To determine the overall ranking of players based on their wins and losses, you can use a directed graph where each player is a node, and a directed edge from player A to player B indicates that A has won against B. You can then perform a topological sort on this graph to establish the ranking order. If there are cycles, it indicates that there are players with circular wins, which complicates the ranking. In such cases, additional methods like using a scoring system or pairwise comparisons may be needed to resolve ties.
To determine the overall ranking of players based on their wins and losses, you can use a directed graph where each player is a node, and a directed edge from player A to player B indicates that A has won against B. You can then perform a topological sort on this graph to establish the ranking order. If there are cycles, it indicates that there are players with circular wins, which complicates the ranking. In such cases, additional methods like using a scoring system or pairwise comparisons may be needed to resolve ties.
```python
def merge_sorted_arrays(arr1, arr2):
m = len(arr1)
n = len(arr2)
# Last index of arr1 and arr2
i = m - 1
j = n - 1
k = m + n - 1
# Merge in reverse order
while i >= 0 and j >= 0:
if arr1[i] > arr2[j]:
arr1[k] = arr1[i]
i -= 1
else:
arr1[k] = arr2[j]
j -= 1
k -= 1
# If there are remaining elements in arr2
while j >= 0:
arr1[k] = arr2[j]
j -= 1
k -= 1
# Example usage:
arr1 = [1, 3, 5, 0, 0, 0
To apply binary search on a 2D array sorted both horizontally and vertically, start from the top-right corner of the array. Compare the target value with the current element:
1. If the current element equals the target, return true.
2. If the current element is greater than the target, move left (decrease the column index).
3. If the current element is less than the target, move down (increase the row index).
Repeat this process until you either find the target or go out of bounds of the array. If you exit the loop without finding the target, return false.
The time complexity of this approach is O(m + n), where m is the number of rows and n is the number of columns in the 2D array.
To find the cousin nodes of a given node in a binary tree, follow these steps:
1. **Find the Level and Parent**: Perform a breadth-first search (BFS) to determine the level of the given node and its parent node.
2. **Collect Cousins**: During the BFS, collect all nodes at the same level as the given node but with a different parent.
Here is a Python implementation:
```python
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def find_cousins(root, target):
if not root:
return []
from collections import deque
queue = deque([(root, None)]) # (node, parent)
target_parent = None
target_level = 0
current_level = 0
cousins = []
while queue:
level_size =
```python
def do_rectangles_intersect(rect1, rect2):
# rect1 and rect2 are tuples of the form (top_left, bottom_right)
# where top_left and bottom_right are tuples of (x, y)
# Unpack the rectangles
(x1_min, y1_max), (x1_max, y1_min) = rect1
(x2_min, y2_max), (x2_max, y2_min) = rect2
# Check for intersection
if (x1_max < x2_min or x2_max < x1_min or
y1_min > y2_max or y2_min > y1_max):
return False
return True
```
```python
def remove_characters(input_string, chars_to_remove):
return ''.join(c for c in input_string if c not in chars_to_remove)
# Test Cases
print(remove_characters("hello world", ['h', 'e'])) # Output: "llo world"
print(remove_characters("abcdef", ['a', 'b', 'c'])) # Output: "def"
print(remove_characters("12345", ['1', '3', '5'])) # Output: "24"
print(remove_characters("openai", ['o', 'p'])) # Output: "enai"
print(remove_characters("", ['a', 'b'])) # Output: ""
```
Best case: O(1)
Worst case: O(n)
```cpp
#include <vector>
#include <limits>
#include <algorithm>
struct Node {
int id;
// Other properties
};
struct Edge {
Node* from;
Node* to;
int weight;
};
std::vector<Edge> find_sec(Node* n1, Node* n2, std::vector<Edge>& edges) {
std::vector<int> minWeights;
std::vector<Edge> paths;
for (const auto& edge : edges) {
if ((edge.from == n1 && edge.to == n2) || (edge.from == n2 && edge.to == n1)) {
minWeights.push_back(edge.weight);
paths.push_back(edge);
}
}
std::sort(minWeights.begin(), minWeights.end());
int secondMin = std::numeric_limits<int>::max();
for (const auto& weight : minWeights) {
if (weight > minWeights[0]) {
secondMin = std
The number of smaller cubes that have paint on them is 26.
To implement the color filling algorithm (similar to the flood fill algorithm), you can use either Depth-First Search (DFS) or Breadth-First Search (BFS). Below is a simple implementation using DFS in Python:
```python
def flood_fill(matrix, x, y, new_color):
if matrix[x][y] == new_color:
return
original_color = matrix[x][y]
fill(matrix, x, y, original_color, new_color)
def fill(matrix, x, y, original_color, new_color):
if x < 0 or x >= len(matrix) or y < 0 or y >= len(matrix[0]):
return
if matrix[x][y] != original_color:
return
matrix[x][y] = new_color
fill(matrix, x + 1, y, original_color, new_color) # Down
fill(matrix, x - 1, y, original_color, new_color
To search a suffix array for the longest match of a new input string, follow these steps:
1. **Binary Search**: Use binary search on the suffix array to find the starting point where the input string could fit.
2. **Compare**: From the found position, compare the input string with the suffixes in the array to find the longest match.
3. **Track Length**: Keep track of the length of the longest match found during the comparisons.
For building a GUI interface for this task:
1. **Input Field**: Create a text box for users to enter the input string.
2. **Search Button**: Add a button to trigger the search operation.
3. **Display Area**: Include a section to display the longest match and its length.
4. **Feedback**: Provide visual feedback (like loading indicators) while searching.
5. **Error Handling**: Implement error messages for invalid inputs or no matches found.
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