }
+
+ return result
+}
+
+
+/// Recursive Spiral Matrix traversal
+/// - Parameter array: 2D array
+/// - Returns: One-dimensional array of all elements from the 2D array traversed in spiral order.
+func spiralTraverseRecursive(array: [[Int]]) -> [Int] {
+ var result = [Int]()
+ spiralHelper(array, 0, array.count - 1, 0, array[0].count - 1, &result)
+ return result
+}
+
+/// Spiral traversal helper function
+/// - Parameters:
+/// - array: 2D array
+/// - startRow: Index of starting row
+/// - endRow: Index of last row
+/// - startCol: Index of first column
+/// - endCol: index of last column
+/// - result: One-dimensional result array.
+func spiralHelper(_ array: [[Int]], _ startRow: Int, _ endRow: Int, _ startCol: Int, _ endCol: Int, _ result: inout [Int]) {
+ if startRow > endRow || startCol > endCol {
+ return
+ }
+
+ for col in stride(from: startCol, through: endCol, by: 1) {
+ result.append(array[startRow][col])
+ }
+
+ for row in stride(from: startRow + 1, through: endRow, by: 1) {
+ result.append(array[row][endCol])
+ }
+
+ for col in stride(from: endCol - 1, through: startCol, by: -1) {
+ if startRow == endRow { break }
+ result.append(array[endRow][col])
+ }
+
+ for row in stride(from: endRow - 1, through: startRow + 1, by: -1) {
+ if startCol == endCol { break }
+ result.append(array[row][startCol])
+ }
+
+ spiralHelper(array, startRow + 1, endRow - 1, startCol + 1, endCol - 1, &result)
+}
+
+
+///Test
+let array = [
+ [1,2,3,4],
+ [12,13,14,5],
+ [11,16,15,6],
+ [10,9,8,7]
+]
+spiralTraverseIterative(array: array)
+spiralTraverseRecursive(array: array)
diff --git a/Spiral Matrix/spiral-matrix.playground/contents.xcplayground b/Spiral Matrix/spiral-matrix.playground/contents.xcplayground
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+
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+
+
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diff --git a/Spiral Matrix/spiral-matrix.playground/playground.xcworkspace/contents.xcworkspacedata b/Spiral Matrix/spiral-matrix.playground/playground.xcworkspace/contents.xcworkspacedata
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+
+
+
+
diff --git a/Spiral Matrix/spiral-matrix.playground/playground.xcworkspace/xcshareddata/IDEWorkspaceChecks.plist b/Spiral Matrix/spiral-matrix.playground/playground.xcworkspace/xcshareddata/IDEWorkspaceChecks.plist
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+
+
+
+
+ IDEDidComputeMac32BitWarning
+
+
+