This problem requires finding the minimum number of adjacent swaps to sort an array containing a permutation of numbers from 1 to n. The cost of each adjacent swap is 1.

The key insight is that each adjacent swap changes the number of inversions in the array by exactly one. To sort the array in ascending order, we aim to eliminate all inversions. The minimum number of swaps needed is precisely the total number of inversions in the original array.

A naive approach to count inversions takes O(n^2) time, which is too slow for the given constraints. A more efficient method is to use a modified merge sort algorithm. During the merge step, when an element from the right subarray is placed before an element from the left subarray, it signifies an inversion. The count of such inversions can be accumulated.

Alternatively, a Fenwick tree (Binary Indexed Tree) can also be used to count inversions in O(n log n) time. This involves iterating through the array and, for each element, querying the number of elements already processed that are greater than the current element.

For this implementation, we will use the merge sort approach.

def merge_and_count_inversions(arr, left, mid, right):
    i = left
    j = mid + 1
    k = 0
    inversion_count = 0
    temp_arr = [0] * (right - left + 1)

    while i <= mid and j <= right:
        if arr[i] <= arr[j]:
            temp_arr[k] = arr[i]
            i += 1
        else:
            temp_arr[k] = arr[j]
            # If arr[j] is smaller, it means it forms inversions with all remaining elements in the left subarray
            inversion_count += (mid - i + 1)
            j += 1
        k += 1

    # Copy remaining elements from the left subarray, if any
    while i <= mid:
        temp_arr[k] = arr[i]
        i += 1
        k += 1

    # Copy remaining elements from the right subarray, if any
    while j <= right:
        temp_arr[k] = arr[j]
        j += 1
        k += 1

    # Copy sorted elements back to the original array slice
    for idx in range(len(temp_arr)):
        arr[left + idx] = temp_arr[idx]

    return inversion_count

def merge_sort_and_count(arr, left, right):
    total_inversions = 0
    if left < right:
        mid = (left + right) // 2
        # Recursively count inversions in left and right halves
        total_inversions += merge_sort_and_count(arr, left, mid)
        total_inversions += merge_sort_and_count(arr, mid + 1, right)
        # Count inversions between the two halves during merge
        total_inversions += merge_and_count_inversions(arr, left, mid, right)
    return total_inversions

def calculate_min_swaps(arr):
    # Create a copy to avoid modifying the original input if needed elsewhere
    arr_copy = list(arr)
    return merge_sort_and_count(arr_copy, 0, len(arr_copy) - 1)

# Example Usage:
# n = int(input())
# permutation_array = list(map(int, input().split()))
# result = calculate_min_swaps(permutation_array)
# print(result)

Finding the Smallest Element Greater Then or Equal to a Target

Given a sequence of n numbers and multiple queries, each query asks for the smallest number in the sequence that is greater than or equal to a given value x. If no such number exists, output -1.

Input Format:

The first line contains an integer n, the length of the sequence. The second line contains n space-separated positive integers, the elements of the sequence. These elements do not exceed 10^9. The third line contains an integer Q, the number of queries. Each of the following Q lines contains an integer x for a query.

Output Format:

Output Q lines, each containing the answer to the corresponding query.

Constraints: For 50% of test cases, n <= 2000. For 100% of test cases, n <= 300,000.

Hint: If the original sequence is sorted first, binary search can be efficiently applied. The lower_bound function in C++ STL is equivalent to finding the first element not less than the target. Its recommended for beginners to implement their own binary search.

To solve this problem efficiently, we can first sort the input sequence. Once sorted, each query can be answered using binary search. The goal is to find the first element in the sorted sequence that is greater than or equal to the query value x.

We can implement a custom binary search function that returns the smallest element found that meets the condition. If the binary search completes without finding such an element, it means no element in the sequence is greater than or equal to x, and we should return -1.

def find_smallest_greater_equal(sorted_arr, target):
    low, high = 0, len(sorted_arr) - 1
    result_val = -1

    while low <= high:
        mid = (low + high) // 2
        if sorted_arr[mid] >= target:
            # Found a potential candidate. Store it and try to find a smaller one to the left.
            result_val = sorted_arr[mid]
            high = mid - 1
        else:
            # The current element is too small, search in the right half.
            low = mid + 1

    return result_val

def process_queries(sequence, queries):
    # Sort the sequence once to enable efficient searching
    sorted_sequence = sorted(sequence)
    
    results = []
    for query_val in queries:
        results.append(find_smallest_greater_equal(sorted_sequence, query_val))
    return results

# Example Usage:
# n = int(input())
# sequence_data = list(map(int, input().split()))
# q = int(input())
# query_list = [int(input()) for _ in range(q)]
# 
# output_results = process_queries(sequence_data, query_list)
# for res in output_results:
#     print(res)

Shortest Path in an Undirected Weighted Graph

Given an undirected weighted graph with n nodes (numbered 1 to n), m edges, a starting node S, and a destination node T, find the length of the shortest path between S and T.

Input Format:

The first line contains four integers: n, m, S, T. n: number of nodes m: number of edges S: starting node T: destination node

The next m lines each describe an edge with three integers: u, v, w, representing an edge between nodes u and v with weight w. The graph is guaranteed to have 1 <= u, v <= n.

Output Format:

A single integer representing the shortest path length from S to T.

Constraints: For the first 10 test cases, n <= 2500, m <= 6200, and edge weights w <= 1000. These test cases have a gradient. For all 12 test cases, n <= 100,000, m <= 250,000.

Hint: This problem is a standard application of Dijkstra's algorithm. A naive implementation of Dijkstra's algorithm (using a simple array for the priority queue) can pass the first 10 test cases. To pass all test cases, an optimized Dijkstra's algorithm using a min-heap (like Python's heapq or C++'s std::priority_queue) is required.

Dijkstra's algorithm finds the shortest paths from a single source node to all other nodes in a graph with non-negative edge weights. It works by maintaining a set of visited nodes and a priority queue of nodes to visit, prioritized by their current shortest distance from the source.

The algorithm proceeds as follows:

1. Initialize the distance to the start node S as 0 and all other distances to infinity.
2. Add the start node S to a min-priority queue with distance 0.
3. While the priority queue is not empty: a. Extract the node u with the smallest distance from the priority queue. b. If this distance is greater than the already known shortest distance to u, skip it (this is a stale entry). c. For each neighbor v of u with edge weight w: i. If the distance to u plus w is less than the current shortest distance to v: - Update the shortest distance to v. - Add v to the priority queue with its new shorter distance.

After the algorithm terminates, the shortest distance to the target node T will be available.

import heapq

def find_shortest_path(graph_adj, start_node, end_node):
    num_nodes = len(graph_adj)
    # Initialize distances: infinity for all nodes except the start node
    distances = [float('inf')] * num_nodes
    distances[start_node - 1] = 0 # Adjusting for 0-based indexing
    
    # Priority queue stores tuples of (distance, node)
    priority_queue = [(0, start_node)]

    while priority_queue:
        current_distance, current_node = heapq.heappop(priority_queue)

        # If we found a shorter path already, skip this one
        if current_distance > distances[current_node - 1]:
            continue

        # Explore neighbors
        if current_node in graph_adj: # Check if the node has outgoing edges
            for neighbor, weight in graph_adj[current_node]:
                distance_through_current = current_distance + weight
                
                # If a shorter path to the neighbor is found
                if distance_through_current < distances[neighbor - 1]:
                    distances[neighbor - 1] = distance_through_current
                    heapq.heappush(priority_queue, (distance_through_current, neighbor))

    # Return the shortest distance to the end node, or infinity if unreachable
    final_distance = distances[end_node - 1]
    return final_distance if final_distance != float('inf') else -1 # Or handle unreachable case as needed

# Example Usage:
# n_nodes, m_edges, start_node, end_node = map(int, input().split())
# adjacency_list = {i: [] for i in range(1, n_nodes + 1)}

# for _ in range(m_edges):
#     u, v, w = map(int, input().split())
#     adjacency_list[u].append((v, w))
#     adjacency_list[v].append((u, w)) # For undirected graph

# shortest_distance = find_shortest_path(adjacency_list, start_node, end_node)
# print(shortest_distance)