java easy solution
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public boolean canReach(String s, int minJump, int maxJump) {\n        if (s.charAt(s.length() - 1) == \'1\') return false;\n        boolean[] avaiable = new boolean[s.length()];\n        avaiable[0] = true;\n        int start;\n        int end = 0;\n        for (int i = 0; i < s.length(); ++i) {\n            if (!avaiable[i]) continue;\n            start = Math.max(i + minJump, end + 1);\n            end = Math.min(i + maxJump, s.length() - 1);\n            for (int j = start; j <= end; ++j) {\n                if (s.charAt(j) == \'0\') {\n                    avaiable[j] = true;\n                }\n                if (avaiable[avaiable.length - 1]) return true;\n            }\n        }\n        return avaiable[s.length() - 1];\n    }\n}\n```
['Java']