Midterm Review

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Midterm Exam

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4 questions + 1 bonus question. Finish previous 4 questions first, work on the bonus one if you have time.
Time limit is 2 hours, 18:30 - 20:30 PST
Copy and paste the NDA above to your txt file and sign. Every candidate should sign the NDA first, if not, the solutions will be regarded as INVALID.
All solutions should be replied to govalley.201612@gmail.com
All solutions should be submitted in a separate file, named as YourName_midterm.txt.
Only one API needs to be implemented for each problem (Java or C++).
Comment the ideas and thoughts above your implementation.
The submitted TXT file should include NDA, 4 + 1 explanations, complexity & implementations.

Midterm Exam

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Sample:
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 = "aabcc",
s2 = "dbbca",
When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

Q1
/*
the idea is that...

Time: O(n)
Space: O(n)
*/

public boolean isInterleave(String s1, String s2, String s3) {

}

Midterm Review

Time Complexity
Data Structure
- Array, Sort
- LinkedList
- Stack
- Queue
- HashMap
Algorithms
- Binary Search
- Recursion

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Time Complexity

Definition
- O
- Best Case, Average Case, Worst Case
How to calculate
- Master Method
- Maths Deduction
  - Recursion
- Solution/Search Space
- From the code
Important Time Complexity
- Sort: O(nlogn)
- Binary Search: O(logn)

\Theta

\Theta

\Omega

\Omega

Arrays - add (to end):

- Best: O(1)

- Worst: O(n)

- Average: O(1)

- T(n) = T(n/2) + c = T(n/4) + 2c = .... = T(1) + (logn)*c

- T(n) = T(n-1) + cn = T(n-2) + cn + c(n-1) = ... = T(1) + c(n + (n-1) + ... + 0)

- Combination, Permutation, ...

- Two Pointers

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Array

Definition
- Same type elements, continuously in memory
- Get any element in constant time.
Basic Operation
- Get, Add, Remove
Implementation
- capacity v.s. size -> resize(ensureSize),
- Element[] data
Two Pointers
- Use two/multiple pointers to iterate the array
- Two Sum, Three Sum, Odd Even Sort, ...

- Get: O(1)

- Add: O(1), O(n)

- Remove: O(1), O(n)

2Sum -> 3Sum -> 4Sum -> ... -> Combination Sum

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Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.

Example Input: [2, 3, 1, 2, 4, 3]; 7

Example Output: 2 ([4, 3])

public int minSubArrayLen(int s, int[] nums) {
    int sum = 0;
    int min_length = Integer.MAX_VALUE;    // Why max?
    for (int start = 0, end = 0; start < nums.length; start++) {
        while (end < nums.length && sum < s) {
            sum += nums[end++];
        }
        if (sum >= s) {
            min_length = Math.min(min_length, end-start);
        }
        sum -= nums[start];
    }
    return min_length == Integer.MAX_VALUE ? 0 : min_length;
}

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Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC".

Notes:

- If there is no such window in S that covers all characters in T, return the empty string "".

- You are guaranteed that there will always be only one unique minimum window in S, if exist.

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HashMap Homework

public String minWindow(String s, String t) {
    int[] expected = new int[256];
    int[] visited = new int[256];
    for (int i = 0; i < t.length(); i++) {
        expected[(int)t.charAt(i)]++;
    }
    int end = 0, minLength = Integer.MAX_VALUE, count = 0;
    String result = "";
    for (int start = 0; start < s.length(); start++) {
        while (end < s.length() && count != t.length()) {
            visited[s.charAt(end)]++;
            if (visited[s.charAt(end)] <= expected[s.charAt(end)]) {
                count++;
            }
            end++;
        }
        if (count == t.length()) {
            if (end - start < minLength) {
                minLength = end - start;
                result = s.substring(start, end);
            }
        }
        if (expected[s.charAt(start)] > 0) {
            visited[s.charAt(start)]--;
            if (visited[s.charAt(start)] < expected[s.charAt(start)]) {
                count--;
            }
        }
    }
    return result;
}

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🤔

public String minWindow(String s, String t) {
    int[] hm = parseToIntArray(t);
    String result = s + t;
    int count = t.length();
    for (int i = 0, j = 0; i < s.length(); i++) {    // move left each outer
        for(; count != 0 && j < s.length(); j++) {   // run right inner
            if (hm[s.charAt(j)] > 0) {
                count--;
            }
            hm[s.charAt(j)]--;
        }
        if (count == 0) {
            result = (j - i) < result.length() ? s.substring(i, j) : result;
        }
        hm[s.charAt(i)]++;
        if (hm[s.charAt(i)] > 0) {
            count++;
        }
    }
    return result.length() == (s.length() + t.length()) ? "" : result;
}

private int[] parseToIntArray(String t) {
    int[] hm = new int[256];
    for (char c : t.toCharArray()) {
        hm[c]++;
    }
    return hm;
}

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Sort

Always try to sort first before your algorithm.
- number order, alphabetical order, linked list order, etc.
- Easy for PRUNING
Two Sum
Search -> Binary Search
Permutation
Combination Sum
Merge Intervals
- Given a collection of intervals, merge all overlapping intervals.
- [1,3],[2,6],[8,10],[15,18] -> [1,6],[8,10],[15,18]

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Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example Input: [1,3],[2,6],[8,10],[15,18]

Example Output: [1,6],[8,10],[15,18]

public List<Interval> merge(List<Interval> intervals) {
    if (intervals.size() == 0) {
        return new ArrayList<Interval>();
    }
    Collections.sort(intervals, new Comparator<Interval>(){
        @Override
        public int compare(Interval a1, Interval a2) {
            return a1.start - a2.start;		//neg, 0, pos
        } 
    });
    
    ArrayList<Interval> results = new ArrayList<Interval>();
    Interval pre = intervals.get(0);
    for (int i = 1; i < intervals.size(); i++) {
        if (intervals.get(i).start <= pre.end) {
            pre.end = Math.max(intervals.get(i).end, pre.end);
        } else {
            results.add(new Interval(pre.start, pre.end));
            pre = intervals.get(i);
        }
    }
    results.add(pre);    // Don't forget the last one
    return results;
}

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class Interval {
    int start;
    int end;
}

Linked List

Java variables & data types
- Primitive Type & Reference Type
- Variable Definition/Assignment, Parameter Passing
Definition
- Linear collection, not continuous, same type
- Accessing one node needs iterating all previous ones.
Basic Operation
- Get, Add, Remove - O(n)
Dummy Node
- Remove specialty for head node when structure change may happen.
Two Pointers

function (ListNode node, int val) {

node.val = 5; // original node changed.

node = new ListNode(5); // no change.

val = 6; // original val will never change.

}

Add: O(1) with tail node.

ListNode dummy = new ListNode(-1);

dummy.next = head

- kth node from the end.

- mid node of the list.

- Whether circle exists

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Linked List

Programming Tips
- Try to simplify edge cases.
- Keep one node's next before it is changed.
- Check NULL when using while
  - while (cur != null) { cur.next }
  - while (pre.next != null) { pre.next; pre.next.next; }
- Test Cases
Make sure you think very clearly before writing code.
Practicing is the most important for Linked List

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Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative

Example Input: 1->2->3->4->5, 2

Example Output: 4->5->1->2->3

public ListNode rotateRight(ListNode head, int n) {
    if (head == null || n == 0)
        return head;
    ListNode tail = head;
    int length = 1;
    while (tail.next != null) {
        tail = tail.next;
        length++;
    }  // Calculate length of the linked list.
    n %= length;
    if (n == 0) {
        return head;
    } // If it is going to do a full rotate, then no need to do.
    
    ListNode newHeadPre = head;
    int index = length - n - 1;
    while (index-- > 0) {
        newHeadPre = newHeadPre.next;
    } // Get the previous node of the new head.
    
    tail.next = head;
    head = newHeadPre.next;
    newHeadPre.next = null;
    return head;
}

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Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

Example Input: 1->2->3->4

Example Output: 2->1->4->3

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

public ListNode swapPairs(ListNode head) {
    ListNode dummy = new ListNode(-1);
    dummy.next = head;
    ListNode pre = dummy;
    while (pre.next != null && pre.next.next != null) {
        ListNode first = pre.next, second = pre.next.next;
        first.next = second.next;
        second.next = first;
        pre.next = second;
        pre = first;
    }
    return dummy.next;
}

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Stack & Queue

Definition
- LIFO v.s. FIFO
Basic Operation
- size, pop, push, peek
Implementation
- Based on Array.
- One pointer for Stack, two pointers for Queue.
When to use Stack & Queue
- Stack: Deep First Search - recursion.
- Queue: Breadth First Search
Mono-increasing stack

Array -> Queue

Array -> Stack

Stack <-> Queue

- Largest Rectangle in Histogram

- Trapping Rain Water

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Stack & Queue

Mono-increasing stack
- Keep the elements in an "increasing" order so that previous elements can always extend to current one.
  - If new element is larger, push.
  - If new element is smaller, pop previous ones until the current one is the largest.

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right: the end of the stack

left: the previous element

Balance Number

Given an array of ints.

Is there a number k in this array that k is larger than all the numbers before it and smaller than all the numbers after it?

[1,2,3,4,5] true

[6,2,3,8,10,9] true

[4,5,3,6] true

[4,3,5,2] false

[1] true

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Extra: return all balance numbers

Balance Number

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public boolean existNumber(int[] nums) {
    Stack<Integer> stack = new Stack<>();
    int max = Integer.MIN_VALUE;
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] > max) {
            stack.push(nums[i]);
            max = nums[i];
        } else {
            while (!stack.isEmpty() && nums[i] <= stack.peek()) {
                stack.pop();
            }
        }
    }
    return !stack.isEmpty();
}

<Key, Value> What to store as K and V
put(key, value)
get(key)
containsKey(key),
Set<Map.Entry<Key, Value>> entrySet(),
keySet()
putAll()

hashcode()

Collision

equals()

HashMap

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public int maxPoints(Point[] points) {
    if (points == null || points.length == 0) {
        return 0;
    }
    int res = 0, n = points.length;
    Map<Pair, Integer> hm = new HashMap<>();
    for (int i = 0; i < n; i++) {
        hm.clear();
        int samePoint = 0, localMax = 0;
        for (int j = i + 1; j < n; j++) {
            int x = points[i].x - points[j].x;
            int y = points[i].y - points[j].y;
            if (x == 0 && y == 0) {
                samePoint++;
                continue;
            }
            int gcd = GCD(x, y);
            Pair p = new Pair(x/gcd, y/gcd);    // why?
            if (!hm.containsKey(p)) {
                hm.put(p, 1);
            } else {
                hm.put(p, hm.get(p) + 1);
            }
            localMax = Math.max(localMax, hm.get(p));
        }
        res = Math.max(res, localMax + samePoint + 1);
    }
    return res;
}

Max Points on a Line

Given n points on a 2D plane

find the maximum number of points that lie on the same straight line.

Cartesian coordinate system

笛卡尔坐标系（直角坐标系）

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class Pair {
    int x, y;
    public Pair(int x, int y) {
        this.x = x;
        this.y = y;
    }
    public int hashCode() {
        String s = x + "," + y;
        return s.hashCode();
    }
    public boolean equals(Object p) {
        if (this == p) {
            return true;
        }
        if (p == null || this.getClass() != p.getClass()) {
            return false;
        }
        Pair pair = (Pair) p;
        return this.x == pair.x && this.y == pair.y;
    }
}

private int GCD(int x, int y) {
    if (y == 0) {
        return x;
    }
    return GCD(y, x%y);
}

Max Points on a Line

Given n points on a 2D plane

find the maximum number of points that lie on the same straight line.

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// Greatest Common Divisor
public int GCD(int a, int b) {
    if (a == 0 || b == 0) {
        return a == 0 ? b : a;
    }
    if (a == b) {
        return a;
    }
    if (a > b) {
        return GCD(a - b, b);
    } else {
        return GCD(a, b - a);
    }
}

// modified version
// work if x and y are both positive
// return negative if y is negative.
// Normal mathematical GCD always return positive.
private int GCD(int x, int y) {
    if (y == 0) {
        return x;
    }
    return GCD(y, x % y);
}

// Least Common Multiple  
public int LCM(int a, int b) {
    return a * b / GCD(a, b);
}

Max Points on a Line

Greatest Common Divisor

GCD 最大公约数

Least Common Multiple

LCM 最小公倍数

Euclidean algorithm

欧几里得算法

辗转相除

Binary Search

The fastest way to search in sorted array. O(logN)
Be careful with INFINITE LOOP.
- int begin = 0, end = nums.length; [begin, end)
- int begin = 0, end = nums.length-1; [begin, end]
- Condition to terminate the loop
- Condition for edge cases
- In each loop iteration, there can be only two cases.
  - The range of [begin, end]/[begin, end) decrease.
  - Jump out of the loop
Make sure all cases are taken care of before writing code.
Insert Position
- Search 2D Matrix.

Find Minimum in Rotate Sorted Array

- Comparing mid with begin/end is different.

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public boolean searchMatrix(int[][] matrix, int target) {
    int begin = 0, end = matrix.length, mid = 0;
    while (begin < end) {
        mid = (begin + end) >> 1;
        if (matrix[mid][0] == target) {
            return true; 
        } else if (matrix[mid][0] > target) {
            end = mid ;
        } else {
            begin = mid + 1;
        }
    }
    int row = begin - 1;
    if (row < 0) return false;
    begin = 0;
    end = matrix[0].length;
    while (begin < end) {
        mid = (begin + end) >> 1;
        if (matrix[row][mid] == target) {
            return true; 
        } else if (matrix[row][mid] > target) {
            end = mid;
        } else {
            begin = mid + 1;
        }
    }
    return false;
}

Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

- Integers in each row are sorted from left to right.

- The first integer of each row is greater than the last integer of the previous row.

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Recursion

Divide problems into smaller instances, solve smaller problems and then solve original problem
Principles
- How to represent the question
  - Parameters that define the question
  - Parameters that store the temporary result or state
- Base case
- Recursion rules
  - Backtracking
  - Pruning strategy
Recursion is very similar to real thinking.
- Pseudo code can help a lot with these questions.

- Pruning before going into next recursion iteration is very important.

- Permutation II

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Permutation II

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Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:

[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

Permutation II

public List<List<Integer>> permuteUnique(int[] nums) {
    List<List<Integer>> result = new ArrayList<>();
    Arrays.sort(nums);
    dfs(result, intToObj(nums), 0);
    return result;
}
private void dfs(List<List<Integer>> result, Integer[] nums, int index) {
    // we don't have to wap the last number with itself.
    if (index == nums.length - 1) {
        result.add(new LinkedList<Integer>(Arrays.asList(nums)));
        return;
    }
    for (int i = index; i < nums.length; i++) {
        if (!(i > index && (nums[i] == nums[index]))) { 
            // (i == index || nums[i] != nums[index])
            swap(nums, index, i);
            dfs(result, Arrays.copyOf(nums, nums.length), index + 1);
            // why not swap back?
        }
    }
}
private void swap(Integer[] nums, int i, int j) {
    int temp = nums[i];
    nums[i] = nums[j];
    nums[j] = temp;
}
private Integer[] intToObj(int[] nums) {
    Integer[] result = new Integer[nums.length];
    for (int i = 0; i < nums.length; i++) {
        result[i] = nums[i];
    }
    return result;
}

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古之欲明明德于天下者，先治其国；欲治其国者，先齐其家；欲齐其家者，先修其身；欲修其身者，先正其心；欲正其心者，先诚其意；欲诚其意者，先致其知，致知在格物。

物格而后知至，知至而后意诚，意诚而后心正，心正而后身修，身修而后家齐，家齐而后国治，国治而后天下平。

Recursion

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https://www.zhihu.com/question/20507130