From e039cc3f02393d962dc0d53666c75061b43c71d7 Mon Sep 17 00:00:00 2001
From: anebz <ane.berasategi@gmail.com>
Date: Sat, 23 Mar 2019 00:56:32 +0100
Subject: [PATCH] big O initial commit

* time complexity
* space complexity
* constants, non-dominant terms
* multi-part algorithms
* amortized time
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+# Chapter 1
+
+## Big O
+
+### 1.1. Time complexity
+
+* Big O: upper bound on the time. if O(N), then O(N^2), O(N^3), etc
+* Big Omega: lower bound. if omega(N), then omega(1), omega(logN)
+* Big theta: an algorithm is big theta(N) if it's both O(N) and omega(N). theta gives a tight bound on runtime
+
+Best case scenarios:
+
+* Best case: with any algorithm, in a special case, we can make O(1)
+* Worst case scenario: quick sort of a reverse ordered array, it might take O(N^2)
+* Expected case: O(NlogN)
+
+For most algorithms, the worst case and the expected case are the same.
+
+### 1.2. Space complexity
+
+Space complexity is a parallel concept to time complexity. Stack space in recursive calls counts, too.
+
+```c
+int sum(int n) { /* ex1 */
+    if (n <= 0) {
+        return 0;
+    }
+    return n + sum(n - 1);
+}
+```
+
+This code takes O(n) time and O(n) space. in sum(4), there are 4 calls to the stack (sum(3), sum(2), sum(1), sum(0)).
+
+```c
+int pairSumSeq(int n) { /* ex2 */
+    int sum = 0;
+    for(int i=0; i < n; i++) {
+        sum += pairSum(i, i+1);
+    }
+}
+int pairSum(int a, int b) {
+    return a + b;
+}
+```
+
+`ex2` takes O(n) time complexity but O(1) space. There are O(n) calls to `pairSum`, but those calls don't happen simultaneously like in the recursive case of `ex1`.
+
+### 1.3. Drop the constants
+
+It's possible for O(N) code to run faster than O(1) code for specific inputs. Big O just describes the rate of increase, so we drop constants in runtime. O(2N) = O(N).
+
+### 1.4. Drop the non-dominant terms
+
+O(N^2 + N) = O(N^2)
+
+### 1.5. Multi-part algorithms
+
+In an algorithm with two steps, when to multiply runtimes and when to add?
+
+* Add runtime: do A chinks of work, then B chunks of work
+
+```c
+for(int a: arrA) {
+    print(a);
+}
+for(int b: arrB) {
+    print(b);
+}
+```
+
+* Multiply runtime: do B chunks of work for each element in A.
+
+```c
+for(int a: arrA) {
+    for(int b: arrB) {
+        print(a + ", " + b);
+    }
+}
+```
+
+### 1.6. Amortized time
+
+With `ArrayList`s, when they reach full capacity, the class creates a new array with double the capacity and copy all the elements over to the new array. Say we want to add a new element to the `ArrayList`.
+
+* If the array is full and contains N elements, adding a new element takes O(N) time because we have to create a new array of size 2N and then copy N elements over, O(2N + N) = O(3N) = O(N).
+* But most of the times the `ArrayList` won't be full, and adding a new element will take O(1).
+
+In the worst case it takes O(N), but in N-1 cases it takes O(1). Once the worst case happens, it won't happen again for so long that the cost is "amortized". If we add X elements to the `ArrayList`, it takes ~2X --> X adds take O(X) time, the amortized time for each adding is O(1). 
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