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Minimum Reverse Operations

Hard
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Topics:
ArraysGraphsDynamic Programming

You are given an integer n and an integer p representing an array arr of length n where all elements are set to 0's, except position p which is set to 1. You are also given an integer array banned containing restricted positions. Perform the following operation on arr:

  • Reverse a subarray with size k if the single 1 is not set to a position in banned.

Return an integer array answer with n results where the i<sup>th</sup> result is the minimum number of operations needed to bring the single 1 to position i in arr, or -1 if it is impossible.

Example 1:

Input: n = 4, p = 0, banned = [1,2], k = 4 Output: [0,-1,-1,1]

Explanation:

  • Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
  • We can never place 1 on the banned positions, so the answer for positions 1 and 2 is -1.
  • Perform the operation of size 4 to reverse the whole array.
  • After a single operation 1 is at position 3 so the answer for position 3 is 1.

Example 2:

Input: n = 5, p = 0, banned = [2,4], k = 3 Output: [0,-1,-1,-1,-1]

Explanation:

  • Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
  • We cannot perform the operation on the subarray positions [0, 2] because position 2 is in banned.
  • Because 1 cannot be set at position 2, it is impossible to set 1 at other positions in more operations.

Example 3:

Input: n = 4, p = 2, banned = [0,1,3], k = 1 Output: [-1,-1,0,-1]

Explanation:

Perform operations of size 1 and 1 never changes its position.

Constraints:

  • 1 <= n <= 10<sup>5</sup>
  • 0 <= p <= n - 1
  • 0 <= banned.length <= n - 1
  • 0 <= banned[i] <= n - 1
  • 1 <= k <= n
  • banned[i] != p
  • all values in banned are unique

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What is the range of values for `n` (the size of the array) and the elements within the array?
  2. Can the blocked cells be assumed to be within the bounds of the array (0 to n-1), and is the starting position always valid?
  3. If it's impossible to reach a cell, what value should I return for its distance?
  4. Are the reverse operations performed in place, modifying the original array, or are they logical and only used to calculate distances?
  5. What is the expected behavior if the starting position is also a blocked cell?

Brute Force Solution

Approach

We want to find the shortest sequence of reversals to reach a target position. The brute force way explores every single possible combination of reversals to see if we can reach the target.

Here's how the algorithm would work step-by-step:

  1. Start at the initial position and consider all possible reversals we can do first.
  2. For each of those reversals, imagine doing all possible reversals next, and so on.
  3. Keep exploring these possibilities, like branching out a tree, until we either reach the target position or have gone too far (reversals exceeding some reasonable limit).
  4. If we reach the target, we remember the number of reversals it took to get there.
  5. We keep searching through all the reversal combinations until we have explored everything.
  6. Finally, we compare all the reversal counts where we found the target, and pick the smallest one. That's our shortest path.

Code Implementation

def minimum_reverse_operations_brute_force(number_count, initial_position, forbidden_positions, target_position):
    queue = [(initial_position, 0)]
    visited = {initial_position}
    minimum_reversals = float('inf')

    while queue:
        current_position, reversals_count = queue.pop(0)

        if current_position == target_position:
            minimum_reversals = min(minimum_reversals, reversals_count)

        # Only search up to a reasonable reversal count to avoid infinite loops
        if reversals_count > number_count:
            continue

        for subarray_length in range(1, number_count + 1):
            for start_position in range(number_count - subarray_length + 1):
                center = start_position + (subarray_length - 1) // 2
                end_position = center * 2 - start_position

                possible_position = current_position

                # Simulate applying a reverse and checking the new location
                if current_position >= start_position and current_position <= end_position:
                    possible_position = start_position + end_position - current_position

                # Check if we have already visited the position and if it is valid
                if (0 <= possible_position < number_count and\
                        possible_position not in forbidden_positions and\
                        possible_position not in visited):
                    queue.append((possible_position, reversals_count + 1))
                    visited.add(possible_position)

    if minimum_reversals == float('inf'):
        return -1
    else:
        return minimum_reversals

Big(O) Analysis

Time Complexity
O(n * 2^n)The described brute-force solution essentially explores a tree where each node represents a state (a particular arrangement of the array after some reversals). From each state, we can potentially perform reversals centered at any position within the array, leading to 'n' possible next states. The depth of this search tree is related to the maximum number of reversals considered. In the worst case, we might explore a significant portion of all possible reversal sequences. Since each level of the search tree branches out by roughly a factor of 'n', and the number of levels can potentially be proportional to n, the total number of nodes explored grows exponentially with 'n'. Therefore, the number of operations can be approximated as n multiplied by an exponential function of n, specifically 2^n, leading to O(n * 2^n) time complexity.
Space Complexity
O(N)The provided solution utilizes a breadth-first search (BFS) approach, implicitly storing states (positions) to explore in a queue. In the worst-case scenario, the algorithm might need to explore a significant portion of the possible positions within the range, requiring a queue of size proportional to the total number of positions, N. Furthermore, a set or similar data structure is used to keep track of visited positions to avoid revisiting them, potentially storing up to N positions as well. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The problem asks for the fewest flips needed to reach a target state on a line. The key idea is to treat the line like a network and to find the shortest path from the starting point to the desired result, considering only allowed moves.

Here's how the algorithm would work step-by-step:

  1. First, figure out which positions we're allowed to reverse, since some are blocked.
  2. Imagine each allowed reverse as a possible step we can take from one line configuration to another.
  3. We want to find the shortest sequence of steps to go from the initial line to the final target line.
  4. Use a method that explores options in layers: first consider all flips one step away, then two steps away, and so on, until we find the target.
  5. As we explore, keep track of the fewest flips it took to reach each line configuration we've seen so far, so we don't waste time revisiting the same configuration with more flips.
  6. If we reach a line configuration we've already seen via a shorter path, we ignore the longer path.
  7. Once we reach the final target, the number of flips recorded is the minimum number needed.

Code Implementation

from collections import deque

def minimum_reverse_operations(number_of_elements, positions, blocked, target):
    blocked_set = set(blocked)
    queue = deque([(positions, 0)])
    visited = {positions: 0}
    distances = [-1] * number_of_elements
    distances[positions] = 0

    while queue:
        current_position, current_distance = queue.popleft()

        for length_of_subsegment in range(1, number_of_elements + 1):

            # Find the possible start positions
            left_bound = max(0, current_position - (length_of_subsegment - 1))
            right_bound = min(number_of_elements - length_of_subsegment,
                              current_position)

            for start_position in range(left_bound,
                                          right_bound + 1):
                new_position = start_position + (length_of_subsegment - 1) - \
                               (current_position - start_position)

                if new_position < 0 or new_position >= number_of_elements:
                    continue

                is_blocked = False

                for element_index in range(start_position,
                                             start_position + length_of_subsegment):
                    if element_index in blocked_set:
                        is_blocked = True
                        break

                if is_blocked:
                    continue

                # Only process unvisited states
                if distances[new_position] == -1:
                    distances[new_position] = current_distance + 1
                    queue.append((new_position, current_distance + 1))

    return distances

Big(O) Analysis

Time Complexity
O(n^2)The algorithm explores possible line configurations using a breadth-first search approach. In the worst-case scenario, we might need to visit each possible configuration of the line. The number of reachable configurations depends on the length of the line n and the allowed reverse operations. Since the maximum number of visited configurations grows quadratically with the line length, the algorithm's time complexity is O(n^2). Each allowed move is considered, and the breadth-first search ensures that each reachable configuration is visited at most once.
Space Complexity
O(N)The algorithm uses a queue to explore possible line configurations in a breadth-first manner. In the worst-case scenario, the queue might contain a significant portion of all possible configurations of the line, each represented by its state (or a unique identifier corresponding to the configuration). Additionally, we need to keep track of the minimum number of flips required to reach each configuration seen so far, typically using a data structure like a hash map or an array of size N, where N is the total number of positions on the line. Therefore, the auxiliary space complexity is dominated by the size of the queue and the data structure used to track visited configurations, resulting in O(N) space complexity.

Edge Cases

CaseHow to Handle
k = 1 (trivial reverse)The solution should handle k=1 correctly, essentially flipping only the selected element, but may lead to infinite loop if not handled correctly.
k > n (reverse beyond array size)Adjust k to min(k, n) to handle reverse operations larger than the array size.
forbidden array contains starting positionIf the starting position is in the forbidden array, the distance is -1.
forbidden array contains consecutive numbers that creates unreachable regionsCheck if the forbidden array disconnects portions of the array, making some nodes unreachable, resulting in -1 for those nodes.
n is very large (potential memory issues)Optimize memory usage by using appropriate data structures like sets or boolean arrays instead of full integer arrays, and avoid unnecessary copies.
All positions are forbidden except the start positionReturn an array with -1 everywhere except start as no other positions are reachable.
Reaching the same index multiple times via different pathsUse a visited set to track visited nodes in BFS to avoid cycles and ensure shortest path is found.
Integer overflow when calculating mid-points during reversal for large indicesUse long type for midpoint calculations to avoid integer overflow, or use alternative calculation methods to prevent exceeding integer limits.