"Optimal Two-Pointer Approach for Minimizing Steps to Eliminate '1...0' Patterns in a Binary String"
Approach\n Describe your approach to solving the problem. \n1) Initialization:\n\nStart with two pointers i (left) and j (right).\nKeep a counter ans to accumul
# Approach\n<!-- Describe your approach to solving the problem. -->\n1) Initialization:\n\nStart with two pointers i (left) and j (right).\nKeep a counter ans to accumulate the minimum number of steps.\n\n2) Two-pointer traversal:\n\nWhile i is less than or equal to j:\nIf s[i] == \'1\' and s[j] == \'0\':\nThese two form the pattern "1...0".\nThe minimum number of steps to remove them is the distance between i and j, i.e., j - i.\nAdd j - i to the total answer.\nMove both pointers inward (i++ and j--).\nIf s[j] == \'1\':\nMove the right pointer leftward (j--) because we need to find a \'0\' for the \'1\'.\nIf s[i] == \'0\':\nMove the left pointer rightward (i++) because we need to find a \'1\' for the \'0\'.\n\n3) Return the result:\n\nOnce the pointers cross, return the accumulated steps (ans), which will be the minimum number of steps needed to eliminate all "1...0" patterns.\n\n# Complexity\n- Time complexity: O ( N )\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O ( 1 )\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    long long minimumSteps(string s) {\n        long long ans =\n            0; // Initialize the variable to hold the total number of steps.\n        int i = 0,\n            j = s.size() - 1; // Initialize two pointers: i points to the start,\n                              // j points to the end of the string.\n\n        // Loop until the two pointers cross each other.\n        while (i <= j) {\n            // Case 1: If there is a \'1\' at the start (i-th index) and a \'0\' at\n            // the end (j-th index).\n            if (s[i] == \'1\' && s[j] == \'0\') {\n                ans +=\n                    j -\n                    i; // Add the distance between i and j to the total steps.\n                i++;   // Move the left pointer (i) one step to the right.\n                j--;   // Move the right pointer (j) one step to the left.\n            }\n            // Case 2: If there is a \'1\' at the end (j-th index), move the right\n            // pointer leftwards.\n            else if (s[j] == \'1\')\n                j--;\n            // Case 3: If there is a \'0\' at the start (i-th index), move the\n            // left pointer rightwards.\n            else if (s[i] == \'0\')\n                i++;\n        }\n        return ans; // Return the total number of steps required.\n    }\n};\n\n```
['Two Pointers', 'String']