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Program to Find the Inverted Inversions in Python
Suppose we have been provided with a list of numbers nums. We have to find the number of existing quadruplets (a, b, c, d) such that a < b < c < d and nums[a] < nums[b] and nums[c] > nums[d].
The array nums is a permutation of the integers 1...N
So, if the input is like nums = [3, 4, 7, 6, 5], then the output will be 5.
From the given input, we have these inverted inversions −
3, 4, 7, 6
3, 4, 6, 5
3, 4, 7, 5
3, 7, 6, 5
4, 7, 6, 5
To solve this, we will follow these steps −
m := 10^9 + 7
if size of nums < 4, then
return 0
n := size of nums
sorted_ds := a new list
Insert the last item of nums in sorted_ds
sort the list sorted_ds
ds_smaller_than_c := [0] * n
for c in range n − 2 to −1, decrease by 1, do
ds_smaller_than_c[c] := return the rightmost position in sorted_ds where nums[c] − 1 can be inserted and sorted order can be maintained
insert nums[c] at the end of sorted_ds
sort the list sorted_ds
quadruplet_count := 0
sorted_as := a new list
insert the first number of nums in sorted_as
sort the list sorted_as
as_smaller_than_b_sum := 0
for b in range 1 to n − 2, do
as_smaller_than_b_sum := as_smaller_than_b_sum + the rightmost position in sorted_as where nums[b] – 1 can be inserted and sorted order can be maintained
sort the list sorted_as
as_smaller_than_b_sum := as_smaller_than_b_sum mod m
insert nums[b] at the end of sorted_as
sort the list sorted_as
quadruplet_count := quadruplet_count + as_smaller_than_b_sum * ds_smaller_than_c[b + 1]
quadruplet_count := quadruplet_count mod m
return quadruplet_count
Let us see the following implementation to get better understanding −
Example
import bisect MOD = 10 ** 9 + 7 class Solution: def solve(self, nums): if len(nums) < 4: return 0 n = len(nums) sorted_ds = list([nums[−1]]) sorted_ds.sort() ds_smaller_than_c = [0] * n for c in range(n − 2, −1, −1): ds_smaller_than_c[c] = bisect.bisect_right(sorted_ds, nums[c] − 1) sorted_ds.append(nums[c]) sorted_ds.sort() quadruplet_count = 0 sorted_as = list([nums[0]]) sorted_as.sort() as_smaller_than_b_sum = 0 for b in range(1, n − 2): as_smaller_than_b_sum += bisect.bisect_right(sorted_as, nums[b] − 1) sorted_as.sort() as_smaller_than_b_sum %= MOD sorted_as.append(nums[b]) sorted_as.sort() quadruplet_count += as_smaller_than_b_sum * ds_smaller_than_c[b + 1] quadruplet_count %= MOD return quadruplet_count ob = Solution() print(ob.solve([3, 4, 7, 6, 5]))
Input
[3, 4, 7, 6, 5]
Output
5
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