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
:
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:
Example 2:
Input: n = 5, p = 0, banned = [2,4], k = 3 Output: [0,-1,-1,-1,-1]
Explanation:
[0, 2]
because position 2 is in banned.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
banned
are uniqueWhen 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:
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:
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
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:
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
Case | How 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 position | If the starting position is in the forbidden array, the distance is -1. |
forbidden array contains consecutive numbers that creates unreachable regions | Check 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 position | Return an array with -1 everywhere except start as no other positions are reachable. |
Reaching the same index multiple times via different paths | Use 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 indices | Use long type for midpoint calculations to avoid integer overflow, or use alternative calculation methods to prevent exceeding integer limits. |