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/*
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    Time Complexity : O(logN), Since we are going through the entire number
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   digit by digit, the time complexity should be O(log10N). The reason behind
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   log10 is because we are dealing with integers which are base 10.
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    Space Complexity : O(1), We are not using any data structure for interim
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   operations, therefore, the space complexity is O(1).
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    Solved using String.
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*/
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class Solution {
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 public:
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  int myAtoi(string s) {
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    int len = s.size();
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    double num = 0;
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    int i = 0;
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    while (s[i] == ' ') {
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      i++;
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    }
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    bool positive = s[i] == '+';
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    bool negative = s[i] == '-';
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    positive == true ? i++ : i;
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    negative == true ? i++ : i;
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    while (i < len && s[i] >= '0' && s[i] <= '9') {
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      num = num * 10 + (s[i] - '0');
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      i++;
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    }
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    num = negative ? -num : num;
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    cout << num << endl;
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    num = (num > INT_MAX) ? INT_MAX : num;
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    num = (num < INT_MIN) ? INT_MIN : num;
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    cout << num << endl;
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    return int(num);
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  }
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};