feat(python): solutions

Author: hongbo-miao

JavaScript/0034. Find First and Last Position of Element in Sorted Array.js

@@ -55,7 +55,6 @@ const searchRange3 = (nums, target) => {
 
   while (l < r) {
     const m = Math.ceil((l + r) / 2);   // note using Math.ceil
-
     if (nums[m] > target) r = m - 1;
     else l = m;
   }
@@ -86,7 +85,6 @@ const searchRange = (nums, target) => {
 
   while (l < r) {
     const m = Math.ceil((l + r) / 2);   // note using Math.ceil
-
     if (nums[m] > target) r = m - 1;
     else l = m;
   }

JavaScript/0038. Count and Say.js

@@ -28,26 +28,26 @@
  * @return {string}
  */
 const countAndSay = (n) => {
-  let res = '1';
-  for (let i = 1; i < n; i++) {
-    res = say(res);
+  let s = '1';
+  for (let i = 0; i < n - 1; i++) {
+    s = say(s);
   }
-  return res;
+  return s;
 };
 
 const say = (s) => {
   let res = '';
   let count = 0;
-  let num = s[0];
+  let n = s[0];
 
   for (let i = 0; i < s.length; i++) {
-    if (s[i] === num) {
+    if (s[i] === n) {
       count++;
     } else {
       res += count + s[i - 1];
       count = 1;
-      num = s[i];
+      n = s[i];
     }
   }
-  return res + count + num;
+  return res + count + n;
 };

JavaScript/0041. First Missing Positive.js

@@ -47,6 +47,7 @@
 // i++
 // i = 3
 // i++
+// i = 4
 const firstMissingPositive = (nums) => {
   const swap = (i, j) => [nums[i], nums[j]] = [nums[j], nums[i]];
 

JavaScript/0069. Sqrt(x).js

@@ -57,8 +57,7 @@ const mySqrt = (x) => {
     const m = ~~((l + r) / 2);
 
     if (m * m === x) return m;
-
-    if (m * m < x) l = m + 1;
+    else if (m * m < x) l = m + 1;
     else r = m - 1;
   }
   return r;

Python/0005. Longest Palindromic Substring.py

@@ -49,10 +49,10 @@ class Solution:
     def longestPalindrome(self, s: str) -> str:
         res = ""
         for i in range(len(s)):
-            s1 = self.expend_from_center(s, i, i)
+            s1 = Solution.expend_from_center(s, i, i)
             if len(s1) > len(res):
                 res = s1
-            s2 = self.expend_from_center(s, i, i + 1)
+            s2 = Solution.expend_from_center(s, i, i + 1)
             if len(s2) > len(res):
                 res = s2
         return res

Python/0029. Divide Two Integers.py

@@ -0,0 +1,26 @@
+# Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator.
+# The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8, and -2.7335 would be truncated to -2.
+# Return the quotient after dividing dividend by divisor.
+#
+# Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For this problem, if the quotient is strictly greater than 2^31 - 1, then return 2^31 - 1, and if the quotient is strictly less than -2^31, then return -2^31.
+#
+# Example 1:
+#
+# Input: dividend = 10, divisor = 3
+# Output: 3
+# Explanation: 10/3 = 3.33333.. which is truncated to 3.
+#
+# Example 2:
+#
+# Input: dividend = 7, divisor = -3
+# Output: -2
+# Explanation: 7/-3 = -2.33333.. which is truncated to -2.
+#
+# Constraints:
+#
+# -2^31 <= dividend, divisor <= 2^31 - 1
+# divisor != 0
+
+
+class Solution:
+    def divide(self, dividend: int, divisor: int) -> int:

Python/0033. Search in Rotated Sorted Array.py

@@ -0,0 +1,98 @@
+# There is an integer array nums sorted in ascending order (with distinct values).
+# Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
+# Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
+# You must write an algorithm with O(log n) runtime complexity.
+#
+# Example 1:
+#
+# Input: nums = [4,5,6,7,0,1,2], target = 0
+# Output: 4
+#
+# Example 2:
+#
+# Input: nums = [4,5,6,7,0,1,2], target = 3
+# Output: -1
+#
+# Example 3:
+#
+# Input: nums = [1], target = 0
+# Output: -1
+#
+# Constraints:
+#
+# 1 <= nums.length <= 5000
+# -10^4 <= nums[i] <= 10^4
+# All values of nums are unique.
+# nums is an ascending array that is possibly rotated.
+# -10^4 <= target <= 10^4
+
+
+# 1) Binary search
+# https://leetcode.com/problems/search-in-rotated-sorted-array/discuss/273622/Javascript-Simple-O(log-N)-Binary-Search-Solution
+#
+# Time O(log n)
+# Space O(1)
+#
+# e.g. [1, 2, 3, 4, 5, 6, 7]
+#
+# When you divide the rotated array into two halves, using mid index, at least one of them should remain sorted ALWAYS.
+#
+# [3, 4, 5] [6, 7, 1, 2] the left side remains sorted
+# [6, 7, 1] [2, 3, 4, 5] the right side remains sorted
+# [1, 2, 3] [4, 5, 6, 7] Both sides remain sorted.
+#
+# If you know one side is sorted, the rest of logic becomes very simple.
+# If one side is sorted, check if the target is in the boundary, otherwise it's on the other side.
+#
+# IF smallest <= target <= biggest
+#   then target is here
+# ELSE
+#   then target is on the other side
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        l = 0
+        r = len(nums) - 1
+        while l <= r:
+            m = (l + r) // 2
+            if nums[m] == target:
+                return m
+
+            # When dividing the rotated array into two halves, one must be sorted
+            # Check if the left side is sorted
+            if nums[l] <= nums[m]:
+                if nums[l] <= target <= nums[m]:
+                    r = m - 1  # target is in the left
+                else:
+                    l = m + 1  # target is in the right
+            else:
+                if nums[m] <= target <= nums[r]:
+                    l = m + 1  # target is in the right
+                else:
+                    r = m - 1  # target is in the left
+        return -1
+
+
+# 2) Binary search, similar to 1)
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        l = 0
+        r = len(nums) - 1
+        while l + 1 < r:
+            m = (l + r) // 2
+            if nums[m] == target:
+                return m
+            if nums[l] < nums[m]:  # e.g. 4, 5, 6, 7
+                if nums[l] <= target <= nums[m]:
+                    r = m
+                else:
+                    l = m
+            else:  # e.g. 7, 0, 1, 2
+                if nums[m] <= target <= nums[r]:
+                    l = m
+                else:
+                    r = m
+        if nums[l] == target:
+            return l
+        if nums[r] == target:
+            return r
+        return -1

Python/0034. Find First and Last Position of Element in Sorted Array.py

@@ -0,0 +1,88 @@
+# Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.
+# If target is not found in the array, return [-1, -1].
+# You must write an algorithm with O(log n) runtime complexity.
+#
+# Example 1:
+#
+# Input: nums = [5,7,7,8,8,10], target = 8
+# Output: [3,4]
+#
+# Example 2:
+#
+# Input: nums = [5,7,7,8,8,10], target = 6
+# Output: [-1,-1]
+#
+# Example 3:
+#
+# Input: nums = [], target = 0
+# Output: [-1,-1]
+#
+# Constraints:
+#
+# 0 <= nums.length <= 10^5
+# -10^9 <= nums[i] <= 10^9
+# nums is a non-decreasing array.
+# -10^9 <= target <= 10^9
+
+
+# 1) Brute force / Linear scan
+# Time O(n)
+# Space O(1)
+
+
+# 2) Binary search
+# Time O(log n)
+# Space O(1)
+class Solution:
+    def searchRange(self, nums: List[int], target: int) -> List[int]:
+        l = Solution.lower_bound(nums, target)
+        r = Solution.upper_bound(nums, target) - 1
+        if l <= r:
+            return [l, r]
+        return [-1, -1]
+
+    @staticmethod
+    def lower_bound(nums: List[int], target: int) -> int:
+        l = 0
+        r = len(nums)
+        while l < r:
+            m = (l + r) // 2
+            if nums[m] < target:
+                l = m + 1
+            else:
+                r = m
+        return l
+
+    @staticmethod
+    def upper_bound(nums: List[int], target: int) -> int:
+        l = 0
+        r = len(nums)
+        while l < r:
+            m = (l + r) // 2
+            if nums[m] <= target:
+                l = m + 1
+            else:
+                r = m
+        return l
+
+
+# 3) Binary search, similar to 2)
+class Solution:
+    def searchRange(self, nums: List[int], target: int) -> List[int]:
+        l = Solution.lower_bound(nums, target)
+        r = Solution.lower_bound(nums, target + 1) - 1
+        if l <= r:
+            return [l, r]
+        return [-1, -1]
+
+    @staticmethod
+    def lower_bound(nums: List[int], target: int) -> int:
+        l = 0
+        r = len(nums)
+        while l < r:
+            m = (l + r) // 2
+            if nums[m] < target:
+                l = m + 1
+            else:
+                r = m
+        return l