New Problem Solution - "1851. Minimum Interval to Include Each Query"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1851|[Minimum Interval to Include Each Query](https://leetcode.com/problems/minimum-interval-to-include-each-query/) | [C++](./algorithms/cpp/minimumIntervalToIncludeEachQuery/MinimumIntervalToIncludeEachQuery.cpp)|Hard|
 |1850|[Minimum Adjacent Swaps to Reach the Kth Smallest Number](https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/) | [C++](./algorithms/cpp/minimumAdjacentSwapsToReachTheKthSmallestNumber/MinimumAdjacentSwapsToReachTheKthSmallestNumber.cpp)|Medium|
 |1849|[Splitting a String Into Descending Consecutive Values](https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/) | [C++](./algorithms/cpp/splittingAStringIntoDescendingConsecutiveValues/SplittingAStringIntoDescendingConsecutiveValues.cpp)|Medium|
 |1848|[Minimum Distance to the Target Element](https://leetcode.com/problems/minimum-distance-to-the-target-element/) | [C++](./algorithms/cpp/minimumDistanceToTheTargetElement/MinimumDistanceToTheTargetElement.cpp)|Easy|

algorithms/cpp/minimumIntervalToIncludeEachQuery/MinimumIntervalToIncludeEachQuery.cpp

@@ -0,0 +1,85 @@
+// Source : https://leetcode.com/problems/minimum-interval-to-include-each-query/
+// Author : Hao Chen
+// Date   : 2021-05-03
+
+/***************************************************************************************************** 
+ *
+ * You are given a 2D integer array intervals, where intervals[i] = [lefti, righti] describes the i^th 
+ * interval starting at lefti and ending at righti (inclusive). The size of an interval is defined as 
+ * the number of integers it contains, or more formally righti - lefti + 1.
+ * 
+ * You are also given an integer array queries. The answer to the j^th query is the size of the 
+ * smallest interval i such that lefti <= queries[j] <= righti. If no such interval exists, the answer 
+ * is -1.
+ * 
+ * Return an array containing the answers to the queries.
+ * 
+ * Example 1:
+ * 
+ * Input: intervals = [[1,4],[2,4],[3,6],[4,4]], queries = [2,3,4,5]
+ * Output: [3,3,1,4]
+ * Explanation: The queries are processed as follows:
+ * - Query = 2: The interval [2,4] is the smallest interval containing 2. The answer is 4 - 2 + 1 = 3.
+ * - Query = 3: The interval [2,4] is the smallest interval containing 3. The answer is 4 - 2 + 1 = 3.
+ * - Query = 4: The interval [4,4] is the smallest interval containing 4. The answer is 4 - 4 + 1 = 1.
+ * - Query = 5: The interval [3,6] is the smallest interval containing 5. The answer is 6 - 3 + 1 = 4.
+ * 
+ * Example 2:
+ * 
+ * Input: intervals = [[2,3],[2,5],[1,8],[20,25]], queries = [2,19,5,22]
+ * Output: [2,-1,4,6]
+ * Explanation: The queries are processed as follows:
+ * - Query = 2: The interval [2,3] is the smallest interval containing 2. The answer is 3 - 2 + 1 = 2.
+ * - Query = 19: None of the intervals contain 19. The answer is -1.
+ * - Query = 5: The interval [2,5] is the smallest interval containing 5. The answer is 5 - 2 + 1 = 4.
+ * - Query = 22: The interval [20,25] is the smallest interval containing 22. The answer is 25 - 20 + 
+ * 1 = 6.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= intervals.length <= 10^5
+ * 	1 <= queries.length <= 10^5
+ * 	queries[i].length == 2
+ * 	1 <= lefti <= righti <= 10^7
+ * 	1 <= queries[j] <= 10^7
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
+        set<vector<int>> s;
+        
+        vector<vector<int>> iQueries;
+        for(int i=0; i < queries.size(); i++) {
+            iQueries.push_back({queries[i], i});
+        }
+        
+        sort(intervals.begin(), intervals.end());
+        sort(iQueries.begin(), iQueries.end());
+        
+        vector<int> result(queries.size(), -1);
+        
+        int i = 0, len = intervals.size();
+        
+        for(auto& iq: iQueries) {
+            int q = iq[0];
+            int idx = iq[1];
+            
+            while( i < len && intervals[i][0] <= q) {
+                s.insert({intervals[i][1] - intervals[i][0] + 1, intervals[i][1]});
+                i++;
+            }
+            while( !s.empty() ) {
+                auto it = s.begin();
+                if ( (*it)[1] >= q ) break;
+                s.erase(s.begin());
+            }
+            if ( !s.empty() ) {
+                auto it = s.begin();
+                result[idx] = (*it)[0];
+            }
+        }
+        
+        return result;
+    }
+};