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

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 = [0]*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

Arnab Chakraborty

Updated on: 16-Oct-2021

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