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array1.py
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array1.py
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import math
def fruits_into_baskets(fruits):
K = 2
if len(fruits) < 1:
return -1
start = 0
max_l = 0
dict_fruits = {}
for i in range(len(fruits)):
c = fruits[i]
if c not in dict_fruits:
dict_fruits[c] = 0
dict_fruits[c] += 1
while len(dict_fruits) > K:
l = fruits[start]
dict_fruits[l] -= 1
if dict_fruits[l] == 0:
del dict_fruits[l]
start += 1
max_l = max(max_l, i - start + 1)
return max_l
def non_repeat_substring(str):
if len(str) < 1:
return -1
max_l = 0
start = 0
dict_set = {}
for i in range(len(str)):
c = str[i]
if c in dict_set:
start = max(max_l, dict_set[c] + 1)
dict_set[c] = i
max_l = max(max_l, i - start + 1)
return max_l
def length_of_longest_substring(str, k):
# TODO: Write your code here
if len(str) < 1:
return -1
start, max_l, max_letter_count = 0,0,0
dict_map = {}
for i in range(len(str)):
l = str[i]
if l not in dict_map:
dict_map[l] = 0
dict_map[l] += 1
max_letter_count = max(max_letter_count, dict_map[l])
if(i- start + 1 - max_letter_count) > k:
f = str[start]
dict_map[f] -= 1
start += 1
max_l = max(max_l, i - start + 1)
return max_l
def length_of_longest_substring1(arr, k):
start, max_l, max_one_count = 0,0,0
for i in range(len(arr)):
if arr[i] == 1:
max_one_count +=1
if i - start + 1 - max_one_count > k:
if arr[start] == 1:
max_one_count -= 1
start += 1
max_l = max(max_l, i - start + 1)
return max_l
def triplet_sum_close_to_target(arr, target_sum):
smallest = math.inf
arr.sort()
for i in range(len(arr) - 2):
left = i+1
right = len(arr) - 1
while left < right:
t_diff = target_sum - arr[i] - arr[left] - arr[right]
if t_diff == 0:
return target_sum - t_diff
if abs(t_diff) < abs(smallest):
smallest = t_diff
if t_diff > 0:
left +=1
else:
right -= 1
return target_sum - smallest
def triplet_with_smaller_sum(arr, target):
arr.sort()
count = 0
for i in range(len(arr) - 2):
count += searchPair(arr, target-arr[i], i)
return count
def searchPair(arr, target_sum, first):
left = first + 1
right = len(arr) - 1
count = 0
while left < right:
if arr[left] + arr[right] < target_sum:
count += right - left
left += 1
else:
right =-1
return count
def find_permutation(str, pattern):
window_start, matched = 0, 0
char_frequency = {}
for chr in pattern:
if chr not in char_frequency:
char_frequency[chr] = 0
char_frequency[chr] += 1
# our goal is to match all the characters from the 'char_frequency' with the current window
# try to extend the range [window_start, window_end]
for window_end in range(len(str)):
right_char = str[window_end]
if right_char in char_frequency:
# decrement the frequency of matched character
char_frequency[right_char] -= 1
if char_frequency[right_char] == 0:
matched += 1
if matched == len(char_frequency):
return True
# shrink the window by one character
if window_end >= len(pattern) - 1:
left_char = str[window_start]
window_start += 1
if left_char in char_frequency:
if char_frequency[left_char] == 0:
matched -= 1
char_frequency[left_char] += 1
return False
def find_string_anagrams(str, pattern):
result_indexes = []
match = 0
start = 0
dict_map = {}
for p in pattern:
if p not in dict_map:
dict_map[p] = 0
dict_map[p] += 1
for i in range(len(str)):
l = str[i]
if l in dict_map:
dict_map[l] -= 1
if dict_map[l] == 0:
match += 1
if match == len(pattern):
result_indexes.append(start)
if i >= len(pattern) - 1:
c = str[start]
start += 1
if c in dict_map:
if dict_map[c] == 0:
match -=1
dict_map[c] += 1
return result_indexes
def find_permutation2(str, pattern):
window_start, matched, substr_start = 0, 0, 0
min_length = len(str) + 1
char_frequency = {}
for chr in pattern:
if chr not in char_frequency:
char_frequency[chr] = 0
char_frequency[chr] += 1
# try to extend the range [window_start, window_end]
for window_end in range(len(str)):
right_char = str[window_end]
if right_char in char_frequency:
char_frequency[right_char] -= 1
if char_frequency[right_char] >= 0: # Count every matching of a character
matched += 1
# Shrink the window if we can, finish as soon as we remove a matched character
while matched == len(pattern):
if min_length > window_end - window_start + 1:
min_length = window_end - window_start + 1
substr_start = window_start
left_char = str[window_start]
window_start += 1
if left_char in char_frequency:
# Note that we could have redundant matching characters, therefore we'll decrement the
# matched count only when the last occurrence of a matched character is going out of the window
if char_frequency[left_char] == 0:
matched -= 1
char_frequency[left_char] += 1
if min_length > len(str):
return ""
return str[substr_start:substr_start + min_length]
#print("Maximum number of fruits: " + str(fruits_into_baskets(['A', 'B', 'C', 'A', 'C'])))
#print("Length of the longest substring: " + str(non_repeat_substring("aabccbb")))
#print('Permutation exist: ' + str(find_permutation("aaacb", "abc")))
#print(find_string_anagrams("ppqp", "pq"))
print(find_permutation2("aabdec", "abc"))