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Program to Find Minimum Jumps Required to Reach a Value with Different Parity in Python
Suppose, we are provided with a list of numbers called nums. Here we can jump to index i + numbers[i] or to i − numbers[i] from index i if the values exist in the list. So we have to find the number of jumps required at least to reach another value with different parity keeping the order in the input unchanged. If we cannot reach another number with different parity, it is set to −1.
So, if the input is like numbers = [7, 3, 4, 5, 6, 9, 6, 7], then the output will be [−1, 1, 2, −1, −1, −1, 1, −1].
To solve this, we will follow these steps −
Define a function bfs() . This will take i
q := a new double ended queue with a pair (i, 0)
seen := a new set
while q is not empty, do
(j, d) := left most element of q and delete the leftmost item from q
add j into seen
if (nums[i] + nums[j]) mod 2 is non−zero, then
return d
for each k in [j + nums[j], j − nums[j]], do
if 0 <= k < size of nums and k is not in seen, then
insert (k, d + 1) at the right end of q
return 10^10
From the main method do the following −
ans := a new list
for i in range 0 to size of nums, do
seen := a new set
x := bfs(i)
append x in ans when x < 10^10 otherwise append −1
return ans
Let us see the following implementation to get better understanding −
Example
from collections import deque class Solution: def solve(self, nums): def bfs(i): q = deque([(i, 0)]) seen = set() while q: j, d = q.popleft() seen.add(j) if (nums[i] + nums[j]) % 2: return d for k in [j + nums[j], j − nums[j]]: if 0 <= k < len(nums) and k not in seen: q.append((k, d + 1)) return 10 ** 10 ans = [] for i in range(len(nums)): seen = set() x = bfs(i) ans.append(x if x < 10 ** 10 else −1) return ans ob = Solution() print(ob.solve([7, 3, 4, 5, 6, 9, 6, 7]))
Input
numbers = [7, 3, 4, 5, 6, 9, 6, 7]
Output
[−1, 1, 2, −1, −1, −1, 1, −1]
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