Simple Solution in O(N) TIME and O(1) Space
Problem Description:\nThe task is to compute the minimum steps required based on the number of 1s and 0s in the string s.\nEvery time you encounter a 0, you cou
# Problem Description:\nThe task is to compute the minimum steps required based on the number of 1s and 0s in the string s.\nEvery time you encounter a 0, you count how many 1s have appeared before that position, which are stored in the white variable. The number of steps for each 0 is incremented by the number of 1s that have already been seen.\n\n# Variables Used:\n\nwhite: This variable tracks the number of 1s encountered so far as we traverse the string.\nres: This stores the total number of "steps" or movements required based on the problem\'s logic.\nj: This is the loop iterator, which tracks the position in the string s.\n# Core Logic:\n\nYou traverse the string s from left to right.\nEach time you encounter a 0, you add the current value of white to res. This means that the more 1s you have seen before a 0, the more steps are added to the result.\nEvery time you encounter a 1, you increment white by 1, keeping track of how many 1s have been encountered.\n\n# Time Complexity:\nO(n): We iterate through the string s once, making the time complexity linear in relation to the length of the string n.\n# Space Complexity:\nO(1): We only use a few variables (white, res, etc.), making the space complexity constant.\nThis approach efficiently computes the result with minimal space and time overhead.\n\n# Code\n```python3 []\nclass Solution:\n    def minimumSteps(self, s: str) -> int:\n        white=0\n        res=0\n        j=0\n        n=len(s)\n        while(j<n):\n            if s[j]==\'0\':\n                res+=white\n            white+=1 if s[j]==\'1\' else 0\n            j+=1\n            \n            \n        return res\n```\n# Code\n```cpp []\nclass Solution {\npublic:\n    int minimumSteps(string s) {\n        int white = 0;  // To count the number of \'1\'s\n        int res = 0;    // To store the result (minimum steps)\n        int n = s.length();\n        \n        for (int j = 0; j < n; ++j) {\n            if (s[j] == \'0\') {\n                res += white;  // Add the number of \'1\'s seen so far to the result\n            } else if (s[j] == \'1\') {\n                white++;  // Increment the count of \'1\'s\n            }\n        }\n        \n        return res;\n    }\n};\n```\n
['Two Pointers', 'String', 'Greedy', 'C++', 'Python3']