cf B #695 doubt

#include <bits/stdc++.h>
using namespace std;
#define ll long long
/* Approach is mentioned below code.
*/
void solve(){ // testcases
    int n;
    scanf("%d",&n);
    int arr[n];
    for(int i=0;i<n;i++) scanf("%d",&arr[i]);
    if(n==1 || n==2){
        printf("0\n"); return;
    }

    bool brr[n];
    for(int i=0;i<n;i++) brr[i]=false;
    int prev=-1,curr=-1,next=-1;
    int cnt=0;
    for(int i=1;i<n-1;i++){
        curr=arr[i]; prev=arr[i-1]; next=arr[i+1];
        if((curr<prev && curr<next) || (curr>prev && curr>next)) brr[i]=true, cnt++;
    }
    int c=0,mc=0;
    for(int i=1;i<n-1;i++){
        if(brr[i]) c++, mc=max(mc,c);
        else c=0;
    }
    if(mc>=3) printf("%d\n",cnt-3);
    else printf("%d\n",cnt-mc);
    return;
}

int main(){
    int testcase=1,z=1;
    scanf("%d",&testcase);
    while(z<=testcase){
        solve(); z++;
    }
    return 0;
}
/*

Approach:

THought: as we know that by changing any one of the values in the array (arr) , we can at maximum reduce
the intimidation by 3 (only when there are >=3 consecutive hills/valleys). (just a greedy approach )

So, if there are >=3 consecutive hills/valleys, then ans is total hills+valleys -3;
else{ (there are no consecutive 3 hills/valleys)
    reduce the ans by max number of consecutive hills/valleys  (i.e. 0,1, or 2)
}


*/