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consequtive number code -python✅
Consecutive numbers in an array -java ✅✅
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fibonacci triangle code python
int Children = 0,Adult = 0;
for(int i=0;i
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
dict1 = dict()
for i in nums1:
if i not in dict1:
dict1[i] = 1
else: @Coding_000
dict1[i] += 1
ret = []
for i in nums2:
if i in dict1 and dict1[i]>0:
ret.append(i)
dict1[i] -= 1
return ret
Python
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minimum product code -keka exam python
def cond1(s):
for i in range(len(s) - 2):
if (s[i] == s[i + 1] and
s[i + 1] == s[i + 2]):
return True
return False
def cond2(s):
for i in range(len(s) - 2):
if ((s[i] < s[i + 1] and
s[i + 1] < s[i + 2]) or
(s[i] > s[i + 1] and
s[i + 1] > s[i + 2])):
return True
return False
def cond3(s):
a = [0] * 10
for i in range(len(s)):
a[s[i] - '0'] = a[s[i] - '0'] + 1
for i in range(len(9)):
if (a[i] >= 4):
return True
return False
def isFancy(s):
if (cond1(s) or cond2(s) or cond3(s)):
return True
else:
return False
s = input(“ “)
if (isFancy(s)):
print("Yes")
else:
print("No")
Language: Python3
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
dict1 = dict()
for i in nums1:
if i not in dict1:
dict1[i] = 1
else:
dict1[i] += 1
ret = []
for i in nums2:
if i in dict1 and dict1[i]>0:
ret.append(i)
dict1[i] -= 1
return ret
Python3
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