# Program to find minimum interval to include each query in Python

PythonServer Side ProgrammingProgramming

Suppose we have a list of intervals, where intervals[i] has a pair (left_i, right_i) represents the ith interval starting at left_i and ending at right_i (both inclusive). We also have another array called queries. The answer to the jth query is the size of the smallest interval i such that left_i <= queries[j] <= right_i. If we cannot find such interval, then return -1. We have to find an array containing the answers to the queries.

So, if the input is like intervals = [[2,5],[3,5],[4,7],[5,5]] queries = [3,4,5,6], then the output will be [3, 3, 1, 4], because The queries are processed as follows −

• for query = 3: The interval [3,5] is the smallest one holding 3. so, 5 - 3 + 1 = 3.

• for query = 4: The interval [3,5] is the smallest one holding 4. so, 5 - 3 + 1 = 3.

• for query = 5: The interval [5,5] is the smallest one holding 5. so, 5 - 5 + 1 = 1.

• for query = 6: The interval [4,7] is the smallest one holding 6. so, 7 - 4 + 1 = 4.

To solve this, we will follow these steps −

• intervals := sort the list intervals in reverse order

• h := a new list

• res := a new map

• for each q in the sorted list of queries, do

• while intervals is not empty and end time of last interval <= q, do

• (i, j) := last interval from intervals, and remove it

• if j >= q, then

• insert pair (j-i+1, j) into heap h

• while h is not empty and end time of top most interval in h < q, do

• delete top element from h

• res[q] := start time of top most interval of h if h is not empty otherwise -1

• return a list of res[q] for all q in queries

## Example

Let us see the following implementation to get better understanding

import heapq
def solve(intervals, queries):
intervals = sorted(intervals)[::-1]
h = []
res = {}
for q in sorted(queries):
while intervals and intervals[-1][0] <= q:
i, j = intervals.pop()
if j >= q:
heapq.heappush(h, [j - i + 1, j])
while h and h[0][1] < q:
heapq.heappop(h)
res[q] = h[0][0] if h else -1
return [res[q] for q in queries]

intervals = [[2,5],[3,5],[4,7],[5,5]]
queries = [3,4,5,6]
print(solve(intervals, queries))

## Input

[[2,5],[3,5],[4,7],[5,5]], [3,4,5,6]

## Output

[3, 3, 1, 4]
Published on 08-Oct-2021 08:38:02