solve problem

leet-code/69-sqrt-x/README.md

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+# 69-sqrt-x
+
+## Problem Statement
+
+Given a non-negative integer `x`, return the square root of `x` **rounded down** to the nearest integer.  
+The returned integer should be non-negative as well.
+
+**Do not use** any built-in exponent function or operator.  
+For example, do not use `pow(x, 0.5)` in C++ or `x ** 0.5` in Python.
+
+### Example 1:
+
+**Input:**  
+`x = 4`
+
+**Output:**  
+`2`
+
+**Explanation:**  
+The square root of 4 is 2, so we return 2.
+
+### Example 2:
+
+**Input:**  
+`x = 8`
+
+**Output:**  
+`2`
+
+**Explanation:**  
+The square root of 8 is approximately 2.82842...  
+Rounding down gives 2.
+
+### Constraints:
+
+- `0 <= x <= 2^31 - 1`

leet-code/69-sqrt-x/index.js

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+function initialGuess(x) {
+  const log2S = Math.log2(x);
+  const exponent = Math.floor(log2S / 2);
+  return Math.pow(2, exponent);
+}
+
+/**
+ * Approximating Square Roots with Newton's Method
+ * f(x)=x^2−S
+ * x_n+1 = 1/2 * (x_n + S / x_n)
+ * 
+ * @param {number} x
+ * @return {number}
+ */
+var mySqrt = function(x) {
+    // short circuit on edge case
+    if (x === 0) return x
+    // speed up convergence with a reasonable guess
+    x0 = initialGuess(x)
+    // store the next tabulated value
+    let xn = x0;
+    // 32bit signed ints only need about 6 iterations to converge
+    for (var i = 0; i < 6;  i++) {
+        // apply newtons method to find the root of the function
+        xn = (xn + (x/xn))/2
+    }
+    // floor the approximation
+    return Math.floor(xn)
+};