Array

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INCEPTION

What is Array?

A data structure

consisting of a collection of

elements (values or variables),

each identified by at least one

array index or key.

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Array in Java?

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Just consider it as an Object

Array in Interview

Continuous space
Same type elements
Identified by index

Linked List

Tree

String

Set

HashMap

Stack/Queue

Heap

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Graph

Array in Code

Java
- int[] a = new int[5];
- System.out.println(a.length); // a.length = 5
C++
- int a[5];
- no way to get length.

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Array in Interview

Continuous space
- addr(elem) = addr(head) + sizeof(elem) * index
Same type elements
- int[] arr = new int[10];
Identified by index
- Accessing any element in the array, using CONSTANT time. O(1) == fast

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What is Array?

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Single-dimension Array

int[] arr = new int[3];

Two-dimensional Array

int[ ][ ] arr = new int[3][ ];
arr[0] = new int[3];
arr[1] = new int[5];
arr[2] = new int[4];

@4e25154f

WHY use array?

To make code simple and clean

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Read 10 numbers and print out reversely

int a0, a1, a2, ..., a9;

a0 = 1;
a1 = 2;
a2 = 3;
...

System.out.println(a9);
System.out.println(a8);
System.out.println(a7);
...

int[] a = new int[10];

for (int i = 0; i < 10; i++) {
    a[i] = i + 1;
}

for (int i = 9; i >= 0; i--) {
    System.out.println(a[i]);
}

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Give me an array of ...

Try it out

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public void moveZeroes(int[] nums) {

}

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example,

given nums = [0, 1, 0, 3, 9], after calling your function,

nums should be [1, 3, 9, 0, 0].

Move Zeros

Try it out

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index:    [0, 1, 2, 3, 4]
input:    [0, 1, 0, 3, 9]
output:   [1, 3, 9, 0, 0]

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example,

given nums = [0, 1, 0, 3, 9], after calling your function,

nums should be [1, 3, 9, 0, 0].

Move Zeros

Try it out

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public void moveZeroes(int[] nums) {
    if (nums == null || nums.length == 0) {
        return;
    }
    int p = 0;
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] != 0) {
            nums[p] = nums[i];
            p++;
        }
    }
    for (; p < nums.length; p++) {
        nums[p] = 0;
    }
}

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example,

given nums = [0, 1, 0, 3, 9], after calling your function,

nums should be [1, 3, 9, 0, 0].

Move Zeros

null & 0

Two pointers

Clean up

Try it out

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public void moveZeroes(int[] nums) {
    if (nums == null || nums.length == 0) {
        return;
    }
    int p = 0;
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] != 0) {
            nums[p++] = nums[i];
        }
    }
    while (p < nums.length) {
        nums[p++] = 0;
    }
}

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example,

given nums = [0, 1, 0, 3, 9], after calling your function,

nums should be [1, 3, 9, 0, 0].

Move Zeros

null & 0

Two pointers

Clean up

Try it out

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public void moveZeroes(int[] nums) {
    if (nums == null || nums.length == 0) return;
    int p = 0;
    for (int i = 0; i < nums.length; i++)
        if (nums[i] != 0) nums[p++] = nums[i];
    while (p < nums.length) nums[p++] = 0;
}

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example,

given nums = [0, 1, 0, 3, 9], after calling your function,

nums should be [1, 3, 9, 0, 0].

Move Zeros

Please Don't

Two-Pointer

Use TWO pointers, instead of one, to traverse the array in the same/opposite direction.

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Two-Pointer

Mostly, SORTED.
- in number order.
- in sequence order.
- linked list order.
To find two numbers, or two set of numbers, which are subject to some conditions.

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Java

Variables

&

Data Types

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Java Variables & Data Types

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8 Primitive Types
- byte
- short
- int
- long
- float
- double
- char
- boolean

Reference Types
- Object obj = new Object();
- int[] array = {1,2,3};
- Interface
- Enum

Java Variables & Data Types

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char

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char c = 'a';
System.out.println(c);
System.out.println(c + 1); 

48 = 32 + 16
65 = 64 + 1
97 = 64 + 32 + 1

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Java Variables & Data Types

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Java Variables & Data Types

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Java Variables & Data Types

int a = 5;
int[] arr = new int[6];
Object obj = new Object();

a = 5

arr = @4e25154f

@4e25154f

obj

int b = a;
int[] arr2 = arr;
Object obj2 = obj;

@9d1c81a8

0 1 2 3 4 5

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Java Variables & Data Types

int a = 5;
int[] arr = new int[6];
Object obj = new Object();

a = 5

arr = @4e25154f

@4e25154f

obj = @9d1c81a8

@9d1c81a8

int b = a;
int[] arr2 = arr;
Object obj2 = obj;

b = 5

arr2 = @4e25154f

obj2 = @9d1c81a8

THINK

a = 10;
// what about b?
arr[0] = 8;
// what about arr1[1]
obj = null;
// what about obj2?

0 1 2 3 4 5

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Java Variables & Data Types

void caller() {
    int a = 5;
    int[] arr = new int[6];
    Human hu = new Human(15);
    change(a, arr, hu);
    System.out.println(a);
    System.out.println(arr[1]);
    System.out.println(hu.age);
}

void change(int b, int[] a_, Human animal) {
    b = 10;
    a[1] = animal.age - 4;
    animal.age = 22;
}

class Human {
    int age;
    public Human(int age) {
        this.age = age;
    }
}

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Java Variables & Data Types

void caller() {
    int a = 5;
    int[] arr = new int[6];
    Human hu = new Human(15);
    change(a, arr, hu);
    System.out.println(a);
    System.out.println(arr[1]);
    System.out.println(hu.age);
}

void change(int b, int[] a_, Human animal) {
    b = 10;
    a[1] = animal.age - 4;
    animal.age = 22;
}

class Human {
    int age;
    public Human(int age) {
        this.age = age;
    }
}

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Java Variables & Data Types

arr

a = 5

animal

age = 15

void caller() {
    int a = 5;
    int[] arr = new int[6];
    Human hu = new Human(15);
    change(a, arr, hu);
    System.out.println(a);
    System.out.println(arr[1]);
    System.out.println(hu.age);
}

void change(int b, int[] a_, Human animal) {
    b = 10;
    a[1] = animal.age - 4;
    animal.age = 22;
}

class Human {
    int age;
    public Human(int age) {
        this.age = age;
    }
}

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Java Variables & Data Types

arr

a = 5

b = 10

animal

age = 22

Primitive Type
Passing the COPY of primitive value.

Reference Type
Passing the COPY of object address.

void caller() {
    int a = 5;
    int[] arr = new int[6];
    Human hu = new Human(15);
    change(a, arr, hu);
    System.out.println(a);
    System.out.println(arr[1]);
    System.out.println(hu.age);
}

void change(int b, int[] a_, Human animal) {
    b = 10;
    a[1] = animal.age - 4;
    animal.age = 22;
}

class Human {
    int age;
    public Human(int age) {
        this.age = age;
    }
}

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Java Variables & Data Types

arr

a = 5

age = 22

Primitive Type
Passing the COPY of primitive value.

Reference Type
Passing the COPY of object address.

void caller() {
    int a = 5;
    int[] arr = new int[6];
    Human hu = new Human(15);
    change(a, arr, hu);
    System.out.println(a);
    System.out.println(arr[1]);
    System.out.println(hu.age);
}

void change(int b, int[] a_, Human animal) {
    b = 10;
    a[1] = animal.age - 4;
    animal.age = 22;
}

class Human {
    int age;
    public Human(int age) {
        this.age = age;
    }
}

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Variable Comparison

Primitive Type
- Compare the value of the variable.
  - int a = 5;
  - int b = 5; // a == b ? True
Reference Type
- Compare the address of the object.
  - MyObject obj1 = new MyObject();
  - MyObject obj2 = obj1; // obj1 == obj2 ? True
  - obj2 = new MyObject(); // obj1 == obj2 ? False

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String a = "abc";
String b = "abc";
System.out.println(a == b); // true 

String c = new String("abc"); 
System.out.println(a == c); // false 
System.out.println(a.equals(c)); // true 

String d = new String("abc"); 
System.out.println(c == d); // false 
System.out.println(d.equals(c)); // true

Variable Comparison

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Java Data Type Summary

Primitive Type
- Always means a literal value
  - 5
  - 's'
  - true
Reference Type
- Always means an address.
  - Object obj = new Object();
  - obj = obj2;
  - int[] array = new int[3];
  - myFunction(array);

Address of the real object.

Let's talk about ArrayList

Array

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ArrayList

Basic Operations

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Operation	Input	Output	Time
Get	<Index>	value	O(1)
Set	<index, value>	void/ArrayList	O(1)
Add	<[index], value>	void/ArrayList	O(n)
Remove	<index/value>	void/ArrayList	O(n)
Find	<value>	boolean/index	O(n)

How to implement Java ArrayList

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public class ArrayList{
    // TODO: implement this class.
    public int get(int index) {
        // TODO: implement this method.
    }
    public void set(int index, int value) {
        // TODO: implement this method.
    }
    public void add(int value) {
        // TODO: implement this method.
    }
    public void add(int index, int value) {
        // TODO: implement this method.
    }
    public void remove(int index) {
        // TODO: implement this method.
    }
    public void remove(int value) {
        // TODO: implement this method.
    }
}

ArrayList - Fields

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public class ArrayList{

    private int capacity;
    private int size;
    private int[] data;

    public ArrayList(int capacity_) {
        capacity = capacity_;
        size = 0;
        data = new int[capacity];
    }
    
}

ArrayList - Get

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public class ArrayList{

    public int get(int index) {
        return data[index];
    }

}

ArrayList - Get

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public class ArrayList{

    public int get(int index) {
        if (index < 0 || index >= size) {
            // throw Exception
        }
        return data[index];
    }

}

ArrayList - Set

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public class ArrayList{

    public void set(int index, int value) {
        if (index < 0 || index >= size) {
            // throw Exception
        }
        data[index] = value;
    }

}

ArrayList - Add

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public class ArrayList{

    public void add(int index, int value) {
        if (index < 0 || index > size) {
            // throw Exception
        }
        size++;
        for (int i = size-1; i >= index+1; i--) {
            data[i] = data[i-1];
        }
        data[index] = value;
    }
}

ArrayList - Add

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public class ArrayList{

    public void add(int index, int value) {
        if (index < 0 || index > size) {
            // throw Exception
        }
        if (size == capacity) {
            resize();
        }
        size++;
        for (int i = size-1; i >= index+1; i--) {
            data[i] = data[i-1];
        }
        data[index] = value;
    }
    private void resize() {
        capacity *= 2;
        int[] temp_data = new int[size];
        for (int i = 0; i < size; i++) {
            temp_data[i] = data[i];
        }
        data = new int[capacity];
        for (int i = 0; i < size; i++) {
            data[i] = temp_data[i];
        }
    }
}

ArrayList - Add

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public class ArrayList{

    public void add(int index, int value) {
        if (index < 0 || index >= size) {
            // throw Exception
        }
        if (size == capacity) {
            resize();
        }
        size++;
        for (int i = size-1; i >= index+1; i--) {
            data[i] = data[i-1];
        }
        data[index] = value;
    }
    private void resize() {
        capacity *= 2;
        int[] new_data = new int[capacity];
        for (int i = 0; i < size; i++) {
            new_data[i] = data[i];
        }
        data = new_data;
    }
}

ArrayList - Remove

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public class ArrayList{

    public void remove(int index) {
        if (index < 0 || index >= size) {
            // throw Exception
        }
        size--;
        for (int i = index; i < size; i++) {
            data[i] = data[i+1];
        }
    }
}

ArrayList - Key Points

Key data storage, data[]
Initialize with capacity
Alway Check Bound
Resize

very very very important!!!

source code for Java ArrayList

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Warmup Questions

Sum of the array numbers
Minimum element of the array
Second minimum element of the array
- kth min/max elem of the array
Swap every two elements in an array
- [1,2,3,4] -> [2,1,4,3]
Mirror k elems in a group for an array. k <= n
- [1,2,3,4,5,6], k = 4 -> [4,3,2,1,5,6]

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Two Sum

Given an array of integers (no duplicate), find two numbers such that they add up to a specific target number.

The function twoSum should return the two numbers such that they add up to the target, where number1 must be less than number2.

You may assume that each input would have exactly one solution.

Example:

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Input: numbers={2, 7, 11, 15}, target=9
Output: {2, 7}
public int[] twoSum(int[] nums, int target) {
    // TODO: implement this function.
}

Always ASK before you write code.

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Two Sum

Given an array of integers (no duplicate), find two numbers such that they add up to a specific target number.

The function twoSum should return the two numbers such that they add up to the target, where number1 must be less than number2.

You may assume that each input would have exactly one solution.

Example:

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Input: numbers={2, 7, 11, 15}, target=9
Output: {2, 7}
public int[] twoSum(int[] nums, int target) {
    // TODO: implement this function.
}

Solution-1

Try one number nums[i], check all other numbers to see if their sum is expected.
After trying all i, we can certainly find the pair.

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// sudo code
for i from 0 to n - 1
    for j from i + 1 to n - 1
        if nums[i] + nums[j] == target
        return

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Solution-1

public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    if (numbers.length < 2) {
        return result;
    }
    for (int i = 0; i < numbers.length-1; i++) {
        for (int j = i+1; j < numbers.length; j++) {
            if (numbers[i] + numbers[j] == target) {
                if (numbers[i] < numbers[j]) {
                    result[0] = numbers[i];
                    result[1] = numbers[j];
                } else {
                    result[0] = numbers[j];
                    result[1] = numbers[i];
                }
                return result;
            }
        }
    }
    return result;
}

C(n, 2)

C(n, 2)

Big O?

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Solution-1

public int[] twoSum(int[] nums, int target) {
    int[] result = new int[2];
    for (int i = 0; i < nums.length-1; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] == target) {
                return (nums[i] < nums[j]) ? new int[] {nums[i], nums[j]} : new int[] {nums[j], nums[i]};
            }
        }
    }
    return result;
}

Solution-1 Summary

Direct, Easy
Time Complexity, O(n^2).
Inefficient
- Numbers : {1, 4, 20, 15, 8, 6, 3 }, Target = 10.
  - First loop, 1 + 20 > 10
  - Second loop, 4 + 20, NO NEED!
Brute Force
Sort

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Why sort helps?

J > 10 => pass
{1, 3, 4, 6, 8, 15, 20}, 10
- 1+20>10, 1+15>10
- for 3, start from 8.

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Solution-2

Sort the array
Try one number (i) from the beginning, check the other one (j) from the end.
- if sum == target, RESOLVED.
- if sum > target, j--.
- if sum < target, i++.
- Numbers : {1, 4, 20, 15, 8, 6, 3 }, Target = 10.
  - After sorting, {1, 3, 4, 6, 8, 15, 20}.
  - There won't even be comparisons like 4+20 ? 10
Time Complexity: O(nlogn) + O(n) = O(nlogn).

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Solution-2

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public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    Arrays.sort(numbers);
    int first = 0, second = numbers.length - 1;
    while (first < second) {
        if (numbers[first] + numbers[second] == target) {
            result[0] = numbers[first];
            result[1] = numbers[second];
            return result;
        }
        if (numbers[first] + numbers[second] > target) {
            second--;
        }
        if (numbers[first] + numbers[second] < target) {
            first++;
        }
    }
    return result;    
}

Three Sum

Given an array of integers (no duplicate), find THREE numbers such that they add up to a specific target number.

The function threeSum should return the three numbers such that they add up to the target, where three numbers are in increasing order.

You may assume that each input would have exactly one solution.

Examples:

COMMUNICATION!!!

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Input: numbers={-1, 0, 1, 2, -4}, target=0
Output: {-1, 0, 1}
public int[] threeSum(int[] nums, int target) {
    // TODO: implement this function.
}

Solution

Try first two numbers,
- check the third number in all other numbers to see if their sum are expected (target-num1-num2).

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Solution

~~Try first two numbers,~~
- ~~check the third number in all other numbers to see if their sum are expected (target-num1-num2).~~
Try the first number,
- check the other two in all other numbers to see if their sum are expected (target-num1).

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Solution

~~Try first two numbers,~~
- ~~check the third number in all other numbers to see if their sum are expected (target-num1-num2).~~
Try the first number,
- check the other two in all other numbers to see if their sum are expected (target-num1).
Time Complexity
- Sort + twoSum for n different first number.
- O(nlogn) + n*O(n) = O(n^2)

twoSum

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Solution

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public int[] threeSum(int[] nums, int target) {
// Sudo code
    int[] result = new int[3];
    if (nums.length < 3) {
        return nums;
    }
    Arrays.sort(nums);
    for (int i = 0; i < nums.length-2; i++) {
        twoSum(nums[i+1..], target-nums[i]);
        if (twoSum has results) {
            result = {nums[i], (twoSum result)}
        }
    }
    return result;
}

method {

m1();

m2();

}

Solution

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public int[] threeSum(int[] nums, int target) {
    int[] result = new int[3];
    if (nums.length < 3) {
        return nums;
    }
    Arrays.sort(nums);
    for (int i = 0; i < nums.length-2; i++) {
        int first = i+1, second = nums.length-1, new_target = target-nums[i];
        while (first < second) {
            if (nums[first] + nums[second] == new_target) {
                result[0] = nums[i];
                result[1] = nums[first];
                result[2] = nums[second];
                return result;
            }
            if (nums[first] + nums[second] > new_target) {
                second--;
            }
            if (nums[first] + nums[second] < new_target) {
                first++;
            }
        }
    }
    return result;
}

Summary for k-Sum

Sort.
Try first number, use (k-1)-Sum.
Time Complexity
- 2-Sum: O(nlogn) + O(n) = O(nlogn)
- 3-Sum: O(nlogn) + O(n^2) = O(n^2)
- 4-Sum: O(nlogn) + O(n^3) = O(n^3)
- k-Sum: O(nlogn) + O(n^(k-1)) = O(n^(k-1))

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Reverse Array

Given an array, reverse all the numbers in the array.

Example:

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Input: {1, 2, 3, 4, 5, 6, 7}
Output: {7, 6, 5, 4, 3, 2, 1}
public void reverseArray(int[] nums) {
    // TODO: implement this function.
}

Inplace?

What is Inplace?

Reverse Array

Given an array, reverse all the numbers in the array.

Example:

http://www.zhitongguigu.com/

Input: {1, 2, 3, 4, 5, 6, 7}
Output: {7, 6, 5, 4, 3, 2, 1}

public void reverseArray(int[] nums) {
    int first = 0, end = nums.length - 1;
    while (first < end) {
        swap(nums, first++, end--);
    }
}

private void swap(int[] nums, int first, int second) {
    int temp = nums[first];
    nums[first] = nums[second];
    nums[second] = temp;
}

Swap without third var?

Reverse Array

Given an array, reverse all the numbers in the array.

Example:

http://www.zhitongguigu.com/

Input: {1, 2, 3, 4, 5, 6, 7}
Output: {7, 6, 5, 4, 3, 2, 1}

public void reverseArray(int[] nums) {
    // TODO: implement this function.
    int first = 0, end = nums.length - 1;
    while (first < end) {
        swap(nums, first++, end--);
    }
}

private void swap(int[] nums, int first, int second) {
    int temp = nums[first];
    nums[first] = nums[second];
    nums[second] = temp;
}

Follow up:

Reverse Number (1234 -> 4321)
Palindrome Number/String
- ('abcd' -> false; 343 -> true)
Odd Even Sort, Pivot Sort

Odd Even Sort

Given an array of integers, sort them so that all odd integers come before even integers.

The order of elements can be changed. The order of sorted odd numbers and even numbers doesn't matter.

Example:

http://www.zhitongguigu.com/

Input: {4, 3, 5, 2, 1, 11, 0, 8, 6, 9}
Output: {9, 3, 5, 11, 1, 2, 0, 8 , 6, 4}
public void oddEvenSort(int[] nums) {
    // TODO: implement this function. After sorting,
    //    nums should start with odd numbers and then
    //    even numbers.
}

Odd Even Sort

Given an array of integers, sort them so that all odd integers come before even integers.

The order of elements can be changed. The order of sorted odd numbers and even numbers doesn't matter.

Example:

http://www.zhitongguigu.com/

Input: {4, 3, 5, 2, 1, 11, 0, 8, 6, 9}
Output: {9, 3, 5, 11, 1, 2, 0, 8 , 6, 4}

public void oddEvenSort(int[] nums) {
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (nums[first] % 2 == 1) {
            first++;
        }
        while (nums[second] % 2 == 0) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
}

Odd Even Sort

Given an array of integers, sort them so that all odd integers come before even integers.

The order of elements can be changed. The order of sorted odd numbers and even numbers doesn't matter.

Example:

http://www.zhitongguigu.com/

Input: {4, 3, 5, 2, 1, 11, 0, 8, 6, 9}
Output: {9, 3, 5, 11, 1, 2, 0, 8 , 6, 4}

public void oddEvenSort(int[] nums) {
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (first < second && nums[first] % 2 == 1) {
            first++;
        }
        while (first < second && nums[second] % 2 == 0) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
}

No inplace

Traverse 1 time

Pivot Sort

Given an array of integers and a target number, sort them so that all numbers that are smaller than the target always come before the numbers that are larger than the target. The order of elements can be changed.

Example:

http://www.zhitongguigu.com/

Input: {4, 9, 5, 2, 1, 11, 0, 8, 6, 3}, 7
Output: {4, 3, 5, 2, 1, 6, 0, 8, 11, 9}

public void pivotSort(int[] nums, int target) {
    // TODO: implement this function.
}

Pivot Sort

Example:

http://www.zhitongguigu.com/

Input: {4, 9, 5, 2, 1, 11, 0, 8, 6, 3}, 7
Output: {4, 3, 5, 2, 1, 6, 0, 8, 11, 9}

public static void pivotSort(int[] nums, int target) {
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (first < second && nums[first] <= target) {
            first++;
        }
        while (first < second && nums[second] > target) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
}

public static void swap(int[] nums, int i, int j) {
    int temp = nums[i];
    nums[i] = nums[j];
    nums[j] = temp;
}

Stability

Pivot Sort

Example:

http://www.zhitongguigu.com/

Input: {4, 9, 5, 2, 1, 11, 0, 8, 6, 3}, 7
Output: {4, 3, 5, 2, 1, 6, 0, 8, 11, 9}

public static void pivotSort(int[] nums, int target) {
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (first < second && nums[first] <= target) {
            first++;
        }
        while (first < second && nums[second] > target) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
}

public static void swap(int[] nums, int i, int j) {
    int temp = nums[i];
    nums[i] = nums[j];
    nums[j] = temp;
}

QuickSort

[Homewrok]

Follow up

Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

http://www.zhitongguigu.com/

Input: {10, 9, 5, 3, 9, 9, 8, 6, 7}, 9
Output: {10, 5, 3, 8, 6, 7, X, X, X}, 6

public int removeElement(int[] nums, int val) {
    // TODO: implement this function. After removing,
    //    nums should contain only instances which
    //    is not equal to val (from 0 to new length).
}

Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

http://www.zhitongguigu.com/

Input: {10, 9, 5, 3, 9, 9, 8, 6, 7}, 9
Output: {10, 5, 3, 8, 6, 7, X, X, X}, 6

public int removeElement(int[] nums, int val) {
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (first < second && nums[first] != val) {
            first++;
        }
        while (first < second && nums[second] == val) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
    return first;
}

[3, 2, 2, 3]

Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

http://www.zhitongguigu.com/

Input: {10, 9, 5, 3, 9, 9, 8, 6, 7}, 9
Output: {10, 5, 3, 8, 6, 7, X, X, X}, 6

public int removeElement(int[] nums, int val) {
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (first < second && nums[first] != val) {
            first++;
        }
        while (first < second && nums[second] == val) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
    return first;
}

[2, 2, 3, 3]

Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

http://www.zhitongguigu.com/

Input: {10, 9, 5, 3, 9, 9, 8, 6, 7}, 9
Output: {10, 5, 3, 8, 6, 7, X, X, X}, 6

public int removeElement(int[] nums, int val) {
    if (nums.length == 0) {
        return 0;
    }
    int first = 0, second = nums.length - 1;
    while (first < second) {
        while (first < second && nums[first] != val) {
            first++;
        }
        while (first < second && nums[second] == val) {
            second--;
        }
        if (first < second) {
            swap(nums, first++, second--);
        }
    }
    return nums[first] != val ? first+1 : first;
}

Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

http://www.zhitongguigu.com/

Input: {10, 9, 5, 3, 9, 9, 8, 6, 7}, 9
Output: {10, 5, 3, 8, 6, 7, X, X, X}, 6

public int removeElement(int[] nums, int val) {
    int index = 0, len = nums.length;
    // len is the valid length of remaining array.
    while (index < len) {
        if (nums[index] == val) {
            len--; // remove one element.
            nums[index] = nums[len]; // Keep the possible valid element.
        } else {
            index++;
        }
    }
    return len;
}

Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

http://www.zhitongguigu.com/

Input: {10, 9, 5, 3, 9, 9, 8, 6, 7}, 9
Output: {10, 5, 3, 8, 6, 7, X, X, X}, 6

public int removeElement(int[] nums, int val) {
    int index = 0, len = nums.length;
    while (index < len) {
        nums[index] = nums[index] == val ? nums[--len] : nums[index++];
    }
    return len;
}

Merge Two Sorted Array

Given two sorted arrays of integer, both with increasing order. Please merge them into one sorted array, with increasing order.

Examples:

Input: {1, 3, 5}, {2, 4, 6}

Output: {1, 2, 3, 4, 5, 6}

http://www.zhitongguigu.com/

Merge Two Sorted Array

Given two sorted arrays of integer, both with increasing order. Please merge them into one sorted array, with increasing order.

Examples:

Input: {1, 3, 5}, {2, 4, 6}

Output: {1, 2, 3, 4, 5, 6}

Solution:

Iterate two arrays at the same time, and always pick the smaller element from two numbers to put into the result array.

http://www.zhitongguigu.com/

public int[] removeElement(int[] arr1, int[] arr2) {
    int[] result = new int[arr1.length + arr2.length];
    int index = 0, index1 = 0, index2 = 0;
    while (index1 < arr1.length && index2 < arr2.length) {
        if (arr1[index1] < arr2[index2]) {
            result[index++] = arr1[index1++];
        } else {
            result[index++] = arr2[index2++];
        }
    }
    for (int i = index1; i < arr1.length; i++) {
        result[index++] = arr1[i];
    }
    for (int i = index2; i < arr2.length; i++) {
        result[index++] = arr2[i];
    }
}

Merge Two Sorted Array

Given two sorted arrays of integer, both with increasing order. Please merge them into one sorted array, with increasing order.

Examples:

Input: {1, 3, 5}, {2, 4, 6}

Output: {1, 2, 3, 4, 5, 6}

Follow up: Homework

- Merge Two Sorted Linked List

- Merge K Sorted Array

http://www.zhitongguigu.com/

Homework

http://www.zhitongguigu.com/

Two Sum - follow up

Given an array of integers ~~(no duplicate)~~, find two numbers such that they add up to a specific target number.

The function twoSum should return the two numbers such that they add up to the target, where number1 must be less than number2.

~~You may assume that each input would have exactly one solution.~~

Java:

public ArrayList<ArrayList<Integer>> twoSum(int[] nums, int target);

C++:

vector<vector<int>> twoSum(vector<int>& nums, int target);

http://www.zhitongguigu.com/

Partition Array

Given an array nums of integers and an int k, partition the array (i.e move the elements in "nums") such that:

All elements < k are moved to the left
All elements >= k are moved to the right

Return the partitioning index, i.e the first index i nums[i] >= k.

Java:

public int partitionArray(int[] nums, int k);

C++:

int partitionArray(vector<int> nums, int k);

http://www.zhitongguigu.com/

Optional Homework

http://www.zhitongguigu.com/