GitHub Action to use codespell to find typos (TheAlgorithms#105)

cclauss · web-flow · commit 9778175f3ccc · 2021-09-03T09:05:03.000-07:00
* GitHub Action to use codespell to find typos https://github.com/codespell-project/codespell * codespell ./en/ * Fix typo discovered by codespell * Fix typo discovered by codespell * Fix typo discovered by codespell * Fix broken URL
diff --git a/.github/workflows/codespell.yml b/.github/workflows/codespell.yml
@@ -0,0 +1,9 @@
+name: codespell
+on: [pull_request, push]
+jobs:
+  codespell:
+    runs-on: ubuntu-latest
+    steps:
+      - uses: actions/checkout@v2
+      - run: python3 -m pip install codespell
+      - run: codespell ./en/  # --ignore-words-list="" --skip=""
diff --git a/en/Dynamic Programming/Coin Change.md b/en/Dynamic Programming/Coin Change.md
@@ -42,7 +42,7 @@ Let's say we have 3 coin types `[1,2,3]` and we want to change for `7` cents. So
 ```
 4 ways to make 7 cents using value of 1 and 2. `{{1,1,1,1,1,1,1}, {1,1,1,1,1,2}, {1,1,1,2,2}, {1,2,2,2}}`
 
-* For the final iteration (3rd iteration), we take a coin that has a value of 3. Like before, now all the columns that can be devided by 3 will store another new way. And the final result will be like:
+* For the final iteration (3rd iteration), we take a coin that has a value of 3. Like before, now all the columns that can be divided by 3 will store another new way. And the final result will be like:
 ```
 [1, 1, 2, 3, 4, 5, 7, 8]
 ```
diff --git a/en/Dynamic Programming/Longest Common Subsequence.md b/en/Dynamic Programming/Longest Common Subsequence.md
@@ -49,7 +49,7 @@ B 0 ? ? ? ?
 C 0 ? ? ? ?
 B 0 ? ? ? ?
 ```
-Now we will start to fill our table from 1st row. Since `S[1] = A` and `T[1] = B`, the `dp[1][1]` will be tha maximal value of `dp[0][1] = 0` and `dp[1][0] = 0`. So `dp[1][1] = 0`. But now `S[2] = B = T[1]`, so `dp[1][2] = dp[0][1] + 1 = 1`. `dp[1][3]` is `1` since `A != C` and we pick `max{dp[1][2], dp[0][3]}`. And `dp[1][4] = dp[0][3] + 1 = 1`.
+Now we will start to fill our table from 1st row. Since `S[1] = A` and `T[1] = B`, the `dp[1][1]` will be the maximal value of `dp[0][1] = 0` and `dp[1][0] = 0`. So `dp[1][1] = 0`. But now `S[2] = B = T[1]`, so `dp[1][2] = dp[0][1] + 1 = 1`. `dp[1][3]` is `1` since `A != C` and we pick `max{dp[1][2], dp[0][3]}`. And `dp[1][4] = dp[0][3] + 1 = 1`.
 ```
 # # A B C B
 # 0 0 0 0 0
diff --git a/en/Image Processing/Harris Detector.md b/en/Image Processing/Harris Detector.md
@@ -15,9 +15,8 @@ Given image $I$, $n\times n$ size Gaussian Kernel $G_{n\times n}$,
 
 ## Code Implementation Links
 
-- [Python](https://github.com/TheAlgorithms/Python/blob/master/digital_image_processing/feature_detectors/harris.py)
+- [Python](https://github.com/TheAlgorithms/Python/blob/master/computer_vision/harriscorner.py)
 
 ## Reference
 
-C. Harris and M. Stephens, “A Combined Corner and Edge Detector,” in *Procedings of the Alvey Vision Conference 1988*, Manchester, 1988, pp. 23.1-23.6.
-
+C. Harris and M. Stephens, “A Combined Corner and Edge Detector,” in *Proceedings of the Alvey Vision Conference 1988*, Manchester, 1988, pp. 23.1-23.6.
