void secret(int arr[], int start_index, int end_index) { if(start_index >= end_index) return; int min_index; int temp; /* Assume that minIndex() returns index of minimum value in array arr[start_index...end_index] */ min_index = minIndex(arr, start_index, end_index); temp = arr[start_index]; arr[start_index] = arr[min_index]; arr[min_index] = temp; secret(arr, start_index + 1, end_index); }
10 / \ 12 15 / \ / \ 25 30 36 42
public static <T> List<T> convertBTreeToArrayList(BinaryTreenode<T> root) {
25, 12, 30, 10, 36, 15, 42
Consider the following directed graph, which is given in adjacency list form:
1: 2, 4, 6 2: 4, 5 3: 1, 2, 6, 9 4: 5 5: 4, 7 6: 1, 5, 7 7: 3, 5 8: 2, 6, 7 9: 1, 7
Part A:
Part B:
39 / \ 25 45 / \ \ 11 33 55 / \ \ / \ 7 17 37 47 57
Show the red-black tree that results from inserting each sequence of integers into a tree that is initially empty. In your answer, you can grapically draw red nodes with their data and use R_ as the prefix and black nodes with B_ as the prefix. An example for this is shown below.
42, 36, 40, 33, 35, 32, 41, 31, 34, 49
2, 1, 4, 5, 9, 3, 6, 7, 11, 10
B_14
/ \
B_7 R_20
/ \ / \
R_1 R_11 B_18 B_23
\
R_29
Consider the following undirected graph, which is given in adjacency matrix form and where vertices have the given labels and edges have non-negative integer weights (no value indicates no edge exists):
v0 v1 v2 v3 v4 v5 v6 +---------------------------- v0 | 3 6 2 | v1 | 3 1 6 | v2 | 6 1 3 4 | v3 | 2 3 8 7 | v4 | 6 4 8 3 2 | v5 | 7 3 2 | v6 | 2 2
Iteration | List of Visited Vertexes and their shortest distances from start |
Priority Queue's items (listed in increasing order) |
0 | - | 0 v0 |
1 | v0 0 | 2 v3, 3 v1, 6 v2 |
2 | v0 0, v3 2 |
public static List<Integer> setUnion(List<Integer> firstList, List<Integer> secondList) {
...
}