EASY !! EXPLAINED SOLUTION 100%
Intuition\nThe given Java code defines a class Solution with a method canReach that aims to determine whether it\'s possible to reach the end of a string contai
# Intuition\nThe given Java code defines a class Solution with a method canReach that aims to determine whether it\'s possible to reach the end of a string containing 0s and 1s by following certain jump rules.\n\n# Approach\n- Check if the input string s is null or empty. If it is, return false immediately, as there\'s nothing to traverse.\n- Retrieve the length of the input string n.\n- Check if the first or last character of the string is \'1\'. If either of them is \'1\', it means you can\'t start from the beginning or reach the end, so return false.\n- Create a boolean array dp of length n, where each element represents whether it\'s possible to reach that position from the starting position.\n- Initialize sb[0] to true since you can start at the first position.\n- Initialize two variables, start and end, which will be used to define a sliding window of indices that you will examine during traversal.\n- Loop through the string from the beginning:\n    - If sb[i] is false, skip to the next iteration since this position cannot be reached.\n    - Calculate the new values for start and end based on the current index i, minJump, and maxJump. This defines the range of positions that can be reached from the current position.\n    - Iterate through positions from start to end. If the character at that position is \'0\', set sb[j] to true, indicating that you can reach that position.\n    - Check if you can already reach the end position. If so, return true.\n- After the loop, return the value of sb[n - 1], indicating whether it\'s possible to reach the last position.\n# Complexity\n- Time complexity:\n**O(n)**\n\n- Space complexity:\n**O(n)**\n\n# Code\n```\nclass Solution {\n    public boolean canReach(String s, int minJump, int maxJump) {\n        if(s == null || s.length() == 0) return false;\n        int n = s.length();\n        if(s.charAt(0) == \'1\' || s.charAt(n - 1) == \'1\') return false;\n\n        boolean[] sb = new boolean[n];\n        sb[0] = true;\n\n        int start = 0; \n        int end = 0;  \n        for(int i = 0; i < n; i++){\n            if(!sb[i]){   \n                continue;  \n            }\n            start = Math.max(end + 1, i + minJump);  \n            end = Math.min(n - 1, i + maxJump);  \n            for(int j = start; j <= end; j++){\n                if(s.charAt(j) == \'0\'){\n                    sb[j] = true; \n                }                \n            }\n            if(sb[n - 1]) return true;\n        }\n        return sb[n-1]; \n    }\n}\n```
['Array', 'String', 'Sliding Window', 'Java']