# 🚀Day 09/180 (Math) 50. Pow(x, n)(Leetcode)

[**50\. Pow(x, n)**](https://leetcode.com/problems/powx-n/description/)

[#180DaysOfDSA#](https://leetcode.com/problems/single-number/)[Da](https://leetcode.com/problems/hamming-distance/)[ilyCodingChallenge](https://leetcode.com/problems/single-number/) #LeetCodeJourney #GeeksforGeeks #CodingNinjas #Codechef #CodeForces #ContinuousLearning #TechCommunity

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1719595787026/8e47c66e-1da2-44be-b083-a427ed9bee24.png align="center")

```cpp
class Solution {
public:
    double myPow(double x, int n) {
        double ans = 1.0;
        if(n == 0 || x == 1) return 1;
        else if (x == 0  ) return 0;
        if(n<0)
        {
            x= 1/x;
            if(n == INT_MIN){
                n= INT_MAX;
                ans *= x;
            }
            else
                n = n *(-1);
        }
        while(n>0){
            if((n&1)==1)
                ans *=x;
            x*=x;
            n>>=1;
        }
        return ans;
        }
};
```

Let's perform a detailed dry run for `x = 2.0` and `n = 10`.

### Initial Setup

* `x = 2.0`
    
* `n = 10`
    
* `ans = 1.0`
    

### Iteration Details

1. **First Iteration:**
    
    * Current `n` value: 10 (even)
        
    * `x` squared: ( x = 2.0 \\times 2.0 = 4.0 )
        
    * `n` right-shifted: ( n = 10 &gt;&gt; 1 = 5 )
        
    * `ans` remains: ( ans = 1.0 )
        
2. **Second Iteration:**
    
    * Current `n` value: 5 (odd)
        
    * Update `ans`: ( ans = 1.0 \\times 4.0 = 4.0 )
        
    * `x` squared: ( x = 4.0 \\times 4.0 = 16.0 )
        
    * `n` right-shifted: ( n = 5 &gt;&gt; 1 = 2 )
        
    * Updated `ans`: ( ans = 4.0 )
        
3. **Third Iteration:**
    
    * Current `n` value: 2 (even)
        
    * `x` squared: ( x = 16.0 \\times 16.0 = 256.0 )
        
    * `n` right-shifted: ( n = 2 &gt;&gt; 1 = 1 )
        
    * `ans` remains: ( ans = 4.0 )
        
4. **Fourth Iteration:**
    
    * Current `n` value: 1 (odd)
        
    * Update `ans`: ( ans = 4.0 \\times 256.0 = 1024.0 )
        
    * `x` squared: ( x = 256.0 \\times 256.0 = 65536.0 )
        
    * `n` right-shifted: ( n = 1 &gt;&gt; 1 = 0 )
        
    * Updated `ans`: ( ans = 1024.0 )
        
5. **End of Loop:**
    
    * Current `n` value: 0
        
    * Exit the loop
        

### Final Output

* `ans = 1024.0`
    

Therefore, the function correctly calculates ( 2^{10} = 1024.0 ).