# Solution For Minimum Interval To Include Each Query

Problem statement:

You are given an array of integers nums and an array queries, where queries[i] is a pair of indices (0-indexed). Find the minimum interval [l, r] such that for all queries[i], the subarray nums[l,r] contains A[queries[i][0]], A[queries[i][1]] and so on. A subarray (l,r) is called minimum if there is no subarray (l’,r’) such that r’-l’ < r-l or l’ < l or r’ > r.

Solution:

The problem can be solved by using binary search and the sliding window algorithm in order to find the minimum interval. The idea of the solution is that we will find the minimum interval based on the maximum and minimum indices of the queries.

The steps are:

1. Create a dictionary that maps each number in the input array nums to a list of indices at which it is present.
2. For each query in the input array queries, create a list of indices for each element in the query. For example, if the query is [2, 4], create two lists [2nd index of element 2] and [4th index of element 4].
3. Find the minimum and maximum indices for each query. These indices will define the range of indices we need to consider.
4. For each index i in the input array nums, find the minimum and maximum indices of all the queries that include i. We can do this using the lists of indices we created in step 2 and the minimum and maximum indices we computed in step 3.
5. Using these minimum and maximum indices for each index i, we can use binary search to find the smallest interval that includes all the queries. We can do this by checking if there exists some j such that all the minimum indices of the queries that include i are less than or equal to j, and all the maximum indices are greater than or equal to j. If such a j exists, then we update our minimum interval. We continue this search until we find the smallest possible interval.
6. Finally, we return the minimum interval.

Time Complexity:

The time complexity of this algorithm is O(nlogn), where n is the length of the input array nums. The time complexity of creating the dictionary of indices for nums is O(n), and the time complexity of finding the minimum and maximum indices for each query is O(q), where q is the length of the input array queries. The time complexity of the binary search process is O(nlogn). Therefore, the total time complexity of the algorithm is O(nlogn).

Space Complexity:

The space complexity of this algorithm is O(n), where n is the length of the input array nums. The dictionary that maps each number to a list of indices has space complexity O(n), and the lists of indices for each query also have space complexity O(n). Therefore, the total space complexity of the algorithm is O(n).

## Step by Step Implementation For Minimum Interval To Include Each Query

/**
* Given a list of queries, find the minimum interval that contains all the queries.
*
* Examples:
*
* Input: [ [1, 3], [2, 6], [8, 10], [15, 18] ]
* Output: [3, 6]
*
* Input: [ [1, 4], [2, 5], [7, 9] ]
* Output: [2, 4]
*
* Note:
*
* 1. There may be multiple valid intervals, return any of them.
* 2. It is guaranteed that there is at least one valid interval.
*/

// Time Complexity: O(NlogN)
// Space Complexity: O(N)
class Solution {
public List interval(int[][] queries) {
List ans = new ArrayList<>();
if (queries == null || queries.length == 0)
return ans;

// sort the queries by start time
Arrays.sort(queries, (a, b) -> a[0] - b[0]);

// use a priority queue to track the end time of each query
PriorityQueue pq = new PriorityQueue<>();
int start = queries[0][0], end = queries[0][1];
pq.offer(end);

// go through all queries
for (int i = 1; i < queries.length; i++) {
// if the current query starts before the end of previous queries, update the end time
// (if necessary) and add it to the priority queue
if (queries[i][0] <= end) {
end = Math.max(end, queries[i][1]);
pq.offer(end);
} else {
// otherwise, means we need to find a new interval (start-end)
// update the answer and reset start and end
start = queries[i][0];
end = queries[i][1];
pq.offer(end);
}
}

// don't forget to add the last interval
return ans;
}
}
Given an array of queries, return the minimum interval that contains all queries.

queries = [[1,3],[2,6],[8,10],[15,18]]
Output: [1,6]
Explanation: Starting from the leftmost query, we check if every query is covered by a bigger interval until we find an interval that covers all queries.

def findMinimumInterval(queries):

# sort the queries by start time
queries.sort(key = lambda x: x[0])

# initialize the minimum interval to be the first query
minInterval = queries[0]

# go through the rest of the queries
for i in range(1, len(queries)):
# if the current query's start time is less than or equal to the end time of the minimum interval,
# update the minimum interval to be the current query
if queries[i][0] <= minInterval[1]:
minInterval[1] = queries[i][1]
# otherwise, the current query is not covered by the minimum interval,
# so we can return the minimum interval
else:
return minInterval

# return the minimum interval
return minInterval
/**
* @param {number} n
* @param {number[][]} queries
* @return {number}
*/
var minInterval = function(n, queries) {
// TODO: Implement this function
};
Given an array of queries where queries[i] = [starti, endi], you must determine the minimum interval, inclusive, that must contain at least one of the endpoints of each interval in queries.

Return the shortest such interval [-1, -1], if no interval satisfies the conditions.

Example 1:

Input: queries = [[1,3],[2,6],[8,10],[15,18]]
Output: [3,6]
Explanation: queries[0].start = 1, queries[0].end = 3, queries[1].start = 2, queries[1].end = 6, queries[0].start < queries[1].end.
Example 2:

Input: queries = [[1,4],[4,5],[3,6],[2,7]]
Output: [5,6]
Explanation: queries[0].start = 1, queries[0].end = 4, queries[1].start = 4, queries[1].end = 5, queries[2].start = 3, queries[2].end = 6, queries[3].start = 2, queries[3].end = 7, a point must exist between queries[0].end and queries[1].start, also between queries[2].start and queries[3].end.
Example 3:

Input: queries = [[1,5],[6,8],[0,6],[5,9],[3,4],[6,7],[1,7],[2,3],[8,9]]
Output: [3,4]
//This is a C# solution for the leetcode problem "Minimum Interval to Include Each Query"

using System;
using System.Collections.Generic;

namespace Minimum_Interval_to_Include_Each_Query
{
class Program
{
static void Main(string[] args)
{
int[][] queries = new int[][] { new int[] { 1, 3 }, new int[] { 2, 5 }, new int[] { 3, 7 }, new int[] { 4, 10 }, new int[] { 5, 11 }, new int[] { 6, 12 } };
Console.WriteLine(MinInterval(queries));
}

//This is the method that finds the minimum interval that contains all the queries
public static int MinInterval(int[][] queries)
{
//We sort the queries by their start time
Array.Sort(queries, (a, b) => a[0].CompareTo(b[0]));

//We initialize the minimum interval to be the interval between the first and last queries
int minInterval = queries[queries.Length - 1][1] - queries[0][0];

//We initialize the start and end of the current interval to be the start and end of the first query
int start = queries[0][0];
int end = queries[0][1];

//We iterate through the queries
for (int i = 1; i < queries.Length; i++)
{
//If the start time of the current query is less than or equal to the end time of the current interval,
//we update the end time of the interval to be the maximum of the current end time and the end time of the current query
if (queries[i][0] <= end)
{
end = Math.Max(end, queries[i][1]);
}
//Otherwise, we update the minimum interval to be the minimum of the current minimum interval and the difference between the start time of the current query and the end time of the previous interval
else
{
minInterval = Math.Min(minInterval, queries[i][0] - end);
//We also update the start and end times of the current interval to be the start and end times of the current query
start = queries[i][0];
end = queries[i][1];
}
}

//We return the minimum interval
return minInterval;
}
}
}
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