A non-empty array A consisting of N integers is given.
A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N − 1 and A[P − 1] < A[P] > A[P + 1].
For example, the following array A:
A[0] = 1 A[1] = 5 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2has exactly four peaks: elements 1, 3, 5 and 10.
You are going on a trip to a range of mountains whose relative heights are represented by array A, as shown in a figure below. You have to choose how many flags you should take with you. The goal is to set the maximum number of flags on the peaks, according to certain rules.
Flags can only be set on peaks. What's more, if you take K flags, then the distance between any two flags should be greater than or equal to K. The distance between indices P and Q is the absolute value |P − Q|.
For example, given the mountain range represented by array A, above, with N = 12, if you take:
- two flags, you can set them on peaks 1 and 5;
- three flags, you can set them on peaks 1, 5 and 10;
- four flags, you can set only three flags, on peaks 1, 5 and 10.
You can therefore set a maximum of three flags in this case.
Write a function:
def solution(A)
that, given a non-empty array A of N integers, returns the maximum number of flags that can be set on the peaks of the array.
For example, the following array A:
A[0] = 1 A[1] = 5 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2the function should return 3, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..400,000];
- each element of array A is an integer within the range [0..1,000,000,000].
def solution(A):
from math import sqrt
A_len = len(A)
next_peak = [-1] * A_len
peaks_count = 0
first_peak = -1
# Generate the information, where the next peak is.
for index in range(A_len-2, 0, -1):
if A[index] > A[index+1] and A[index] > A[index-1]:
next_peak[index] = index
peaks_count += 1
first_peak = index
else:
next_peak[index] = next_peak[index+1]
if peaks_count < 2:
# There is no peak or only one.
return peaks_count
max_flags = 1
max_min_distance = int(sqrt(A_len))
for min_distance in range(max_min_distance + 1, 1, -1):
# Try for every possible distance.
flags_used = 1
flags_have = min_distance-1 # Use one flag at the first peak
pos = first_peak
while flags_have > 0:
if (pos + min_distance >= A_len-1) or (next_peak[pos+min_distance] == -1):
break
pos = next_peak[pos+min_distance]
flags_used += 1
flags_have -= 1
max_flags = max(max_flags, flags_used)
return max_flags
def solution(A):
from math import sqrt
A_len = len(A)
next_peak = [-1] * A_len
peaks_count = 0
first_peak = -1
# Generate the information, where the next peak is.
for index in range(A_len-2, 0, -1):
if A[index] > A[index+1] and A[index] > A[index-1]:
next_peak[index] = index
peaks_count += 1
first_peak = index
else:
next_peak[index] = next_peak[index+1]
if peaks_count < 2:
# There is no peak or only one.
return peaks_count
max_flags = 1
max_min_distance = int(sqrt(A_len))
for min_distance in range(max_min_distance + 1, 1, -1):
# Try for every possible distance.
flags_used = 1
flags_have = min_distance-1 # Use one flag at the first peak
pos = first_peak
while flags_have > 0:
if (pos + min_distance >= A_len-1) or (next_peak[pos+min_distance] == -1):
break
pos = next_peak[pos+min_distance]
flags_used += 1
flags_have -= 1
max_flags = max(max_flags, flags_used)
return max_flags
def solution(A):
from math import sqrt
A_len = len(A)
next_peak = [-1] * A_len
peaks_count = 0
first_peak = -1
# Generate the information, where the next peak is.
for index in range(A_len-2, 0, -1):
if A[index] > A[index+1] and A[index] > A[index-1]:
next_peak[index] = index
peaks_count += 1
first_peak = index
else:
next_peak[index] = next_peak[index+1]
if peaks_count < 2:
# There is no peak or only one.
return peaks_count
max_flags = 1
max_min_distance = int(sqrt(A_len))
for min_distance in range(max_min_distance + 1, 1, -1):
# Try for every possible distance.
flags_used = 1
flags_have = min_distance-1 # Use one flag at the first peak
pos = first_peak
while flags_have > 0:
if (pos + min_distance >= A_len-1) or (next_peak[pos+min_distance] == -1):
break
pos = next_peak[pos+min_distance]
flags_used += 1
flags_have -= 1
max_flags = max(max_flags, flags_used)
return max_flags
The solution obtained perfect score.