# Program to check there is any forward path in circular cyclic list or not in Python

PythonServer Side ProgrammingProgramming

Suppose we have a circular list called nums. So the first and the last elements are neighbors. So starting from any index say i, we can move nums[i] number of steps forward if nums[i] is a positive value, otherwise backwards if it's a negative value. We have to check whether there is a cycle whose length is greater than one such that the path only goes forwards or only goes backwards.

So, if the input is like nums = [-1, 2, -1, 1, 2], then the output will be True, because there is a forward path [1 -> 3 -> 4 -> 1]

To solve this, we will follow these steps −

• n := size of nums
• if n is same as 0, then
• return False
• seen := an array of size n and fill with 0
• update nums by getting x mod n for each element x in nums
• iter := 0
• for i in range 0 to n - 1, do
• if nums[i] is same as 0, then
• go for next iteration
• iter := iter + 1
• pos := True
• neg := True
• curr := i
• Do the following repeatedly, do
• if nums[curr] and seen[curr] is same as iter, then
• return True
• if seen[curr] is non-zero, then
• come out from loop
• if nums[curr] > 0, then
• neg := False
• otherwise,
• pos := False
• if neg and pos both are false, then
• come out from the loop
• seen[curr] := iter
• curr := (curr + nums[curr] + n) mod n
• if nums[curr] is same as 0, then
• come out from the loop
• return False

## Example

Let us see the following implementation to get better understanding −

def solve(nums):
n = len(nums)
if n == 0:
return False
seen = *n
nums = [x % n for x in nums]
iter = 0
for i in range(n):
if nums[i] == 0:
continue
iter += 1
pos = True
neg = True
curr = i
while True:
if nums[curr] and seen[curr] == iter:
return True
if seen[curr] :
break
if nums[curr] > 0:
neg = False
else:
pos = False
if not neg and not pos:
break
seen[curr] = iter
curr = (curr + nums[curr] + n) % n
if nums[curr] == 0:
break
return False

nums = [-1, 2, -1, 1, 2]
print(solve(nums))

## Input

[-1, 2, -1, 1, 2]

## Output

True