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# Program to check how many queries finds valid arithmetic sequence in Python

Suppose we have a list of numbers called nums and also have list of queries. Where each query element contains [i, j]. So this query is asking whether the sublist of nums from [i, j] (both inclusive), is an arithmetic sequence or not. So finally we have to find the count of queries that return true.

So, if the input is like nums = [2, 4, 6, 8, 7, 6, 5, 2] queries = [[3, 4],[0, 3],[2, 4]], then the output will be 2, because [2, 4, 6, 8] is an arithmetic sequence, so query [0, 3] is true. and for [8, 7] is also an arithmetic sequence, so query [3, 4] is also true. but [6, 8, 7] is not an arithmetic sequence, so [2, 4] is not true.

To solve this, we will follow these steps −

- if nums is empty, then
- return 0

- n := size of nums
- diff := a list with elements (nums[i + 1] - nums[i]) for each i in range 0 to n - 2
- rle := a list of size (n - 1) and fill with 0
- for i in range 0 to n - 2, do
- if i > 0 and diff[i] is same as diff[i - 1], then
- rle[i] := rle[i - 1] + 1

- otherwise,
- rle[i] := 1

- if i > 0 and diff[i] is same as diff[i - 1], then
- ans := 0
- for each (i, j) in queries, do
- if i is same as j, then
- ans := ans + 1

- otherwise,
- ans := ans + (1 if rle[j - 1] >= (j - i), otherwise 0)

- if i is same as j, then
- return ans

## Example

Let us see the following implementation to get better understanding −

def solve(nums, queries): if not nums: return 0 n = len(nums) diff = [nums[i + 1] - nums[i] for i in range(n - 1)] rle = [0] * (n - 1) for i in range(n - 1): if i > 0 and diff[i] == diff[i - 1]: rle[i] = rle[i - 1] + 1 else: rle[i] = 1 ans = 0 for i, j in queries: if i == j: ans += 1 else: ans += rle[j - 1] >= (j - i) return ans nums = [2, 4, 6, 8, 7, 6, 5, 2] queries = [[3, 4],[0, 3],[2, 4]] print(solve(nums, queries))

## Input

[2, 4, 6, 8, 7, 6, 5, 2], [[3, 4],[0, 3],[2, 4]]

## Output

2