## find all pairs in an array, that sum up to particular number with best complexity

Here we discuss two possible algorithms

• algorithm with the time complexity O(n) and with no additional space complexity
• algorithm which uses additional hashmap data structure which reduces the time complexity to O(2n) at the cost of additional space complexity O(n)
• brute force algorithm with time complexity of O(n^2)

Algorithm with the time complexity O(n) and with no additional space complexity

```        /**
* Find the pair in an array whose sum with complexity O(n). This assumes
* the array passed is sorted array and there are no duplicates in the array
*
* @param arr
*            input array of elements
* @param k
*            sum for which pair of array elements needs to be searched
*/
public void getPair2(int[] arr, int k) {
int start = 0;
int end = arr.length - 1;
int sum = 0;

// output array to record matching pairs
StringBuilder arrs = new StringBuilder();

while (start < end) {
sum = arr[start] + arr[end];
if (sum == k) {
// we have found one pair, add it to our output array
arrs.append(arr[start] + "," + arr[end] + ";");
start++;
end--;
} else if (sum < k) {
// if the sum of the pair is less than the sum we are searching
// then increment the start pointer which would give us the sum
// more than our current sum as the array is sorted
start++;
} else {
// if the sum of the pair is greater than the sum we are searching
// then decrement the end pointer which would give us the sum
// less than our current sum as the array is sorted
end--;
}
}
System.out.println(arrs.toString());
}
```

Output: 1,5;2,4;

Algorithm which uses additional hashmap data structure which reduces the time complexity to O(2n) at the cost of additional space complexity O(n)

```/**
* Find the pair in an array whose sum with complexity O(2n)
*
* @param arr
*            input array of elements
* @param k
*            sum for which pair of array elements needs to be searched
* @return string of combination of array pairs
*/
public String getPair(int[] arr, int k) {
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
String result = "";

/**
* First store array elements into hashmap with key as the value of the
* array This has time complexity of O(n) and space complexity of O(n)
*/
for (int i = 0; i < arr.length; i++) {
hm.put(arr[i], arr[i]);
}

/**
* Try getting the hashmap value at key (sum-array_element) If it
* exists, it means array_element and sum-array_element is the pair
* thats forms sum 'k' This has time complexity of O(n)
*/
for (int i = 0; i < arr.length; i++) {
if (hm.get(k - arr[i]) != null) {
result += arr[i] + "," + (k - arr[i]) + ";";
}
}
return result;
}
```

Output: Pair of elements at O(2n) time complexity and O(n) space complexity: 4,2;2,4;5,1;1,5;

Brute force algorithm with time complexity of O(n^2)

```/**
* Find the pair in an array whose sum with complexity O(n^2)
*
* @param arr
*            input array of elements
* @param k
*            sum for which pair of array elements needs to be searched
* @return string of combination of array pairs
*/
public String findPair(int[] arr, int k) {
int i = 0, j = arr.length - 1;
String result = "";
while (i < j) {
while (i < j) {
if (arr[i] + arr[j] == k) {
result += arr[i] + "," + arr[j] + ";";
}
j--;
}
i++;
j = arr.length - 1;
}
return result;
}
```

Output: Pair of elements at O(n^2) time complexity and no additional space complexity: 4,2;5,1;

Here decimal to binary/octa/hexa decimal conversions are shown.

Algorithm is simple provided we should make use of the stack (stack implementation) to put the coefficients.

```public class DecimalConversion {
TNStack st = new TNStack(30);
private static HashMap<Integer, String> hm = new HashMap<Integer, String>();

/**
* converts number to given base
* @param number
* @param base
*/
public void decimalToXXXConversion(int number, int base) {
System.out.println("Number "+number+" in base "+base+" is: ");
while(number>0) {
st.push(number%base);
number = number/base;
}

while(!st.isEmpty()) {
int n = st.pop();
if(n>9) {
String str = hm.get(n);
System.out.print(str);
} else {
System.out.print(n);
}
}
System.out.println();
}

/**
* @param args
*/
public static void main(String[] args) {
DecimalConversion dc = new DecimalConversion();

// decimal to binary conversion
dc.decimalToXXXConversion(100, 2);

// decimal to octal conversion
dc.decimalToXXXConversion(100, 8);

hm.put(10, "A");
hm.put(11, "B");
hm.put(12, "C");
hm.put(13, "D");
hm.put(14, "E");
hm.put(15, "F");
dc.decimalToXXXConversion(10012, 16);
}
}
```

Output:
Number 100 in base 2 is:
1100100
Number 100 in base 8 is:
144
Number 10012 in base 16 is:
271C

## Queue implementation in Java

Queues operate in FIFO model.

Queues are also used in various other data structures, some of them are

• searching graphs
• printer queues in our machines
• keystroke data that is typed onto the keyboard

Efficiency of queue is that it performs insert and remove operations in O(1) complexity.

```package algorithm.queue;

/**
* @author ntallapa
*
*/
public class TNQueue {
private int size;
private int[] queueArr;
private int front = -1;
private int rear = -1;
private int totalItems;

public TNQueue(int s) {
size = s;
queueArr = new int[s];
}

public void insert(int i) {
rear++;
System.out.println("Inserting "+i);
queueArr[rear] = i;
totalItems++;
}

public int remove() {
front++;
totalItems--;
System.out.println("Removing "+queueArr[front]);
return queueArr[front];
}

public boolean isFull() {
return (totalItems == size);
}

public boolean isEmpty() {
return (totalItems == 0);
}
}

package algorithm.queue;
/**
* @author ntallapa
*
*/
public class TNQueueClient {

/**
* @param args
*/
public static void main(String[] args) {
TNQueue tnq = new TNQueue(3);
if(!tnq.isFull())
tnq.insert(1);
if(!tnq.isFull())
tnq.insert(2);
if(!tnq.isFull())
tnq.insert(3);
if(!tnq.isFull())
tnq.insert(4);
else
System.out.println("Queue is full, cannot insert element");

if(!tnq.isEmpty())
tnq.remove();
if(!tnq.isEmpty())
tnq.remove();
if(!tnq.isEmpty())
tnq.remove();
if(!tnq.isEmpty())
tnq.remove();
else
System.out.println("Queue is empty, cannot remove element");
}
}
```

Output:
Inserting 1
Inserting 2
Inserting 3
Queue is full, cannot insert element
Removing 1
Removing 2
Removing 3
Queue is empty, cannot remove element

## Stack implementation in Java

Stacks operate in LIFO model.

Stack plays vital role in many data structures, some of them are

• in parsing arithmetic expressions
• to help traverse nodes of binary tree
• searching vertices of a graph
• in java, every method’s return type and arguments are pushed on to a stack and when method returns they are popped off.

Efficiency of stack is that it performs push and pop operations in O(1) complexity.

```package algorithm.stack;
/**
* @author ntallapa
*
*/
public class TNStack {
private int size;
private int[] stackArr;
private int top = -1;

public TNStack(int size) {
this.size = size;
stackArr = new int[size];
}

/**
* increment the ctr and push element into stack
* @param i element to be pushed
*/
public void push(int i) {
top++;
System.out.println("Pushing "+i);
stackArr[top] = i;
}

/**
* pop the element from stack and decrement the ctr
* @return the popped element
*/
public int pop() {
int i = stackArr[top];
top--;
System.out.println("Popping "+i);
return i;
}

public int peek() {
System.out.println("Peek "+stackArr[top]);
return stackArr[top];
}

public boolean isFull() {
return (top == size-1);
}

public boolean isEmpty() {
return (top == -1);
}
}

package algorithm.stack;
/**
* @author ntallapa
*
*/
public class TNStackClient {

/**
* @param args
*/
public static void main(String[] args) {
TNStack tns = new TNStack(3);
// push some elements
if(!tns.isFull())
tns.push(4);
if(!tns.isFull())
tns.push(5);
if(!tns.isFull())
tns.push(3);
if(!tns.isFull())
tns.push(6);
else
System.out.println("Stack is full, cannot push element");

// pop some elements
if(!tns.isEmpty())
tns.pop();
if(!tns.isEmpty())
tns.pop();
if(!tns.isEmpty())
tns.pop();
if(!tns.isEmpty())
tns.pop();
else
System.out.println("Stack is empty, cannot pop element");

//reinsert to verify peek method
if(!tns.isFull())
tns.push(6);

// peek couple of times; result should be same
tns.peek();
tns.peek();
}
}

```

Output:
Pushing 4
Pushing 5
Pushing 3
Stack is full, cannot push element
Popping 3
Popping 5
Popping 4
Stack is empty, cannot pop element
Pushing 6
Peek 6
Peek 6

## Creating our own hashmap in java

This is an attempt to come up with my own hashmap in java. It serves all basic needs of original java.util.HashMap with O(1) complexity in read operations. Full source code can be downloaded from here

To deep dive into the fundamentals of java.util.HashMap refer to this article

STEP1: Create a simple data structure with key, value and which can also extend as linked list

• line #2,#3: references to key and value
• line #4: link to itself (used when there is collision in the hashmap)
```    class Entry {
Employee key;
String value;
Entry next;

Entry(Employee k, String v) {
key = k;
value = v;
}

public String getValue() {
return value;
}

public void setValue(String value) {
this.value = value;
}

public Employee getKey() {
return key;
}
}
```

STEP2: Couple of important utility methods

• getSupplementalHash(): Supplemental hash function used to defend against the poor quality hash given by the user
• getBucketNumber(): It makes sure the bucket number falls within the size of the hashmap based on the value of hash
```    private int getSupplementalHash(int h) {
// This function ensures that hashCodes that differ only by
// constant multiples at each bit position have a bounded
// number of collisions (approximately 8 at default load factor).
h ^= (h >>> 20) ^ (h >>> 12);
return h ^ (h >>> 7) ^ (h >>> 4);
}

private int getBucketNumber(int hash) {
return hash & (SIZE - 1);
}
```

STEP3: PUT Method

• line #3: get the user defined hashcode
• line #4: defend against the poor quality hash functions defined by the user (if Employee.hashcode() is poor, this call will do a better job)
• line #8: get the bucket index
• line #12: If the control reaches into for loop, it means that it should either be a duplicate or a collision
• line #14-15: Its a duplicate, it will be replaced with old one
• line #20: Its a collision, new pair will be appended to the list
• line #28: Its either a new_pair/collision, add it to the map
```    public void put(Employee key, String value) {
// get the hashcode and regenerate it to be optimum
int userHash = key.hashCode();
int hash = getSupplementalHash(userHash);

// compute the bucket number (0-15 based on our default size)
// this always results in a number between 0-15
int bucket = getBucketNumber(hash);
Entry existingElement = table[bucket];

for (; existingElement != null; existingElement = existingElement.next) {

if (existingElement.key.equals(key)) {
System.out
.println("duplicate key value pair, replacing existing key "
+ key + ", with value " + value);
existingElement.value = value;
return;
} else {
System.out.println("Collision detected for key " + key
+ ", adding element to the existing bucket");

}
}

//
System.out.println("PUT adding key:" + key + ", value:" + value
+ " to the list");
Entry entryInOldBucket = new Entry(key, value);
entryInOldBucket.next = table[bucket];
table[bucket] = entryInOldBucket;
}
```

STEP4: GET Method

• line #3: defend against the poor quality hash functions defined by the user (if Employee.hashcode() is poor, this call will do a better job)
• line #7: get the bucket index
• line #14: This loop is iterated as many times as the number of collisions for every key
• line #23: If nothing is found
```    public Entry get(Employee key) {
// get the hashcode and regenerate it to be optimum
int hash = getSupplementalHash(key.hashCode());

// compute the bucket number (0-15 based on our default size)
// this always results in a number between 0-15
int bucket = getBucketNumber(hash);

// get the element at the above bucket if it exists
Entry existingElement = table[bucket];

// if bucket is found then traverse through the linked list and
// see if element is present
while (existingElement != null) {
System.out
.println("Traversing the list inside the bucket for the key "
+ existingElement.getKey());
if (existingElement.key.equals(key)) {
return existingElement;
}
existingElement = existingElement.next;
}

// if nothing is found then return null
return null;
}
```

STEP5: Employee object as the key to our custom map (TESTING)
hashCode(): Make sure the hash code falls with 0-10 so that we can reproduce collision easily.
equals(): based on id and name of the person

```	static class Employee {
private Integer id;
private String name;

Employee(Integer id, String name) {
this.id = id;
this.name = name;
}

@Override
public int hashCode() {
// this ensures all hashcodes are between 0 and 15
return id % 10;
}

@Override
public boolean equals(Object obj) {
Employee otherEmp = (Employee) obj;
return this.name.equals(otherEmp.name);
}

@Override
public String toString() {
return this.id + "-" + name;
}
}
```

STEP6: Test Code

• line #13-14: Demonstrate duplicate key replacement in hashmap
• line #30-42 and #31-35: Demonstrate collision in hashmap
• line #44-49: Demonstrate collision along with duplication in hashmap
```		TMHashMap tmhm = new TMHashMap();

Employee e1 = new Employee(100, "Niranjan");
tmhm.put(e1, "dept1");

// duplicate
Employee e1_dup = new Employee(100, "Niranjan");
tmhm.put(e1_dup, "dept12");
// the above value "dept12" should replace the old value "dept1"
Entry e = tmhm.get(e1_dup);
System.out.println("GET element - " + e.getKey() + "::" + e.getValue());

Employee e2 = new Employee(102, "Sravan");
tmhm.put(e2, "dept3");

Employee e3 = new Employee(104, "Ananth");
tmhm.put(e3, "dept2");

Employee e4 = new Employee(106, "Rakesh");
tmhm.put(e4, "dept5");

Employee e5 = new Employee(108, "Shashi");
tmhm.put(e5, "dept2");

// collision with e2
Employee e2_collision = new Employee(112, "Chandu");
tmhm.put(e2_collision, "dept16");
e = tmhm.get(e2_collision);
System.out.println("GET element - " + e.getKey() + "::" + e.getValue());

// collision with e3
Employee e3_collision = new Employee(114, "Santhosh");
tmhm.put(e3_collision, "dept9");
e = tmhm.get(e3_collision);
System.out.println("GET element - " + e.getKey() + "::" + e.getValue());

System.out
.println("============== Adding Duplicate in Collision ===================");
Employee e3_collision_dupe = new Employee(124, "Santhosh");
tmhm.put(e3_collision_dupe, "dept91");
e = tmhm.get(e3_collision_dupe);
System.out.println("GET element - " + e.getKey() + "::" + e.getValue());
```

OUTPUT: of this program

```============== Adding Element ===================
PUT adding key:100-Niranjan, value:dept1 to the list
duplicate key value pair, replacing existing key 100-Niranjan, with value dept12
Traversing the list inside the bucket for the key 100-Niranjan
GET element - 100-Niranjan::dept12
PUT adding key:102-Sravan, value:dept3 to the list
PUT adding key:104-Ananth, value:dept2 to the list
PUT adding key:106-Rakesh, value:dept5 to the list
PUT adding key:108-Shashi, value:dept2 to the list
Collision detected for key 112-Chandu, adding element to the existing bucket
PUT adding key:112-Chandu, value:dept16 to the list
Traversing the list inside the bucket for the key 112-Chandu
GET element - 112-Chandu::dept16
Collision detected for key 114-Santhosh, adding element to the existing bucket
PUT adding key:114-Santhosh, value:dept9 to the list
Traversing the list inside the bucket for the key 114-Santhosh
GET element - 114-Santhosh::dept9
============== Adding Duplicate in Collision ===================
duplicate key value pair, replacing existing key 124-Santhosh, with value dept91
Traversing the list inside the bucket for the key 114-Santhosh
GET element - 114-Santhosh::dept91
```

After receiving a lot of requests/questions related to hashmap behavior on collisions and element duplication, I updated the code. I hope one can take it further from here 🙂

## Towers of Hanoi Recursive Ananlysis

Recursive program to towers of hanoi problem

```package algorithms.recursion;

/**
* @author ntallapa
*
*/
public class TowersOfHanoi {

/**
* This recursive algorithm takes (2^n-1) iterations to complete the task
*
* @param n number of disks
* @param startPole
* @param endPole
*/
public static void move(int n, int startPole, int endPole) {
if (n== 0){
return;
}
// here 6 is summation of poles, i.e, sigma(3) = 3+2+1 = 6
int intermediatePole = 6 - startPole - endPole;

// Move n–1 disks from disk 1 to disk 2, using disk 3 as a temporary holding area.
move(n-1, startPole, intermediatePole);

// Move the last disk (the largest) from disk 1 to disk 3.
System.out.println("Move " +n + " from " + startPole + " to " +endPole);

//  Move n–1 disks from disk 2 to disk 3, using disk 1 as a temporary holding area.
move(n-1, intermediatePole, endPole);
}

public static void main(String[] args) {
move(4, 1, 3);
}
}
``` For 2 discs we need 3 iterations, 3 discs we need 7 iterations, 4 discs we need 15 iterations, etc… from this it can be realized that to move ‘n’ discs we need (2^n)-1 iterations.

## how to print leaf nodes in a binary search tree

The logic here is simple. Recursively go through left and then right elements, when the base condition is met (i.e, when r==null), add a check to see if the node’s left and right wings are null and if they are null print the value of the node.

```    // print leaf nodes
public void leafNodes(BinaryNode r) {
if(r!= null) {
leafNodes(r.left);
leafNodes(r.right);

if(r.left == null && r.right == null) {
System.out.println(r.element);
}
}
}
```

Note: This logic is same is recursive post order traversal, but instead of just printing the node value after left&right traversals we check to see if left&right children are null and then print the node value.

## find intersection of elements in two arrays

Assume two arrays, ‘A’ of size ‘m’ and ‘B’ of size ‘n’

## case 1: Two arrays are unsorted

Here we pick one of the array and load it into hash implemented data structure, i.e, HashSet and then proceeds further to find intersection of elements.

Since hashed data structure’s complexity is O(1), the total complexity to find intersection of elements the complexity would become

Algorithm Time Complexity: O(m) + O(n)*O(1)
Algorithm Space Complexity: O(m)

```	public void intersect1(int[] a, int[] b) {
HashSet<Integer> hs = new HashSet<Integer>();
for (int i = 0; i < b.length; i++) {
}

for (int i = 0; i < a.length; i++) {
if(hs.contains(a[i])) {
System.out.println(a[i]+" is present in both arrays");
}
}
}
```

pros:
best algorithm when compared to all others provided one implements appropriate hashcode method
cons:
when the size of the data structure grows too high, it might lead to hash collisions

## Case 2: Two arrays are sorted

For every element in array A, do a binary search in array B (instead of linear search as shown in case-1), so here for every value in ‘A’ we go through log(n) interations in ‘B’ to find out if element in ‘A’ exists in ‘B’

Algorithm Time Complexity: O(m)*O(logn)

```
public void intersect(int[] a, int[] b) {
for(int i=0; i<a.length; i++) {
int val = binarySearch(b, a[i], 0, a.length);
if(val == b[j]) {
System.out.println(a[i]+" is present in both arrays");
break;
}
}
}
```

## case 3: Two arrays are unsorted

Brute force algorithm: For every element in array A, traverse through all the elements of array B and find out if the element exists or not.
Algorithm Time Complexity: O(m)*O(n)

In Java

```		public void intersect(int[] a, int[] b) {
for(int i=0; i<a.length; i++) {
for(int j=0; j<b.length; j++) {
if(a[i] == b[j]) {
System.out.println(a[i]+" is present in both arrays");
break;
}
}
}
}
```

pros:
works well for smaller arrays
works for unsorted arrays
cons:
order/complexity is directly proportional to the product of size of arrays, this can take huge time for arrays of larger lengths

## types of traversals in a binary tree

There are three types of traversals in binary tree: pre-order, in-order and post-order

Lets take an example of BST

```            20
15              25
12      18      22      28
```

pre-order: visit each node before its children

```public void preorder(BinaryNode root) {
if(root !=  null) {
System.out.println(root.element);
preorder(root.left);
preorder(root.right);
}
}
```

Output: 20, 15, 12, 18, 25, 22, 28

in-order (applied only for binary trees): visit left sub-tree, node and then right sub-tree

```public void inorder(BinaryNode root) {
if(root !=  null) {
inorder(root.left);
System.out.println(root.element);
inorder(root.right);
}
} ```

Output: 12, 15, 18, 20, 22, 25, 28
Note: If we notice the above elements they are all sorted which means the inorder traversal of any BST gives us the sorted elements.

post-order: visit each node after its children (left and right sub trees)

```public void postorder(BinaryNode root) {
if(root !=  null) {
postorder(root.left);
postorder(root.right);
System.out.println(root.element);
}
} ```

Output: 12, 18, 15, 22, 28, 25, 20

## What is BIG(O) NOTATION

BIG O Notation is used to represent complexity of an algorithm in worst case scenario.

Complexity in computer programming can be defined in terms of time and space. Space complexity tells us how much space is going to be taken for a particular algorithm. Time complexity tells us number of iterations (indirectly time) that are required for the completion of an algorithm.

O(1) means It takes constant space/time irrespective of the size of the data structure

O(n) means the space/time complexity is proportional to the length of the data structure

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