Clean, Fast Python3 | 2 Pointer | O(n) Time & Space
Please upvote if it helps!\n```\nclass Solution:\n    def canReach(self, s: str, minJump: int, maxJump: int) -> bool:\n        if s[-1] == \'1\':\n            r
Please upvote if it helps!\n```\nclass Solution:\n    def canReach(self, s: str, minJump: int, maxJump: int) -> bool:\n        if s[-1] == \'1\':\n            return False\n        n, end = len(s), minJump\n        reach = [True] + [False] * (n - 1)\n        for i in range(n):\n            if reach[i]:\n                start, end = max(i + minJump, end), min(i + maxJump + 1, n)\n                for j in range(start, end):\n                    if s[j] == \'0\':\n                        reach[j] = True\n                if end == n:\n                    return reach[-1]\n        return reach[-1]
['Python3']