Merge pull request #65 from junnengsoo/branch-binarySearch

docs: Branch binary search

Author: 4ndrelim

src/main/java/algorithms/binarySearch/README.md

@@ -1,137 +1,22 @@
 # Binary Search
 
+## Background
+
 Binary search is a search algorithm that finds the position of a target value within a sorted array or list. It compares
 the target value to the middle element of the search range, then, based on the comparison, narrows the search to the
 upper or lower half of the current search range.
 
-Two versions of binary search has been implemented in this repository - BinarySearch and BinarySearchTemplated.
-
-## BinarySearch
-
-![binary search img](../../../../../docs/assets/images/BinarySearch.png)
-Image Source: GeeksforGeeks
-
-BinarySearch is a more straightforward and intuitive version of the binary search algorithm. In this approach, after the
-mid-value is calculated, the high or low pointer is adjusted by just one unit. From the above example, after mid points
-to index 4 in the first search, the low pointer moves to index 5 (+1 from 4) when narrowing the search. Similarly, when
-mid points to index 7 in the second search, the high pointer shifts to index 6 (-1 from 7) when narrowing the search.
-This prevents any possibility of infinite loops. During the search, the moment mid-value is equal to the target value,
-the search ends prematurely. Note that there is no need for a "condition" method as the condition is already captured
-in the predicates of the if-else blocks.
-
-## BinarySearchTemplated
-
-BinarySearchTemplated removes the condition that checks if the current mid-value is equal to the target (which helps to
-end the search the moment the target is found). The template adds a "condition" method which will be modified based on
-the requirements of the implementation.
-
-The narrowing of the search space differs from BinarySearch - only the high pointer will be adjusted by one unit.
-
-This template will work for most binary search problems and will only require the following changes:
-
-- Search space (high and low)
-- Condition method
-- Returned value (low or low - 1)
-
-### Search Space (Requires change)
-
-Simply modify the initialisation of the high and low pointer according to the [search space](#search-space-adjustment).
-
-### Condition (Requires change)
-
-We assume that when the condition returns true, the current value "passes" and when the condition returns false, the
-current value "fails".
-
-Note that in this template, the conditional blocks
-
-```
-if (condition(x)) {
-   high = mid;
-} else {
-   low = mid + 1;
-}
-```
-
-requires elements that "fail" the condition to be on the left of the elements that "pass" the condition, see below, in a
-sorted array due to the way the high and low pointers are reassigned.
-
-![binary search templated 1 img](../../../../../docs/assets/images/BinarySearchTemplated1.jpeg)
-
-Hence, we will need to implement a condition method that is able to discern between arrays that "pass" and "fail"
-accurately and also place them in the correct relative positions i.e. "fail" on the left of "pass". Suppose we change
-the condition method implementation in BinarySearchTemplated from `value >= target` to `value <= target`, what will
-happen?
-<details>
-<summary> <b>what will happen?</b> </summary>
-The array becomes "P P F F F F" and the low and high pointers are now reassigned wrongly.
-</details>
-
-### Returned Value (Requires change)
-
-In this implementation of BinarySearchTemplated, we return the first "pass" in the array with `return low`. This is
-because our condition method implementation encompasses the target value that we are finding i.e. when
-`value == target`.
-
-```java
-public static boolean condition(int value, int target) {
-    return value >= target;
-}
-```
-
-![binary search templated 1 img](../../../../../docs/assets/images/BinarySearchTemplated2.jpeg)
-
-However, if we want to return the last "fail" in the array, we will `return low - 1`.
-
-Suppose now we modify the condition to be `value > target`, how can we modify our BinarySearchTemplated to still work as
-expected?
-<details>
-<summary> <b>value > target?</b> </summary>
-Replace `return low` with `return low - 1` and replace arr[low] with arr[low - 1] as now the target value is the last 
-"fail".
-</details>
-
-### Search Space Adjustment
-
-What should be the search space adjustment? Why is only low reassigned with an increment and not high?
-
-Due to the nature of floor division in Java's \ operator, if there are two mid-values within the search range, which is
-when the number of elements is even, the first mid-value will be selected. Suppose we do not increment the low pointer
-during reassignment, `low = mid`, let us take a look at the following example:
-
-![binary search templated 1 img](../../../../../docs/assets/images/BinarySearchTemplated3.jpeg)
-
-The search space has been narrowed down to the range of index 1 (low) to 2 (high). The mid-value is calculated,
-`mid = (1 + 2) / 2`, to be 1 due to floor division. Since `2 < 5`, we enter the else block where there is reassignment
-of `low = mid`. This means that the low pointer is still pointing to index 1 and the high pointer remains unchanged at
-index 2. This results in an infinite loop as the search range is not narrowed down.
-
-To resolve this issue, we need `low = mid + 1`, which will result in the low pointer pointing to index 2 in this
-scenario. We still ensure correctness because the mid-value is not the target value, as the mid-value < target, and we
-can safely exclude it from the search range.
-
-Why do we not need to increment the high pointer during reassignment? This is because the mid-value could be the target
-as the condition implemented is `value >= target`, hence, we cannot exclude it from the search range.
-
-See [here](./binarySearchTemplatedExamples/README.md) to use the template for other problems
-
-Credits: [Powerful Ultimate Binary Search Template](https://leetcode.com/discuss/general-discussion/786126/python-powerful-ultimate-binary-search-template-solved-many-problems)
-
-## Complexity Analysis
+## Implementation Invariant
 
-**Time**:
+At the end of each iteration, the target value is either within the search range or does not exist in the search space.
 
-- Worst case: O(log n)
-- Average case: O(log n)
-- Best case:
-    - BinarySearch O(1)
-    - BinarySearchTemplated O(log n)
+## BinarySearch and BinarySearchTemplate
 
-BinarySearch:
-In the worst case, the target is either in the first index or does not exist in the array at all.
-In the best case, the target is the middle (odd number of element) or the first middle element (even number of elements)
-if floor division is used to determine the middle.
+We will discuss more implementation-specific details and complexity analysis in the respective folders. In short,
+1. The [binarySearch](binarySearch) method is a more straightforward and intuitive version of the binary search 
+algorithm.
+2. The [binarySearchTemplate](binarySearchTemplated) method provides a more generalised template that can be used for 
+most binary search problems by introducing a condition method that can be modified based on the requirements of the 
+implementation.
 
-BinaryTemplated:
-In all cases, O(log n) iterations will be required as there is no condition to exit the loop prematurely.
 
-**Space**: O(1) since no new data structures are used and searching is only done within the array given

…lgorithms/binarySearch/BinarySearch.java …arySearch/binarySearch/BinarySearch.javasrc/main/java/algorithms/binarySearch/BinarySearch.java renamed to src/main/java/algorithms/binarySearch/binarySearch/BinarySearch.java src/main/java/algorithms/binarySearch/BinarySearch.java renamed to src/main/java/algorithms/binarySearch/binarySearch/BinarySearch.java

@@ -1,4 +1,4 @@
-package algorithms.binarySearch;
+package algorithms.binarySearch.binarySearch;
 
 /**
  * Here, we are implementing BinarySearch where we search an array for a target value at O(log n) time complexity.
@@ -8,6 +8,8 @@
  * All elements in the array are unique. (to allow for easy testing)
  * <p>
  * Brief Description and Implementation Invariant:
+ *
+ * Brief Description:
  * With the assumption that the array is sorted in ascending order, BinarySearch reduces the search range by half or
  * half + 1 (due to floor division) after every loop. This is done by reassigning the max (high) or min (low) of the
  * search range to the middle of the search range when the target value is smaller than or larger than the current

src/main/java/algorithms/binarySearch/binarySearch/README.md

@@ -0,0 +1,31 @@
+# BinarySearch
+
+## Background
+
+BinarySearch is a more straightforward and intuitive version of the binary search algorithm. In this approach, after the
+mid-value is calculated, the high or low pointer is adjusted by just one unit.
+
+### Illustration
+
+![binary search img](../../../../../../docs/assets/images/BinarySearch.png)
+
+Image Source: GeeksforGeeks
+
+From the above example, after mid points to index 4 in the first search, the low pointer moves to index 5 (+1 from 4) 
+when narrowing the search. Similarly, when mid points to index 7 in the second search, the high pointer shifts to index 
+6 (-1 from 7) when narrowing the search. This prevents any possibility of infinite loops. During the search, the moment 
+mid-value is equal to the target value, the search ends prematurely.
+
+## Complexity Analysis
+**Time**:
+- Worst case: O(log n)
+- Average case: O(log n)
+- Best case: O(1)
+
+In the worst case, the target is either in the first or last index or does not exist in the array at all.
+In the best case, the target is the middle (odd number of element) or the first middle element (even number of elements)
+if floor division is used to determine the middle.
+
+**Space**: O(1) 
+
+Since no new data structures are used and searching is only done within the array given.

…/binarySearch/BinarySearchTemplated.java …archTemplated/BinarySearchTemplated.javasrc/main/java/algorithms/binarySearch/BinarySearchTemplated.java renamed to src/main/java/algorithms/binarySearch/binarySearchTemplated/BinarySearchTemplated.java src/main/java/algorithms/binarySearch/BinarySearchTemplated.java renamed to src/main/java/algorithms/binarySearch/binarySearchTemplated/BinarySearchTemplated.java

@@ -1,4 +1,4 @@
-package algorithms.binarySearch;
+package algorithms.binarySearch.binarySearchTemplated;
 
 /**
  * Here, we are implementing BinarySearchTemplated where we search an array for a target value at O(log n) time

src/main/java/algorithms/binarySearch/binarySearchTemplated/README.md

@@ -0,0 +1,118 @@
+# BinarySearchTemplated
+
+## Background
+
+BinarySearchTemplated is a more generalised algorithm of [BinarySearch](../binarySearch) that removes the condition that
+checks if the current mid-value is equal to the target (which helps to end the search the moment the target is found). 
+The template adds a "condition" method which will be modified based on the requirements of the implementation.
+
+The narrowing of the search space differs from BinarySearch - only the high pointer will be adjusted by one unit.
+
+This template will work for most binary search problems and will require the following changes when used for different 
+problems:
+- Search space (high and low pointers)
+- Condition method
+- Returned value (low or low - 1)
+
+### Search Space (Requires change)
+Initialise the boundary values of the high and low pointers to include all possible elements in the search space.
+
+### Condition (Requires change)
+We assume that when the condition returns true, the current value "passes" and when the condition returns false, the
+current value "fails".
+
+Note that in this template, the conditional blocks
+```
+if (condition(x)) {
+   high = mid;
+} else {
+   low = mid + 1;
+}
+```
+requires elements that "fail" the condition to be on the left of the elements that "pass" the condition, see below, in a
+sorted array due to the way the high and low pointers are reassigned.
+
+![binary search templated 1 img](../../../../../../docs/assets/images/BinarySearchTemplated1.jpeg)
+
+Hence, we will need to implement a condition method that is able to discern between arrays that "pass" and "fail"
+accurately and also place them in the correct relative positions i.e. "fail" on the left of "pass". Suppose we change
+the condition method implementation by inverting the inputs that "pass" the condition, i.e, inputs that "fail" now
+"pass" and vice-versa, what will happen?
+<details>
+<summary> <b>what will happen?</b> </summary>
+The array becomes "P P F F F F" and our current template will not work as expected. This is because the low and high
+pointers are now reassigned wrongly - the loop invariant is broken as the search space is narrowed down in the wrong
+direction. 
+
+To resolve this issue, there are two fixes:
+1. Swap the conditional blocks of low = mid + 1 and high = mid.
+**OR**
+2. Simply add a "not" in front of the condition, converting "P P F F F F" back to "F F P P P P".
+
+Note that some conditions may be easier to define with "pass" elements being on the left of "fail" elements, i.e.
+"P P F F F F", hence, it is important to adjust the code with the above fixes accordingly.
+</details>
+
+### Returned Value (Requires change)
+In this implementation of BinarySearchTemplated, we return the first "pass" in the array with `return low`. This is
+because our condition method implementation encompasses the target value that we are finding i.e. when
+`value == target`.
+
+```java
+public static boolean condition(int value, int target) {
+    return value >= target;
+}
+```
+![binary search templated 2 img](../../../../../../docs/assets/images/BinarySearchTemplated2.jpeg)
+
+However, if we want to return the last "fail" in the array, we will `return low - 1`.
+
+Suppose now we modify the condition to be `value > target`, how can we modify our BinarySearchTemplated to still work as
+expected?
+<details>
+<summary> <b>value > target?</b> </summary>
+Replace `return low` with `return low - 1` and replace arr[low] with arr[low - 1] as now the target value is the last 
+"fail".
+</details>
+
+
+### Search Space Adjustment
+What should be the search space adjustment? Why is only low reassigned with an increment and not high?
+
+Due to the nature of floor division in Java's \ operator, if there are two mid-values within the search range, which is
+when the number of elements is even, the first mid-value will be selected. Suppose we do not increment the low pointer
+during reassignment, `low = mid`, let us take a look at the following example:
+
+![binary search templated 1 img](../../../../../docs/assets/images/BinarySearchTemplated3.jpeg)
+
+The search space has been narrowed down to the range of index 1 (low) to 2 (high). The mid-value is calculated,
+`mid = (1 + 2) / 2`, to be 1 due to floor division. Since `2 < 5`, we enter the else block where there is reassignment
+of `low = mid`. This means that the low pointer is still pointing to index 1 and the high pointer remains unchanged at
+index 2. This results in an infinite loop as the search range is not narrowed down.
+
+To resolve this issue, we need `low = mid + 1`, which will result in the low pointer pointing to index 2 in this
+scenario. We still ensure correctness because the mid-value is not the target value, as the mid-value < target, and we
+can safely exclude it from the search range.
+
+Why should we not decrement the high pointer during reassignment? This is because the mid-value could be the target
+as the condition implemented is `value >= target`, hence, we cannot exclude it from the search range.
+
+Note that if ceiling division is used instead, `high = mid - 1` will be required to prevent an infinite loop. Both
+`low = mid + 1` and `low = mid` will work in this case.
+
+See [here](binarySearchTemplatedExamples/README.md) to use the template for other problems.
+
+Credits: [Powerful Ultimate Binary Search Template](https://leetcode.com/discuss/general-discussion/786126/python-powerful-ultimate-binary-search-template-solved-many-problems)
+
+
+## Complexity Analysis
+**Time**:
+- Worst case: O(log n)
+- Average case: O(log n)
+- Best case: O(log n)
+
+In all cases, O(log n) iterations will be required as there is no condition to exit the loop prematurely.
+
+**Space**: O(1)
+
+Since no new data structures are used and searching is only done within the array given.

…/binarySearchTemplatedExamples/README.md …/binarySearchTemplatedExamples/README.mdsrc/main/java/algorithms/binarySearch/binarySearchTemplatedExamples/README.md renamed to src/main/java/algorithms/binarySearch/binarySearchTemplated/binarySearchTemplatedExamples/README.md src/main/java/algorithms/binarySearch/binarySearchTemplatedExamples/README.md renamed to src/main/java/algorithms/binarySearch/binarySearchTemplated/binarySearchTemplatedExamples/README.md

@@ -72,7 +72,7 @@ right of those that fail! Since "all the versions after a bad version are also b
 
 Since we want to return the first bad version, we will return low.
 
-![first bad version img](../../../../../../docs/assets/images/firstBadVersion.jpeg)
+![first bad version img](../../../../../../../docs/assets/images/firstBadVersion.jpeg)
 
 <b> Full Solution: </b>
 

src/test/java/algorithms/binarySearch/BinarySearchTest.java

@@ -1,12 +1,11 @@
 package algorithms.binarySearch;
 
-import static org.junit.Assert.assertEquals;
-
+import algorithms.binarySearch.binarySearch.BinarySearch;
+import algorithms.binarySearch.binarySearchTemplated.BinarySearchTemplated;
 import org.junit.Test;
 
-/**
- * Test cases for {@link BinarySearch}.
- */
+import static org.junit.Assert.assertEquals;
+
 public class BinarySearchTest {
     @Test
     public void test_binarySearch() {
@@ -37,12 +36,22 @@ public void test_binarySearchTemplated() {
         int[] secondArray = {1, 5, 10, 11, 12};
         int secondResult = BinarySearchTemplated.search(secondArray, 11);
 
-        // Test 3: target not in array
+        // Test 3: target not in array but could exist within search space
         int[] thirdArray = {1, 5, 10, 11, 12};
         int thirdResult = BinarySearchTemplated.search(thirdArray, 3);
 
+        // Test 4: target not in array but could exist on the right of search space
+        int[] fourthArray = {1, 5, 10, 11, 12};
+        int fourthResult = BinarySearchTemplated.search(thirdArray, 13);
+
+        // Test 3: target not in array but could exist on the left of search space
+        int[] fifthArray = {1, 5, 10, 11, 12};
+        int fifthResult = BinarySearchTemplated.search(thirdArray, 0);
+
         assertEquals(0, firstResult);
         assertEquals(3, secondResult);
         assertEquals(-1, thirdResult);
+        assertEquals(-1, fourthResult);
+        assertEquals(-1, fifthResult);
     }
 }