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authorAndrew Schneider <andrewschneider@Jessicas-Air.fios-router.home>2024-05-31 20:23:35 -0400
committerAndrew Schneider <andrewschneider@Jessicas-Air.fios-router.home>2024-05-31 20:23:35 -0400
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fix README formatting
-rw-r--r--challenge-271/atschneid/README.md96
1 files changed, 48 insertions, 48 deletions
diff --git a/challenge-271/atschneid/README.md b/challenge-271/atschneid/README.md
index cdefaf7b38..a590aed159 100644
--- a/challenge-271/atschneid/README.md
+++ b/challenge-271/atschneid/README.md
@@ -14,32 +14,32 @@ Now onto the code.
## Task 1: Maximum Ones
-> You are given a m x n binary matrix.<\br>
-<\br>
-> Write a script to return the row number containing maximum ones, in case of more than one rows then return smallest row number.<\br>
-<\br>
-> Example 1<\br>
-> Input: $matrix = [ [0, 1],<\br>
-> [1, 0],<\br>
-> ]<\br>
-> Output: 1<\br>
-<\br>
-> Row 1 and Row 2 have the same number of ones, so return row 1.<\br>
-> Example 2<\br>
-> Input: $matrix = [ [0, 0, 0],<\br>
-> [1, 0, 1],<\br>
-> ]<\br>
-> Output: 2<\br>
-<\br>
-> Row 2 has the maximum ones, so return row 2.<\br>
-> Example 3<\br>
-> Input: $matrix = [ [0, 0],<\br>
-> [1, 1],<\br>
-> [0, 0],<\br>
-> ]<\br>
-> Output: 2<\br>
-<\br>
-> Row 2 have the maximum ones, so return row 2.<\br>
+> You are given a m x n binary matrix.<br/>
+<br/>
+> Write a script to return the row number containing maximum ones, in case of more than one rows then return smallest row number.<br/>
+<br/>
+> Example 1<br/>
+> Input: $matrix = [ [0, 1],<br/>
+> [1, 0],<br/>
+> ]<br/>
+> Output: 1<br/>
+<br/>
+> Row 1 and Row 2 have the same number of ones, so return row 1.<br/>
+> Example 2<br/>
+> Input: $matrix = [ [0, 0, 0],<br/>
+> [1, 0, 1],<br/>
+> ]<br/>
+> Output: 2<br/>
+<br/>
+> Row 2 has the maximum ones, so return row 2.<br/>
+> Example 3<br/>
+> Input: $matrix = [ [0, 0],<br/>
+> [1, 1],<br/>
+> [0, 0],<br/>
+> ]<br/>
+> Output: 2<br/>
+<br/>
+> Row 2 have the maximum ones, so return row 2.<br/>
One thing that surprised me here, we want the 1 indexed row. For example, I would have expected the first solution to be 0, the zeroth row. But, I'll solve the problem I'm given.
@@ -64,28 +64,28 @@ And that's about it. I return that `$idx` value, then increment it (0 to 1 index
## Task 2: Sort by 1 bits
-> You are give an array of integers, @ints.A<\br>
-<\br>
-> Write a script to sort the integers in ascending order by the number of 1 bits in their binary representation. In case more than one integers have the same number of 1 bits then sort them in ascending order.<\br>
-<\br>
-> Example 1<\br>
-> Input: @ints = (0, 1, 2, 3, 4, 5, 6, 7, 8)<\br>
-> Output: (0, 1, 2, 4, 8, 3, 5, 6, 7)<\br>
-<\br>
-> 0 = 0 one bits<\br>
-> 1 = 1 one bits<\br>
-> 2 = 1 one bits<\br>
-> 4 = 1 one bits<\br>
-> 8 = 1 one bits<\br>
-> 3 = 2 one bits<\br>
-> 5 = 2 one bits<\br>
-> 6 = 2 one bits<\br>
-> 7 = 3 one bits<\br>
-> Example 2<\br>
-> Input: @ints = (1024, 512, 256, 128, 64)<\br>
-> Output: (64, 128, 256, 512, 1024)<\br>
-<\br>
-> All integers in the given array have one 1-bits, so just sort them in ascending order.<\br>
+> You are give an array of integers, @ints.A<br/>
+<br/>
+> Write a script to sort the integers in ascending order by the number of 1 bits in their binary representation. In case more than one integers have the same number of 1 bits then sort them in ascending order.<br/>
+<br/>
+> Example 1<br/>
+> Input: @ints = (0, 1, 2, 3, 4, 5, 6, 7, 8)<br/>
+> Output: (0, 1, 2, 4, 8, 3, 5, 6, 7)<br/>
+<br/>
+> 0 = 0 one bits<br/>
+> 1 = 1 one bits<br/>
+> 2 = 1 one bits<br/>
+> 4 = 1 one bits<br/>
+> 8 = 1 one bits<br/>
+> 3 = 2 one bits<br/>
+> 5 = 2 one bits<br/>
+> 6 = 2 one bits<br/>
+> 7 = 3 one bits<br/>
+> Example 2<br/>
+> Input: @ints = (1024, 512, 256, 128, 64)<br/>
+> Output: (64, 128, 256, 512, 1024)<br/>
+<br/>
+> All integers in the given array have one 1-bits, so just sort them in ascending order.<br/>
At first I thought I was going to end up reusing some pieces of my solution to Task 1 for this, but it turned out to be just different enough that I didn't think it was worth it.