New Problem "Count of Range Sum"

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|327|[Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/) | [C++](./algorithms/cpp/countOfRangeSum/CountOfRangeSum.cpp)|Hard|
 |326|[Power of Three](https://leetcode.com/problems/power-of-three/) | [C++](./algorithms/cpp/powerOfThree/PowerOfThree.cpp)|Easy|
 |322|[Coin Change](https://leetcode.com/problems/coin-change/) | [C++](./algorithms/cpp/coinChange/coinChange.cpp)|Medium|
 |319|[Bulb Switcher](https://leetcode.com/problems/bulb-switcher/) | [C++](./algorithms/cpp/bulbSwitcher/bulbSwitcher.cpp)|Medium|

algorithms/cpp/countOfRangeSum/CountOfRangeSum.cpp

@@ -0,0 +1,163 @@
+// Source : https://leetcode.com/problems/count-of-range-sum/
+// Author : Hao Chen
+// Date   : 2016-01-15
+
+/*************************************************************************************** 
+ *
+ * Given an integer array nums, return the number of range sums that lie in [lower, 
+ * upper] inclusive.
+ * 
+ *     Range sum S(i, j) is defined as the sum of the elements in nums between indices 
+ * i and 
+ *     j (i ≤ j), inclusive.
+ * 
+ *     Note:
+ *     A naive algorithm of O(n2) is trivial. You MUST do better than that.
+ * 
+ *     Example:
+ *     Given nums = [-2, 5, -1], lower = -2, upper = 2,
+ *     Return 3.
+ *     The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.
+ * 
+ * Credits:Special thanks to @dietpepsi for adding this problem and creating all test 
+ * cases.
+ *               
+ ***************************************************************************************/
+
+
+/*
+ *  At first of all, we can do preprocess to calculate the prefix sums 
+ * 
+ *      S[i] = S(0, i), then S(i, j) = S[j] - S[i]. 
+ *  
+ *  Note: S(i, j) as the sum of range [i, j) where j exclusive and j > i. 
+ *
+ *  With these prefix sums, it is trivial to see that with O(n^2) time we can find all S(i, j) 
+ *  in the range [lower, upper]
+ *
+ *      int countRangeSum(vector<int>& nums, int lower, int upper) {
+ *        int n = nums.size();
+ *        long[] sums = new long[n + 1];
+ *        for (int i = 0; i < n; ++i) {
+ *            sums[i + 1] = sums[i] + nums[i];
+ *        }
+ *        int ans = 0;
+ *        for (int i = 0; i < n; ++i) {
+ *            for (int j = i + 1; j <= n; ++j) {
+ *                if (sums[j] - sums[i] >= lower && sums[j] - sums[i] <= upper) {
+ *                    ans++;
+ *                }
+ *            }
+ *        }
+ *        delete []sums;
+ *        return ans;
+ *      }
+ * 
+ *  The above solution would get time limit error.
+ *
+ *  Recall `count smaller number after self` where we encountered the problem
+ *
+ *      count[i] = count of nums[j] - nums[i] < 0 with j > i
+ *
+ *  Here, after we did the preprocess, we need to solve the problem
+ *
+ *      count[i] = count of a <= S[j] - S[i] <= b with j > i   
+ *
+ *  In other words, if we maintain the prefix sums sorted, and then are able to find out 
+ *  - how many of the sums are less than 'lower', say num1, 
+ *  - how many of the sums are less than 'upper + 1', say num2, 
+ *  Then 'num2 - num1' is the number of sums that lie within the range of [lower, upper].
+ *
+ */
+
+class Node{
+    public:
+        long long val;
+        int cnt; //amount of the nodes
+        Node *left, *right;
+        Node(long long v):val(v), cnt(1), left(NULL), right(NULL) {}
+};
+
+// a tree stores all of prefix sums
+class Tree{
+    public:
+        Tree():root(NULL){ }
+        ~Tree() { freeTree(root); }
+
+        void Insert(long long val) {
+            Insert(root, val);
+        }
+        int LessThan(long long sum, int val) {
+            return LessThan(root, sum, val, 0);
+        }
+
+    private:
+        Node* root;
+
+        //general binary search tree insert algorithm
+        void Insert(Node* &root, long long val) {
+            if (!root) {
+                root = new Node(val);
+                return;
+            }
+
+            root->cnt++;
+
+            if (val < root->val ) {
+                Insert(root->left, val);
+            }else if (val > root->val) {
+                Insert(root->right, val);
+            }
+        }
+        //return how many of the sums less than `val`
+        //  -  `sum` is the new sums which hasn't been inserted
+        //  -  `val` is the `lower` or `upper+1`
+        int LessThan(Node* root, long long sum, int val, int res) {
+
+            if (!root) return res;
+
+            if ( sum - root->val < val) {
+                //if (sum[j, i] < val), which means all of the right branch must be less than `val` 
+                //so we add the amounts of sums in right branch, and keep going the left branch.
+                res += (root->cnt - (root->left ? root->left->cnt : 0) );
+                return LessThan(root->left, sum, val, res);
+            }else if ( sum - root->val > val) {
+                //if (sum[j, i] > val), which means all of left brach must be greater than `val`
+                //so we just keep going the right branch.
+                return LessThan(root->right, sum, val, res);
+            }else {
+                //if (sum[j,i] == val), which means we find the correct place, 
+                //so we just return the the amounts of right branch.]
+                return res + (root->right ? root->right->cnt : 0);
+            }
+        }
+        void freeTree(Node* root){
+            if (!root) return;
+            if (root->left) freeTree(root->left);
+            if (root->right) freeTree(root->right);
+            delete root;
+        }
+
+};
+
+
+
+class Solution {
+public:
+    int countRangeSum(vector<int>& nums, int lower, int upper) {
+        Tree tree;
+        tree.Insert(0);
+        long long sum = 0;
+        int res = 0;
+
+        for (int n : nums) {
+            sum += n;
+            int lcnt = tree.LessThan(sum, lower);
+            int hcnt = tree.LessThan(sum, upper + 1);
+            res += (hcnt - lcnt);
+            tree.Insert(sum);
+        }
+
+        return res;
+    }
+};