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	<title>reversed pairs Archives - Huahua&#039;s Tech Road</title>
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		<title>花花酱 LeetCode 315. Count of Smaller Numbers After Self</title>
		<link>https://zxi.mytechroad.com/blog/algorithms/array/leetcode-315-count-of-smaller-numbers-after-self/</link>
					<comments>https://zxi.mytechroad.com/blog/algorithms/array/leetcode-315-count-of-smaller-numbers-after-self/#respond</comments>
		
		<dc:creator><![CDATA[zxi]]></dc:creator>
		<pubDate>Thu, 04 Jan 2018 05:59:12 +0000</pubDate>
				<category><![CDATA[Array]]></category>
		<category><![CDATA[binary indexed tree]]></category>
		<category><![CDATA[BST]]></category>
		<category><![CDATA[fenwick tree]]></category>
		<category><![CDATA[hard]]></category>
		<category><![CDATA[reversed pairs]]></category>
		<guid isPermaLink="false">http://zxi.mytechroad.com/blog/?p=1487</guid>

					<description><![CDATA[<p>题目大意：给你一个数组，对于数组中的每个元素，返回一共有多少在它之后的元素比它小。 Problem: You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to&#8230;</p>
<p>The post <a rel="nofollow" href="https://zxi.mytechroad.com/blog/algorithms/array/leetcode-315-count-of-smaller-numbers-after-self/">花花酱 LeetCode 315. Count of Smaller Numbers After Self</a> appeared first on <a rel="nofollow" href="https://zxi.mytechroad.com/blog">Huahua&#039;s Tech Road</a>.</p>
]]></description>
										<content:encoded><![CDATA[<p><iframe width="500" height="375" src="https://www.youtube.com/embed/2SVLYsq5W8M?feature=oembed" frameborder="0" allow="autoplay; encrypted-media" allowfullscreen></iframe></p>
<p>题目大意：给你一个数组，对于数组中的每个元素，返回一共有多少在它之后的元素比它小。</p>
<p><strong>Problem:</strong></p>
<p>You are given an integer array <i>nums</i> and you have to return a new <i>counts</i> array. The <i>counts</i> array has the property where <code>counts[i]</code> is the number of smaller elements to the right of <code>nums[i]</code>.</p>
<p><b>Example:</b></p><pre class="crayon-plain-tag">Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.</pre><p>Return the array <code>[2, 1, 1, 0]</code>.</p>
<p><strong>Idea:</strong></p>
<p>Fenwick Tree / Binary Indexed Tree</p>
<p>BST</p>
<p><img class="alignnone size-full wp-image-1515" src="http://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-1.png" alt="" width="960" height="540" srcset="https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-1.png 960w, https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-1-300x169.png 300w, https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-1-768x432.png 768w" sizes="(max-width: 960px) 100vw, 960px" /></p>
<p><img class="alignnone size-full wp-image-1514" src="http://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-2.png" alt="" width="960" height="540" srcset="https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-2.png 960w, https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-2-300x169.png 300w, https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-2-768x432.png 768w" sizes="(max-width: 960px) 100vw, 960px" /></p>
<p><img class="alignnone size-full wp-image-1513" src="http://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-3.png" alt="" width="960" height="540" srcset="https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-3.png 960w, https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-3-300x169.png 300w, https://zxi.mytechroad.com/blog/wp-content/uploads/2018/01/315-ep149-3-768x432.png 768w" sizes="(max-width: 960px) 100vw, 960px" /></p>
<h1><strong>Solution 1: Binary Indexed Tree (Fenwick Tree)</strong></h1>
<p><div class="responsive-tabs">
<h2 class="tabtitle">C++</h2>
<div class="tabcontent">
</p><pre class="crayon-plain-tag">// Author: Huahua
// Runnning time: 32 ms
// Time complexity: O(nlogn)
// Space complexity: O(k), k is the number unique elements
class FenwickTree {    
public:
    FenwickTree(int n): sums_(n + 1, 0) {}
    
    void update(int i, int delta) {
        while (i &lt; sums_.size()) {
            sums_[i] += delta;
            i += lowbit(i);
        }
    }
    
    int query(int i) const {        
        int sum = 0;
        while (i &gt; 0) {
            sum += sums_[i];
            i -= lowbit(i);
        }
        return sum;
    }
private:
    static inline int lowbit(int x) { return x &amp; (-x); }
    vector&lt;int&gt; sums_;
};

class Solution {
public:
    vector&lt;int&gt; countSmaller(vector&lt;int&gt;&amp; nums) {
        // Sort the unique numbers
        set&lt;int&gt; sorted(nums.begin(), nums.end());
        // Map the number to its rank
        unordered_map&lt;int, int&gt; ranks;
        int rank = 0;
        for (const int num : sorted)
            ranks[num] = ++rank;
        
        vector&lt;int&gt; ans;
        FenwickTree tree(ranks.size());
        // Scan the numbers in reversed order
        for (int i = nums.size() - 1; i &gt;= 0; --i) {
            // Chechk how many numbers are smaller than the current number.
            ans.push_back(tree.query(ranks[nums[i]] - 1));
            // Increse the count of the rank of current number.
            tree.update(ranks[nums[i]], 1);
        }
        
        std::reverse(ans.begin(), ans.end());
        return ans;
    }
};</pre><p></div><h2 class="tabtitle">Java</h2>
<div class="tabcontent">
</p><pre class="crayon-plain-tag">// Author: Huahua
// Running time: 28 ms
class Solution {
  private static int lowbit(int x) { return x &amp; (-x); }
  
  class FenwickTree {    
    private int[] sums;    
    
    public FenwickTree(int n) {
      sums = new int[n + 1];
    }

    public void update(int i, int delta) {    
      while (i &lt; sums.length) {
          sums[i] += delta;
          i += lowbit(i);
      }
    }

    public int query(int i) {       
      int sum = 0;
      while (i &gt; 0) {
          sum += sums[i];
          i -= lowbit(i);
      }
      return sum;
    }    
  };
  
  public List&lt;Integer&gt; countSmaller(int[] nums) {
    int[] sorted = Arrays.copyOf(nums, nums.length);
    Arrays.sort(sorted);
    Map&lt;Integer, Integer&gt; ranks = new HashMap&lt;&gt;();
    int rank = 0;
    for (int i = 0; i &lt; sorted.length; ++i)
      if (i == 0 || sorted[i] != sorted[i - 1])
        ranks.put(sorted[i], ++rank);
    
    FenwickTree tree = new FenwickTree(ranks.size());
    List&lt;Integer&gt; ans = new ArrayList&lt;Integer&gt;();
    for (int i = nums.length - 1; i &gt;= 0; --i) {
      int sum = tree.query(ranks.get(nums[i]) - 1);      
      ans.add(tree.query(ranks.get(nums[i]) - 1));
      tree.update(ranks.get(nums[i]), 1);
    }
    
    Collections.reverse(ans);
    return ans;
  }
}</pre><p></div></div></p>
<h1><strong>Solution 2: BST</strong></h1>
<p><div class="responsive-tabs">
<h2 class="tabtitle">C++</h2>
<div class="tabcontent">
</p><pre class="crayon-plain-tag">// Author: Huahua
// Running time: 26 ms
struct BSTNode {
    int val;
    int count;
    int left_count;
    BSTNode* left;
    BSTNode* right;    
    BSTNode(int val): val(val), count(1), left_count(0), left{nullptr}, right{nullptr} {}
    ~BSTNode() { delete left; delete right; }
    int less_or_equal() const { return count + left_count; }
};
class Solution {
public:
    vector&lt;int&gt; countSmaller(vector&lt;int&gt;&amp; nums) {
        if (nums.empty()) return {};
        std::reverse(nums.begin(), nums.end());
        std::unique_ptr&lt;BSTNode&gt; root(new BSTNode(nums[0]));
        vector&lt;int&gt; ans{0};
        for (int i = 1; i &lt; nums.size(); ++i)
            ans.push_back(insert(root.get(), nums[i]));
        std::reverse(ans.begin(), ans.end());
        return ans;
    }
private:
    // Returns the number of elements smaller than val under root.
    int insert(BSTNode* root, int val) {
        if (root-&gt;val == val) {
            ++root-&gt;count;
            return root-&gt;left_count;
        } else if (val &lt; root-&gt;val) {
            ++root-&gt;left_count;
            if (root-&gt;left == nullptr) {
                root-&gt;left = new BSTNode(val);            
                return 0;
            } 
            return insert(root-&gt;left, val);
        } else  {
            if (root-&gt;right == nullptr) {
                root-&gt;right = new BSTNode(val);
                return root-&gt;less_or_equal();
            }
            return root-&gt;less_or_equal() + insert(root-&gt;right, val);
        }
    }
};</pre><p></div><h2 class="tabtitle">Java</h2>
<div class="tabcontent">
</p><pre class="crayon-plain-tag">// Author: Huahua
// Running time: 10 ms
class Solution {
  class Node {
    int val;
    int count;
    int left_count;
    Node left;
    Node right;
    public Node(int val) { this.val = val; this.count = 1; }
    public int less_or_equal() { return count + left_count; }
  }
  
  public List&lt;Integer&gt; countSmaller(int[] nums) {
    List&lt;Integer&gt; ans = new ArrayList&lt;&gt;();
    if (nums.length == 0) return ans;
    int n = nums.length;
    Node root = new Node(nums[n - 1]);
    ans.add(0);
    for (int i = n - 2; i &gt;= 0; --i)
      ans.add(insert(root, nums[i]));
    Collections.reverse(ans);
    return ans;
  }
  
  private int insert(Node root, int val) {
    if (root.val == val) {
      ++root.count;
      return root.left_count;
    } else if (val &lt; root.val) {
      ++root.left_count;
      if (root.left == null) {
        root.left = new Node(val);            
        return 0;
      } 
      return insert(root.left, val);
    } else  {
      if (root.right == null) {
        root.right = new Node(val);
        return root.less_or_equal();
      }
      return root.less_or_equal() + insert(root.right, val);
    }
  }
}</pre><p></div></div></p>
<p>The post <a rel="nofollow" href="https://zxi.mytechroad.com/blog/algorithms/array/leetcode-315-count-of-smaller-numbers-after-self/">花花酱 LeetCode 315. Count of Smaller Numbers After Self</a> appeared first on <a rel="nofollow" href="https://zxi.mytechroad.com/blog">Huahua&#039;s Tech Road</a>.</p>
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