Java Sliding Window Solution | Time Complexity O(n) | Space Complexity O(1)
Problem Explanation\n\nYou are given a binary string s. You can perform two types of operations on this string in any sequence:\n\n1. Type-1: Remove the charact
# Problem Explanation\n\nYou are given a binary string `s`. You can perform two types of operations on this string in any sequence:\n\n1. Type-1: Remove the character at the start of the string `s` and append it to the end of the string.\n2. Type-2: Pick any character in `s` and flip its value (0 becomes 1, and 1 becomes 0).\n\nThe goal is to find the minimum number of Type-2 operations required to make `s` an alternating string, meaning that no two adjacent characters in the string are equal.\n\n# Approach Explanation\n\n**Understanding Why `t = s + s`**\n\nOne key insight is that after performing `n` Type-1 operations on the string `s`, where `n` is the length of the string, we will return to the original string `s`. This is because performing Type-1 operations effectively shifts characters from the left end to the right end in a circular fashion. So, after `n` operations, we have explored all possible circular combinations of the original string.\n\nTo efficiently explore all these combinations, we create a new string `t` by concatenating `s` with itself (`t = s + s`). This effectively doubles the length of the string, ensuring that we have all possible combinations available for comparison.\n\n**Comparing Combinations**\n\nNext, we need to compare every possible combination of the string `s` with two resultant alternating strings: one where odd-indexed characters are \'1\' and even-indexed characters are \'0\' (let\'s call this Resultant String 1), and the other where odd-indexed characters are \'0\' and even-indexed characters are \'1\' (Resultant String 2).\n\nWe iterate through the characters of `t` while maintaining a sliding window of length `s.length()`. The sliding window technique helps us efficiently calculate the differences between the current combination and the two resultant strings.\n\n- We maintain two counters, `diff1` and `diff2`, to keep track of the differences from Resultant String 1 and Resultant String 2, respectively.\n- As we slide the window, we check if the current character at the right end (`rightChar`) is different from what it should be in both Resultant String 1 and Resultant String 2. If so, we increment `diff1` and `diff2` accordingly.\n- When the window size reaches `s.length()`, we update the `min` variable with the minimum of `diff1` and `diff2`, representing the minimum differences observed for this combination.\n\n**Sliding Window to Save Work**\n\nThe sliding window technique helps us save work by avoiding the need to recompute the entire combination. We only need to compare and update differences for the elements that enter and exit the window.\n\nFinally, we return the minimum value stored in the `min` variable, which represents the minimum number of Type-2 operations needed to make the string alternating.\n\nThis approach has a time complexity of O(n) as it efficiently explores all possible circular combinations without the need for nested loops, resulting in a faster and more memory-efficient solution.\n\n# Complexity Analysis\n\n- Time complexity: O(n)\n  - The algorithm iterates through the string `t` once, where `t` is twice the length of the input string `s`.\n\n- Space complexity: O(1)\n  - The algorithm uses a constant amount of extra space for variables and comparisons.\n\n# Code\n\n```java\nclass Solution {\n    public int minFlips(String s) {\n        // Create a doubled string to explore all circular combinations\n        String t = s + s;\n        \n        // Initialize variables for the sliding window and differences\n        int left = 0; // Left pointer of the window\n        int diff1 = 0; // Differences from Resultant String 1\n        int diff2 = 0; // Differences from Resultant String 2\n        int min = Integer.MAX_VALUE; // Initialize min to a large value\n        \n        // Iterate through the characters of the doubled string t\n        for (int right = 0; right < t.length(); right++) {\n            char rightChar = t.charAt(right);\n            \n            // Check if the right character at even index should be \'0\' and at odd index should be \'1\'\n            if ((right % 2 == 0 && rightChar != \'0\') || (right % 2 != 0 && rightChar != \'1\')) {\n                diff1++; // Increment differences for Resultant String 1\n            }\n\n            // Check if the right character at even index should be \'1\' and at odd index should be \'0\'\n            if ((right % 2 == 0 && rightChar != \'1\') || (right % 2 != 0 && rightChar != \'0\')) {\n                diff2++; // Increment differences for Resultant String 2\n            }\n\n            // Check if the sliding window size is equal to the length of the original string s\n            if (right - left + 1 >= s.length()) {\n                // Update min with the minimum of diff1 and diff2\n                min = Integer.min(min, Math.min(diff1, diff2));\n                \n                // Move the left pointer, and update differences for the character that leaves the window\n                char leftChar = t.charAt(left);\n                if ((left % 2 == 0 && leftChar != \'0\') || (left % 2 != 0 && leftChar != \'1\')) {\n                    diff1--;\n                }\n                if ((left % 2 == 0 && leftChar != \'1\') || (left % 2 != 0 && leftChar != \'0\')) {\n                    diff2--;\n                }\n                left++; // Increment the left pointer to slide the window\n            }\n        }\n        \n        return min; // Return the minimum number of Type-2 operations\n    }\n}\n
['Sliding Window', 'Java']