5.18

Author: JiayangWu

0042.trapping-rain-water/trapping-rain-water.md

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+<p>Given <em>n</em> non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.</p>
+
+<p><img src="https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png" style="width: 412px; height: 161px;" /><br />
+<small>The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. <strong>Thanks Marcos</strong> for contributing this image!</small></p>
+
+<p><strong>Example:</strong></p>
+
+<pre>
+<strong>Input:</strong> [0,1,0,2,1,0,1,3,2,1,2,1]
+<strong>Output:</strong> 6</pre>

0042.trapping-rain-water/trapping-rain-water.py

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+class Solution(object):
+    def trap(self, height):
+        """
+        :type height: List[int]
+        :rtype: int
+        """
+        left_max = [0 for _ in height]
+        right_max = [0 for _ in height]
+        water = [0 for _ in height]
+        
+        for i in range(len(height)):
+            if i - 1 >= 0:
+                left_max[i] = max(left_max[i - 1], height[i])
+            else:
+                left_max[i] = height[i]
+            
+        for i in range(len(height) - 1, -1, -1):
+            if i  < len(height) - 1:
+                right_max[i] = max(right_max[i + 1], height[i])
+            else:
+                right_max[i] = height[i]
+                
+        for i in range(len(height)):
+            tmp = min(left_max[i], right_max[i]) - height[i]
+            if tmp > 0:
+                water[i] = tmp
+        # print height     
+        # print water
+        # print left_max
+        # print right_max
+        return sum(water)

0080.remove-duplicates-from-sorted-array-ii/remove-duplicates-from-sorted-array-ii.md

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+<p>Given a sorted array <em>nums</em>, remove the duplicates <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank"><strong>in-place</strong></a> such that duplicates appeared at most&nbsp;<em>twice</em> and return the new length.</p>
+
+<p>Do not allocate extra space for another array, you must do this by <strong>modifying the input array <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in-place</a></strong> with O(1) extra memory.</p>
+
+<p><strong>Example 1:</strong></p>
+
+<pre>
+Given <em>nums</em> = <strong>[1,1,1,2,2,3]</strong>,
+
+Your function should return length = <strong><code>5</code></strong>, with the first five elements of <em><code>nums</code></em> being <strong><code>1, 1, 2, 2</code></strong> and <strong>3</strong> respectively.
+
+It doesn&#39;t matter what you leave beyond the returned length.</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+Given <em>nums</em> = <strong>[0,0,1,1,1,1,2,3,3]</strong>,
+
+Your function should return length = <strong><code>7</code></strong>, with the first seven elements of <em><code>nums</code></em> being modified to&nbsp;<strong><code>0</code></strong>, <strong>0</strong>, <strong>1</strong>, <strong>1</strong>, <strong>2</strong>, <strong>3</strong> and&nbsp;<strong>3</strong> respectively.
+
+It doesn&#39;t matter what values are set beyond&nbsp;the returned length.
+</pre>
+
+<p><strong>Clarification:</strong></p>
+
+<p>Confused why the returned value is an integer but your answer is an array?</p>
+
+<p>Note that the input array is passed in by <strong>reference</strong>, which means modification to the input array will be known to the caller as well.</p>
+
+<p>Internally you can think of this:</p>
+
+<pre>
+// <strong>nums</strong> is passed in by reference. (i.e., without making a copy)
+int len = removeDuplicates(nums);
+
+// any modification to <strong>nums</strong> in your function would be known by the caller.
+// using the length returned by your function, it prints the first <strong>len</strong> elements.
+for (int i = 0; i &lt; len; i++) {
+&nbsp; &nbsp; print(nums[i]);
+}
+</pre>

0080.remove-duplicates-from-sorted-array-ii/remove-duplicates-from-sorted-array-ii.py

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+class Solution(object):
+    def removeDuplicates(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        i = 0
+        for num in nums:
+            if i < 2 or num != nums[i - 2]:
+                nums[i] = num
+                i += 1
+        return i

0084.largest-rectangle-in-histogram/largest-rectangle-in-histogram.md

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+<p>Given <em>n</em> non-negative integers representing the histogram&#39;s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.</p>
+
+<p>&nbsp;</p>
+
+<p><img src="https://assets.leetcode.com/uploads/2018/10/12/histogram.png" style="width: 188px; height: 204px;" /><br />
+<small>Above is a histogram where width of each bar is 1, given height = <code>[2,1,5,6,2,3]</code>.</small></p>
+
+<p>&nbsp;</p>
+
+<p><img src="https://assets.leetcode.com/uploads/2018/10/12/histogram_area.png" style="width: 188px; height: 204px;" /><br />
+<small>The largest rectangle is shown in the shaded area, which has area = <code>10</code> unit.</small></p>
+
+<p>&nbsp;</p>
+
+<p><strong>Example:</strong></p>
+
+<pre>
+<strong>Input:</strong> [2,1,5,6,2,3]
+<strong>Output:</strong> 10
+</pre>

0084.largest-rectangle-in-histogram/largest-rectangle-in-histogram.py

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+class Solution(object):
+    def largestRectangleArea(self, heights):
+        """
+        :type heights: List[int]
+        :rtype: int
+        """
+        res = 0
+        stack = list()
+        heights = [0] + heights + [0]
+        
+        for i in range(len(heights)):
+            while stack and heights[stack[-1]] > heights[i]:
+                top = stack.pop()
+                res = max(res, (i - stack[-1] - 1) * heights[top])
+                
+            stack.append(i)
+            
+        return res

0217.contains-duplicate/contains-duplicate.md

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+<p>Given an array of integers, find if the array contains any duplicates.</p>
+
+<p>Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.</p>
+
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> [1,2,3,1]
+<strong>Output:</strong> true</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input: </strong>[1,2,3,4]
+<strong>Output:</strong> false</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input: </strong>[1,1,1,3,3,4,3,2,4,2]
+<strong>Output:</strong> true</pre>

0217.contains-duplicate/contains-duplicate.py

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+class Solution(object):
+    def containsDuplicate(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: bool
+        """
+        if len(nums) <= 1:
+            return False
+        nums.sort()
+        for index in range(0,len(nums)-1):
+            if nums[index] == nums[index +1]:#or nums[index] == nums[index-1]:
+                return True
+        return False

0219.contains-duplicate-ii/contains-duplicate-ii.md

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+<p>Given an array of integers and an integer <i>k</i>, find out whether there are two distinct indices <i>i</i> and <i>j</i> in the array such that <b>nums[i] = nums[j]</b> and the <b>absolute</b> difference between <i>i</i> and <i>j</i> is at most <i>k</i>.</p>
+
+<div>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input: </strong>nums = <span id="example-input-1-1">[1,2,3,1]</span>, k = <span id="example-input-1-2">3</span>
+<strong>Output: </strong><span id="example-output-1">true</span>
+</pre>
+
+<div>
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input: </strong>nums = <span id="example-input-2-1">[1,0,1,1]</span>, k = <span id="example-input-2-2">1</span>
+<strong>Output: </strong><span id="example-output-2">true</span>
+</pre>
+
+<div>
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input: </strong>nums = <span id="example-input-3-1">[1,2,3,1,2,3]</span>, k = <span id="example-input-3-2">2</span>
+<strong>Output: </strong><span id="example-output-3">false</span>
+</pre>
+</div>
+</div>
+</div>

0219.contains-duplicate-ii/contains-duplicate-ii.py

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+class Solution(object):
+    def containsNearbyDuplicate(self, nums, k):
+        """
+        :type nums: List[int]
+        :type k: int
+        :rtype: bool
+        """
+        record = dict()
+        for i, num in enumerate(nums):
+            # print record
+            if record.get(num, -1) != -1:
+                if i - record[num] <= k:
+                    return True
+            record[num] = i
+        return False
+