sqrt x: done

Author: rampatra

src/main/java/com/leetcode/arrays/binarysearch/PowXN.java

@@ -63,7 +63,7 @@ public static double myPowNaive(double x, int n) {
      */
     public static double myPow(double x, int n) {
         double res = 1;
-        long absN = Math.abs((long) n);
+        long absN = Math.abs((long) n); // used a long so that `absN / 2` doesn't overflow
 
         while (absN > 0) {
             if (absN % 2 == 1) res *= x; // store whatever we have calculated so far in the final result

src/main/java/com/leetcode/arrays/binarysearch/SearchInsertPosition.java

@@ -26,7 +26,7 @@
  * Example 4:
  * Input: [1,3,5,6], 0
  * Output: 0
- *
+ * <p>
  * Similar question: {@link SmallestLetterGreaterThanTarget}.
  *
  * @author rampatra
@@ -59,5 +59,6 @@ public static int searchInsert(int[] nums, int target) {
 
     public static void main(String[] args) {
         assertEquals(2, searchInsert(new int[]{1, 2}, 3));
+        assertEquals(1, searchInsert(new int[]{1, 3, 5, 6}, 2));
     }
 }

src/main/java/com/leetcode/arrays/binarysearch/SqrtX.java

@@ -0,0 +1,62 @@
+package com.leetcode.arrays.binarysearch;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+/**
+ * Level: Easy
+ * Link: https://leetcode.com/problems/sqrtx/
+ * Description:
+ * Implement int sqrt(int x).
+ *
+ * Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
+ *
+ * Since the return type is an integer, the decimal digits are truncated and only the integer part
+ * of the result is returned.
+ *
+ * Example 1:
+ * Input: 4
+ * Output: 2
+ *
+ * Example 2:
+ * Input: 8
+ * Output: 2
+ * Explanation: The square root of 8 is 2.82842..., and since
+ *              the decimal part is truncated, 2 is returned.
+ *
+ * @author rampatra
+ * @since 2019-08-19
+ */
+public class SqrtX {
+
+    /**
+     * Runtime: <a href="https://leetcode.com/submissions/detail/253282868/">1 ms</a>.
+     *
+     * @param x
+     * @return
+     */
+    public static int mySqrt(int x) {
+        if (x == 0 || x == 1) {
+            return x;
+        }
+        long low = 1;
+        long high = x / 2;
+
+        while (low <= high) {
+            long mid = low + (high - low) / 2;
+            if (mid * mid == x) {
+                return (int) mid;
+            } else if (mid * mid < x) {
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+        return (int) high;
+    }
+
+    public static void main(String[] args) {
+        assertEquals(2, mySqrt(8));
+        assertEquals(3, mySqrt(9));
+        assertEquals(46339, mySqrt(2147395599));
+    }
+}

src/main/java/com/leetcode/strings/AnagramsInString.java

@@ -5,8 +5,40 @@
 import java.util.List;
 
 /**
- * Level: Easy
+ * Level: Medium
  * Problem: https://leetcode.com/problems/find-all-anagrams-in-a-string/
+ * Description:
+ * Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
+ *
+ * Strings consists of lowercase English letters only and the length of both strings s and p will not be larger
+ * than 20,100.
+ *
+ * The order of output does not matter.
+ *
+ * Example 1:
+ *
+ * Input:
+ * s: "cbaebabacd" p: "abc"
+ *
+ * Output:
+ * [0, 6]
+ *
+ * Explanation:
+ * The substring with start index = 0 is "cba", which is an anagram of "abc".
+ * The substring with start index = 6 is "bac", which is an anagram of "abc".
+ *
+ * Example 2:
+ *
+ * Input:
+ * s: "abab" p: "ab"
+ *
+ * Output:
+ * [0, 1, 2]
+ *
+ * Explanation:
+ * The substring with start index = 0 is "ab", which is an anagram of "ab".
+ * The substring with start index = 1 is "ba", which is an anagram of "ab".
+ * The substring with start index = 2 is "ab", which is an anagram of "ab".
  *
  * @author rampatra
  * @since 2019-04-13