Efficient Backward Traversal to Count Minimum Steps Between '1's and '0's
Intuition\nThe problem seems to involve counting the number of steps between certain characters in the string. After thinking about it, a backward traversal mig
# Intuition\nThe problem seems to involve counting the number of steps between certain characters in the string. After thinking about it, a backward traversal might help in efficiently counting the minimum number of steps needed.\n\n# Approach\n- We traverse the string from right to left.\n- We maintain a `count` to keep track of how many `\'0\'` characters we have encountered so far.\n- For each `\'1\'`, we increment the `out` value by the current `count` because each `\'1\'` contributes steps equal to the number of `\'0\'`s to its right.\n- This way, we can calculate the minimum number of steps in a single pass from right to left.\n\n# Complexity\n- Time complexity:  \n  $$O(n)$$ where `n` is the length of the string. We only need a single pass through the string.\n\n- Space complexity:  \n  $$O(1)$$ since we are only using a few variables for counting and summing.\n\n# Code\n```javascript []\n/**\n * @param {string} s\n * @return {number}\n */\nvar minimumSteps = function(s) {\n    let count =0 , out = 0;\n    for( let i = s.length-1; i >= 0; i-- ){\n        if( s[i] == \'0\') count++\n        else out+=count\n    }\n    return out \n};\n```\n\n```TypeScript []\nfunction minimumSteps(s: string): number {\n    let count : number = 0;\n    let out : number = 0;\n    for( let i = s.length-1; i >= 0; i-- ){\n        if( s[i] == \'0\') count++\n        else out += count\n    }\n    return out \n};\n```
['TypeScript', 'JavaScript']