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Program to count number of nice subarrays in Python
Suppose we have an array called nums and another value k. We have to find the number of nice sub-arrays. A subarray is said to be nice subarray if there are k odd numbers on it.
So, if the input is like nums = [1,1,2,1,1], k = 3, then the output will be 2 because there are two subarrays [1,1,2,1] and [1,2,1,1].
To solve this, we will follow these steps −
odd_i := a new list
for i in range 0 to size of nums - 1, do
if nums[i] mod 2 is same as 1, then
insert i at the end of odd_i
start := 0, end := k - 1
i := 0
count := 0
while end < size of odd_i, do
if end is same as size of odd_i - 1, then
j := size of nums - 1
otherwise,
j := odd_i[end + 1] - 1
count := count +(odd_i[start] - i + 1) *(j - odd_i[end] + 1)
i := odd_i[start] + 1
start := start + 1
end := end + 1
return count
Example
Let us see the following implementation to get better understanding −
def solve(nums, k):
odd_i = []
for i in range(len(nums)):
if nums[i] % 2 == 1:
odd_i.append(i)
start = 0
end = k - 1
i = 0
count = 0
while end < len(odd_i):
if end == len(odd_i) - 1:
j = len(nums) - 1
else:
j = odd_i[end + 1] - 1
count = count + (odd_i[start] - i + 1) * (j - odd_i[end] + 1)
i = odd_i[start] + 1
start = start + 1
end = end + 1
return count
nums = [1,1,2,1,1]
k = 3
print(solve(nums, k))Input
[1,2,3,4,5,6,7,8]
Output
2
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