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# Sqrt(x) LeetCode Solution

## Problem – Sqrt(x)

Given a non-negative integer `x`, compute and return the square root of `x`.

Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.

Note: You are not allowed to use any built-in exponent function or operator, such as `pow(x, 0.5)` or `x ** 0.5`.

Example 1:

``````Input: x = 4
Output: 2
``````

Example 2:

``````Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.``````

Constraints:

• `0 <= x <= 231 - 1`

### Sqrt(x) LeetCode Solution in Python

``````    r = x
while r*r > x:
r = (r + x/r) / 2
return r``````

### Sqrt(x) LeetCode Solution in Java

``````public int mySqrt(int x) {
if (x == 0) return 0;
int start = 1, end = x;
while (start < end) {
int mid = start + (end - start) / 2;
if (mid <= x / mid && (mid + 1) > x / (mid + 1))// Found the result
return mid;
else if (mid > x / mid)// Keep checking the left part
end = mid;
else
start = mid + 1;// Keep checking the right part
}
return start;
}
``````

### Sqrt(x) LeetCode Solution in C++

``````long long s=0, e=x, ans, mid;   //long long due to some of test cases overflows integer limit.
while(s<=e){
mid=(s+e)/2;
if(mid*mid==x) return mid;     //if the 'mid' value ever gives the result, we simply return it.
else if(mid*mid<x){
s=mid+1;        //if 'mid' value encounterted gives lower result, we simply discard all the values lower than mid.
ans=mid;        //an extra pointer 'ans' is maintained to keep track of only lowest 'mid' value.
}
else e=mid-1;       //if 'mid' value encountered gives greater result, we simply discard all the values greater than mid.
}
return ans;
``````
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