Codeforces 1638 B. odd swap sort - tree array, no, simple thinking

This way

The question ：

Give you a length of n Array of a, If a[i]+a[i+1] Is odd , Then you can exchange a[i] and a[i+1], Ask if you can finally make the array a Non falling .

Answer key ：

I didn't sleep at noon … The afternoon was sombre , Not clear thinking .
My idea is , Sort from small to large , Each number must reach the sorted position , Naturally, all the numbers smaller than him should be exchanged . Therefore, it depends on whether all the numbers less than it after each number are the same odd or even with it .
My approach is , To the current position , Check whether all the numbers less than it at the same level or even at this position have been taken .
Of course I'm very stupid here , Why should we see if we have finished taking it ？ Just look at whether the odd sequence and the even sequence are incremented respectively ？ I don't know what I'm doing, wasting my time

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,M=1e9+5;
int a[N],b[N],od[N];
struct Array{
    
    unordered_map<int,int>v;
    int lowbit(int x){
    return x&(-x);}
    void add(int x){
    
        for(int i=x;i<M;i+=lowbit(i))
            v[i]++;
    }
    int ask(int x){
    
        int ans=0;
        for(int i=x;i;i-=lowbit(i))
            ans+=v[i];
        return ans;
    }
}tr[2];
int main()
{
    
    int t;
    scanf("%d",&t);
    for(int tim=1;tim<=t;tim++){
    
        tr[0].v.clear(),tr[1].v.clear();
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i];
        sort(b+1,b+1+n);
        for(int i=1;i<=n;i++)od[i]=od[i-1]+(b[i]%2);
        int f=0;
        for(int i=1;i<=n;i++){
    
            tr[a[i]%2].add(a[i]);

            int p=lower_bound(b+1,b+1+n,a[i])-b;
            int num=a[i]%2?od[p]:p-od[p];
            if(tr[a[i]%2].ask(a[i]-1)-num+1)f=1;
        }
        printf("%s\n",f?"NO":"YES");
    }
    return 0;
}