Inversion Graph (Codeforces 1638C)
Tags: Math, dsu

You are given a permutation $$$p_1, p_2, \dots, p_n$$$. Then, an undirected graph is constructed in the following way: add an edge between vertices $$$i$$$, $$$j$$$ such that $$$i < j$$$ if and only if $$$p_i > p_j$$$. Your task is to count the number of connected components in this graph.Two vertices $$$u$$$ and $$$v$$$ belong to the same connected component if and only if there is at least one path along edges connecting $$$u$$$ and $$$v$$$.A permutation is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the permutation.The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le n$$$) — the elements of the permutation.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print one integer $$$k$$$ — the number of connected components.

Note: Each separate test case is depicted in the image below. The colored squares represent the elements of the permutation. For one permutation, each color represents some connected component. The number of distinct colors is the answer.  

Sample Inputs:
6
3
1 2 3
5
2 1 4 3 5
6
6 1 4 2 5 3
1
1
6
3 2 1 6 5 4
5
3 1 5 2 4

Sample Outputs:
3
3
1
1
2
1

Solution (Python):
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    n = int(input())
    c = mx = i = int()
    for x in input().split():
        mx = max(mx, int(x))
        if mx == i + 1:
            c += 1
        i += 1
    print(c)

Solution (Python):
import sys
input = sys.stdin.readline
from heapq import *
for i in range(int(input())):
    n = int(input())
    a = [*map(int, input().split())]
    k = 1
    mx = a[0]
    last = a[0] - 1
    for i in range(1, n):
        if a[i] > mx:
            k += last == 0
            last += a[i] - mx - 1
            mx = a[i]
        else:
            last -= 1
    print(k)

Solution (Python):
from collections import defaultdict, deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, itertools, math
sys.setrecursionlimit(10 ** 5)
input = sys.stdin.readline
sqrt = math.sqrt

def LI():
    return list(map(int, input().split()))

def LF():
    return list(map(float, input().split()))

def LI_():
    return list(map(lambda x: int(x) - 1, input().split()))

def II():
    return int(input())

def IF():
    return float(input())

def S():
    return input().rstrip()

def LS():
    return S().split()

def IR(n):
    return [II() for _ in range(n)]

def LIR(n):
    return [LI() for _ in range(n)]

def FR(n):
    return [IF() for _ in range(n)]

def LFR(n):
    return [LI() for _ in range(n)]

def LIR_(n):
    return [LI_() for _ in range(n)]

def SR(n):
    return [S() for _ in range(n)]

def LSR(n):
    return [LS() for _ in range(n)]
mod = 1000000007
inf = 10000000000.0

def solve():
    t = II()
    for _ in range(t):
        n = II()
        p = LI()
        ans = [0, 0]
        nn = 0
        for pi in p:
            if ans[0] == ans[1]:
                ans[0] = pi
                nn += 1
            if ans[0] < pi:
                ans[0] = pi
            ans[1] += 1
        print(nn)
    return
if __name__ == '__main__':
    solve()