Java APIs

Question 1

String

Answer 1

• "..."
• new String(char[])
• .charAt(int index)
• .length()
• .split(String regex)
• .toCharArray()

Question 2

StringBuilder

Answer 2

• .append(char c)
• .append(char[] str)
• .append(String str)
• charAt(int index)
• deleteCharAt(int index)
• insert(int offset, char/char[]/String)
• .length()
• .toString()

Question 3

java.util.Arrays

Answer 3

• .equals(T[] arr1, T[] arr2)
• .fill(T[] arr, T value)
• .sort(T[] arr)
• .asList(E e, ...)

Question 4

Array

Answer 4

• {value, value, value, ...}
• new T[](int size)
• arr[int index]
• .length

Question 5

Collection to int[]

Answer 5

.stream().mapToInt(Integer::intValue).toArray()

Question 6

Map

Answer 6

• Map<K, V> x = new HashMap<>();
• new HashMap<>(int capacity)
• .containsKey(K key)
• .get(K key)
• .getOrDefault(K key, V defaultValue)
• .put(K key, V value)
• .remove(K key)
• .size()
• .isEmpty()
• .entrySet() -> Set<Map.Entry<K,V>>
• .keySet() -> Set<K>
• .values() -> Collection<V>

Question 7

Set

Answer 7

• Set<E> x = new HashSet<>()
• .add(E e)
• .contains(E e)
• .isEmpty()
• .remove(E e)
• .size()

Question 8

List

Answer 8

• .add(E e)
• .contains(E e)
• .get(int index)
• .isEmpty()
• .remove(int index)
• .size()
• .toArray()
• Collections.sort(List<T> list)

Question 9

Queue

Answer 9

• Queue<E> queue = new LinkedList<>()
• .offer(E e)
• .poll()
• .peek()
• .isEmpty()
• .size()

Question 10

Stack

Answer 10

• Stack<E> stack = new Stack<>()
• .push(E e)
• .pop()
• .peek()
• .size()
• .empty()

Question 11

Monotonic

Answer 11

Stack or queue that is increasing or decreasing in value

If the next item would break the sorting of the collection, remove elements until the next item can be added

Question 12

Formats for Graph Input

Answer 12

• Array of edges [x,y] - [[3,4][2,5][2,4][4,3]]
• Adjacency list (outgoing edges from i) - [[1], [2], [0,3]]
• Adjacency matrix (1 means connection) - [[1, 0, 1], [1, 1, 1], [0, 0, 1]]
• Matrix (like a map)

Question 13

Heap

Answer 13

• PriorityQueue<T> q = new PriorityQueue<>();</T>
• min heap by default
• .add(T)
• .peek()
• .remove()
• .size()
• maxHeap = new PriorityQueue<>(Comparator.reverseOrder());

Question 14

Binary Search

Answer 14

int left = 0;
int right = arr.length - 1;
while (left <= right) {
		int mid = left + (right - left) / 2;
		if (arr[mid] == target) {
				return mid;
		}
		if (arr[mid] > target) {
				right = mid - 1;
		} else {
				left = mid + 1;
		}
}

// target is not in arr, but left is at the insertion point
return left;

Question 15

Binary Search with Duplicates

Answer 15

int left = 0;
int right = arr.length;
while (left < right) {
    int mid = left + (right - left) / 2;
    if (arr[mid] >= target) {
        right = mid;
    } else {
        left = mid + 1;
    }
}

    return left;

Finds left most index. To find right most index change line 5 to >