<p>Given two integers <code>dividend</code> and <code>divisor</code>, divide two integers <strong>without</strong> using multiplication, division, and mod operator.</p>

<p>The integer division should truncate toward zero, which means losing its fractional part. For example, <code>8.345</code> would be truncated to <code>8</code>, and <code>-2.7335</code> would be truncated to <code>-2</code>.</p>

<p>Return <em>the <strong>quotient</strong> after dividing </em><code>dividend</code><em> by </em><code>divisor</code>.</p>

<p><strong>Note: </strong>Assume we are dealing with an environment that could only store integers within the <strong>32-bit</strong> signed integer range: <code>[&minus;2<sup>31</sup>, 2<sup>31</sup> &minus; 1]</code>. For this problem, if the quotient is <strong>strictly greater than</strong> <code>2<sup>31</sup> - 1</code>, then return <code>2<sup>31</sup> - 1</code>, and if the quotient is <strong>strictly less than</strong> <code>-2<sup>31</sup></code>, then return <code>-2<sup>31</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> dividend = 10, divisor = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> 10/3 = 3.33333.. which is truncated to 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> dividend = 7, divisor = -3
<strong>Output:</strong> -2
<strong>Explanation:</strong> 7/-3 = -2.33333.. which is truncated to -2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>-2<sup>31</sup> &lt;= dividend, divisor &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>divisor != 0</code></li>
</ul>

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (dividend==divisor) return 1;
        int sign=1;
        long long ans=0;
        if ((dividend>0 && divisor<0) || (dividend<0 && divisor>0)) sign=-1;
        long long n=abs(dividend);
        long long d=abs(divisor);
        while (n>=d)
        {
            int cnt=0;
            while(n>=d<<(cnt+1)){
                cnt++;
            }
            n=n-(d<<cnt);
            ans=ans+(1<<cnt);
        }
        cout<<-ans;
        if (ans==(1<<31) && sign==1) return INT_MAX;
        if (ans==(1<<31) && sign==-1) return INT_MIN;
        // return sign ? ans : -ans;
        if (sign==-1) return -ans;
        else return ans;
    }
};
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