New Problem Solution - "1862. Sum of Floored Pairs"

Author: haoel

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1862|[Sum of Floored Pairs](https://leetcode.com/problems/sum-of-floored-pairs/) | [C++](./algorithms/cpp/sumOfFlooredPairs/SumOfFlooredPairs.cpp)|Hard|
 |1861|[Rotating the Box](https://leetcode.com/problems/rotating-the-box/) | [C++](./algorithms/cpp/rotatingTheBox/RotatingTheBox.cpp)|Medium|
 |1860|[Incremental Memory Leak](https://leetcode.com/problems/incremental-memory-leak/) | [C++](./algorithms/cpp/incrementalMemoryLeak/IncrementalMemoryLeak.cpp)|Medium|
 |1859|[Sorting the Sentence](https://leetcode.com/problems/sorting-the-sentence/) | [C++](./algorithms/cpp/sortingTheSentence/SortingTheSentence.cpp)|Easy|

algorithms/cpp/sumOfFlooredPairs/SumOfFlooredPairs.cpp

@@ -0,0 +1,68 @@
+// Source : https://leetcode.com/problems/sum-of-floored-pairs/
+// Author : Hao Chen
+// Date   : 2021-05-22
+
+/***************************************************************************************************** 
+ *
+ * Given an integer array nums, return the sum of floor(nums[i] / nums[j]) for all pairs of indices 0 
+ * <= i, j < nums.length in the array. Since the answer may be too large, return it modulo 10^9 + 7.
+ * 
+ * The floor() function returns the integer part of the division.
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [2,5,9]
+ * Output: 10
+ * Explanation:
+ * floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
+ * floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
+ * floor(5 / 2) = 2
+ * floor(9 / 2) = 4
+ * floor(9 / 5) = 1
+ * We calculate the floor of the division for every pair of indices in the array then sum them up.
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [7,7,7,7,7,7,7]
+ * Output: 49
+ * 
+ * Constraints:
+ * 
+ * 	1 <= nums.length <= 10^5
+ * 	1 <= nums[i] <= 10^5
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    int sumOfFlooredPairs(vector<int>& nums) {
+        const int MAX_NUM = 100001;
+        int cnt[MAX_NUM] = {0};
+        int maxn = 0;
+        for(auto& n : nums) {
+            cnt[n]++;
+            maxn = max(maxn, n);
+        }
+        
+        vector<vector<int>> stats;
+        for(int i=1; i<MAX_NUM; i++) {
+            if (cnt[i] > 0) {
+                stats.push_back({i, cnt[i]});
+            }
+            cnt[i] += cnt[i-1];
+        }
+        
+        const int MOD = 1e9+7;
+        int result = 0;
+        for(int i=0; i < stats.size(); i++) {
+            int n = stats[i][0];
+            int c = stats[i][1];
+
+            for(int x=2; x <= maxn/n+1; x++) {
+                int pre = (x-1) * n - 1; 
+                int cur = min( x * n - 1, MAX_NUM-1);
+                result = (result + (cnt[cur] - cnt[pre]) * long(x-1) * c) % MOD;      
+            }
+        }
+        return result;
+    }
+};