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+# Big O Cheatsheet
+
+This is just a cheat sheet containing some observations about common for loop
+patterns taken from Abdul Bari's video series on algorithms.
+
+## For Loops
+
+---
+$$ O(n) $$
+
+```c
+for (i = 0; i < n; i++) {} // O(n)
+for (i = 0; i < n; i = i + 2) {} // = n/2 = O(n)
+for (i = n; i > 1; i--) {} // O(n)
+```
+---
+
+$$ O(n^2) $$
+
+```c
+for (i = 0; i < n; i++) {
+    for (j = 0; j < n; j++) {} // O(n^2)
+}
+```
+
+---
+$$ O(log_2(n)log_2(p)) $$
+
+```c
+for (i = 1; i < n; i = i * 2) {
+    p++; // log_2(n)
+}
+for (j = 1; j < p; j = j * 2) {
+    // log_2(p)
+}
+// log_2(n) + log_2(p) = O(log_2(n)log_2(p))
+```
+---
+
+$$ O(log_2(n)) $$
+
+```c
+for (i = 1; i < n; i = i * 2) {} // O(log_2(n))
+for (i = n; i > 1; i = i / 2) {} // O(log_2(n))
+```
+
+---
+
+$$ O(log_3(n)) $$
+
+```c
+for (i = 1; i < n; i = i * 3) {} // O(log_3(n)
+```
+
+## While Loops
+
+$$ O(n) $$
+
+```c
+i = 0;
+while (i < n) {
+    i++;
+} // O(n)
+```
+
+---
+
+$$ O(log_2(n)) $$
+
+```c
+a = 1;
+while (a < b) {
+    a = a * 2;
+} // O(log_2(n))
+
+/* NOTE:
+    a >= b
+    a = 2^k // where k represents unknown value of b
+    2^k >= b // condition where loop stops
+    2^k = b // by making condition where loop stops true...
+    k = log_2(b) we can determine that unknown value b stops when k is equal to log_2 of b
+    which evaluates to: O(log_2(n))
+*/
+```
+
+```c
+i = n;
+while (i > 1) {
+    i = i / 2;
+}
+```
+
+---
+
+$$ O(\sqrt n) $$
+
+```c
+i = 1;
+k = 1;
+while (k < n) {
+    k = k + i;
+    i++;
+}
+```
+
+Analyzing the above with some dummy data:
+
+| i  | k             |
+| :- | :------------ |
+| 1  | 1             |
+| 2  | 1 + 1         |
+| 3  | 2 + 2         |
+| 4  | 2 + 2 + 3     |
+| 5  | 2 + 2 + 3 + 4 |
+| m  | ...m          |
+
+Which in mathematical notation would be:
+
+$$ \frac{m(m + 1)}{2} $$
+
+We can further analyze this like so:
+
+$$ k \geq n $$
+
+$$ \frac{m(m + 1)}{2} \geq n $$
+
+$$ m^2 \geq n $$
+
+$$ m = \sqrt n $$
+
+Thusly the big O notation of this is:
+
+$$ O(\sqrt n) $$
+
+Note that this is <em>not</em> the same as log_2(n). For an explanation of this,
+please see
+[this stack overflow answer](https://stackoverflow.com/questions/42038294/is-complexity-ologn-equivalent-to-osqrtn).
+According to other sources (GPT), the time complexity of $$ O(\sqrt n) $$
+
+is often unavoidable when a matrix of values is unsorted.
+
+## While loops with If Statements
+
+$$ O(n) $$
+
+The following demonstrates evaluating the GCD(Greatest Common Divisor) between
+two numbers:
+
+```c
+while (m != n)
+{
+    if (m > n)
+        m = m - n;
+    else
+        n = n - m;
+}
+```
+
+Let's take a look at an example where both `m` and `n` start off as equal:
+
+| m | n |
+| - | - |
+| 5 | 5 |
+
+Since they are equal the condition:
+
+```c
+while (m != n)
+```
+
+Will never execute, thus the amount of execution steps is 0. Let's look at
+another example:
+
+| m  | n |
+| -- | - |
+| 14 | 2 |
+
+In this case, the condition:
+
+```c
+if (m > n)
+    m = m - n;
+```
+
+Evaluates to `true` and the value of `m` is equal to `m - n`, replaced with `14
+
+- 2`, and thusly`m`is equal to`12`. This continually evaluates down the line
+  like so:
+
+| m  | n |
+| -- | - |
+| 14 | 2 |
+| 12 | 2 |
+| 10 | 2 |
+| 8  | 2 |
+| 6  | 2 |
+| 4  | 2 |
+| 2  | 2 |
+
+At which point the loop stops because the condition:
+
+```c
+while (m != n)
+```
+
+Evaluates to `false`, and the loop ends, resulting in the loop having executed 7
+times. We add 1 to the amount of times it is executed by nature of the initial
+loop, and it is 8. This evaluates to
+
+$$ \frac{16}{2} $$
+
+or:
+
+$$ \frac{n}{2} $$
+
+Which, in big O notation, evaluates to:
+
+$$ O(n) $$
+
+The <em>minimum</em> time to execute is at least one time, thusly the minimum
+time complexity is:
+
+$$ O(1) $$
+
+## Best Case vs. Worst Case
+
+```c
+void some_algo_test(int n)
+{
+    if (n < 5)
+        printf("%d\n", n); // 1
+    else
+        for (int i = 0; i < n; i++)
+            printf("%d\n", i); // n
+}
+```
+
+This evaluates in best case to:
+
+$$ O(1) $$
+
+This happens when the condition `n < 5` evaluates to `true`, because all it does
+is print the value of `n` to the console.
+
+Otherwise, if `n < 5` evaluates to false (i.e. n is a value greater than or
+equal to 5), then the worst case time complexity occurs, which is:
+
+$$ O(n) $$
+
+In essence:
+
+**BEST CASE**:
+
+$$ O(1) $$
+
+**WORST CASE**:
+
+$$ O(n) $$
+
+---
+
+Now let's adjust our function a bit:
+
+```c
+void some_algo_test(int n)
+{
+    if (n < 5)
+        printf("%d\n", n); // 1
+    for (int i = 0; i < n; i++)
+        printf("%d\n", i); // n
+}
+```
+
+Note here, that now, the `for` loop executes regardless of the `if (n < 5)`
+condition. Thusly the best and worst case scenario become the same:
+
+$$ O(n) $$