Fix up leetcode

leetcode/solutions.org

@@ -3,15 +3,11 @@
 #+description: Description
 #+date: 2024-04-29
 
-* Prelude
-#+begin_src c++ :session c++
-test
-#+end_src
 * Arrays
 ** DONE (#217) Contains duplicate
 If there are any duplicates, then when the list is sorted they would
 be consecutive.
-#+begin_src c++ :tangle 217.cpp :session c++
+#+begin_src c++ :tangle 217.cpp :session c++ :comments link
 #include <bits/stdc++.h>
 
 using namespace std;
@@ -34,7 +30,7 @@ disregarding ordering.
 Sorting the two strings would force the same
 order for both strings.
 
-#+begin_src c++ :tangle 242-sort.cpp
+#+begin_src c++ :tangle 242-sort.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 bool isAnagram(string s, string t)
@@ -45,7 +41,7 @@ bool isAnagram(string s, string t)
 }
 #+end_src
 *** Simple table
-#+begin_src c++ :tangle 242-table.cpp
+#+begin_src c++ :tangle 242-table.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 bool isAnagram(string s, string t)
@@ -64,7 +60,7 @@ bool isAnagram(string s, string t)
 #+end_src
 ** DONE (#1) Two sum
 Given a target find two elements which sum to it.
-#+begin_src c++ :tangle 1.cpp
+#+begin_src c++ :tangle 1.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 vector<int> twoSum(vector<int> &nums, int target)
@@ -81,7 +77,7 @@ vector<int> twoSum(vector<int> &nums, int target)
 }
 #+end_src
 ** DONE (#49) Group anagrams
-#+begin_src c++ :tangle 49.cpp
+#+begin_src c++ :tangle 49.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 vector<vector<string>> groupAnagrams(vector<string> &strs)
@@ -106,7 +102,7 @@ hash tables.
 *** Sorting the map
 Once you've generated the map, sort it by frequency (then by number)
 then take the top k off it.
-#+begin_src c++ :tangle 347-sort.cpp
+#+begin_src c++ :tangle 347-sort.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 vector<int> topKFrequent(vector<int> &nums, int k)
@@ -138,7 +134,7 @@ value.  The priority sorts the members.
 Hence, if we generate the map then place them all in a priority queue,
 we just need to get the k highest priority items to finish the
 problem.
-#+begin_src c++ :tangle 347-queue.cpp
+#+begin_src c++ :tangle 347-queue.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 
@@ -174,7 +170,7 @@ i: [x from 0 to i - 1] * i is the product [x from 0 to i].  Same the
 other way.  So we can iterate once to produce the products before i
 and once for the products after i while only doing a set once!
 
-#+begin_src c++ :tangle 238.cpp
+#+begin_src c++ :tangle 238.cpp :comments link
 #include <bits/stdc++.h>
 
 using namespace std;
@@ -201,11 +197,9 @@ Determine if a 9x9 grid is a valid sudoku.
 You can use a set of hashsets (where membership testing is O(1)) to
 just check if you've seen a number before.  In this case, since we
 know the exact input data and the properties we want we can use
-boolean tables to speed up the process.
+boolean arrays which are smaller and easier for the CPU to work with.
 
-I also use a little maths technique to produce a correct ordering of
-the boxes via integer division.
-#+begin_src c++ :tangle 36.cpp
+#+begin_src c++ :tangle 36.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 
@@ -240,7 +234,7 @@ bool isValidSudoku(vector<vector<char>> &board)
 #+end_src
 ** DONE (#128) Longest consecutive sequence
 
-#+begin_src c++ :tangle 128.cpp
+#+begin_src c++ :tangle 128.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 
@@ -260,9 +254,74 @@ int longestConsecutive(vector<int> &nums)
   return size;
 }
 #+end_src
+** DONE (#53) Maximum Subarray
+Find the maximal sum of a subarray in some array.  This is actually a
+dynamic programming problem.
+*** Brute force
+If there's only one element then that is obviously the maximal sum.
+A brute force approach would compute it every subarray sum and then
+return the best one.
+
+#+begin_src c++ :tangle 53-bf.cpp :comments link
+#include <bits/stdc++.h>
+using namespace std;
+
+int maxSubArray(vector<int> &nums)
+{
+  int max_sum = nums[0];
+  for (int i = 1; i < nums.size(); ++i)
+  {
+    // Is nums[i] alone better than what we've made so far?
+    max_sum = std::max(nums[i], max_sum);
+    int cur_sum = 0;
+    for (int j = 0; j <= i; ++j)
+      cur_sum += nums[i];
+    // Is the sum nums[0..i] better than what we've done so far?
+    max_sum = std::max(std::max(cur_sum, max_sum));
+  }
+  return max_sum;
+}
+#+end_src
+
+O(n^2) but can we do better?
+*** Dynamic programming
+Look at the tree of cur_sum's we need to
+compute:
++ at i = 1, cur_sum is sum[0..1]
++ at i = 2, cur_sum is sum[0..2]
++ at i = k, cur_sum is sum[0..k] = sum[0..k-1] + k
+
+At each iteration we're essentially checking if nums[i] is better or
+the sum is better than what we've seen before.
+
+Obviously the sum is stupidly expensive.  Furthermore, we may not even
+need to do the entire sum; if the best sum starts at k and ends at k'
+there's literally no point in summing numbers before k in cur_sum.  So
+we may as well introduce that optimisation.
+#+begin_src c++ :tangle 53-dp.cpp :comments link
+#+end_src
+*** Kadane's formulation
+#+begin_src c++ :tangle 53-kadane.cpp :comments link
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int maxSubArray(vector<int> &nums)
+{
+  if (nums.size() == 1)
+    return nums[0];
+  int max_sum = nums[0], cur_sum = nums[0];
+  for (size_t i = 1; i < nums.size(); ++i)
+  {
+    cur_sum = std::max(cur_sum + nums[i], nums[i]);
+    max_sum = std::max(cur_sum, max_sum);
+  }
+  return max_sum;
+}
+#+end_src
 * Stack
 ** DONE (#20) Valid Parentheses
-#+begin_src c++ :tangle 20.cpp
+#+begin_src c++ :tangle 20.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 
@@ -306,7 +365,7 @@ functions in O(1) (what?!).
 In exchange for O(n) memory we can implement most of them.
 Unfortunately all the high performance implementations also use stacks
 (what's the point!).
-#+begin_src c++ :tangle 155.cpp
+#+begin_src c++ :tangle 155.cpp :comments link
 #include <bits/stdc++.h>
 
 using namespace std;
@@ -360,7 +419,7 @@ Encountered an issue with C/C++'s semantics when it comes prefix
 decrement: it doesn't have an effect until after the statement so the
 other side of the assignment still considers ~ptr~ to be the same
 value.  Bit interesting, good to note.
-#+begin_src c++ :tangle 150.cpp
+#+begin_src c++ :tangle 150.cpp :comments link
 #include <bits/stdc++.h>
 using namespace std;
 
@@ -398,7 +457,7 @@ parentheses.
 *** Naive solution
 I generate all 2^n bit strings (where n = 2 * number of pairs) then
 filer all the invalid bit strings.
-#+begin_src c++ :tangle 22-naive.cpp
+#+begin_src c++ :tangle 22-naive.cpp :comments link
 #include <bits/stdc++.h>
 
 using namespace std;
@@ -474,7 +533,7 @@ string (add an open brace or add a closed brace) but only generate the
 second one based on a condition on the first.  As we do these
 simultaneously we can generate the two possible next steps given an
 input string.
-#+begin_src c++ :tangle 22-backtrack.cpp
+#+begin_src c++ :tangle 22-backtrack.cpp :comments link
 #include <bits/stdc++.h>
 
 using namespace std;
@@ -531,3 +590,4 @@ vector<int> dailyTemperatures(vector<int> &temperatures)
   return days;
 }
 #+end_src
+* Blind 75