2 line Python solution, 100%

Could be even 1, but split into 2 for readability :)

segs = [len(list(j)) for i,j in groupby(a, lambda x:x>k) if not i]
return math.comb(len(a)+1,2) - sum(math.comb(i+1,2) for i in segs)

Notes:

• answer = count of all subarrays - count of subarrays where all numbers <=k.
• count of all subarrays is just N+1 choose 2
• segs (funny name ikr) array is the lengths of all subarrays where elements are <=k.

include

using namespace std;

define ll long long

define no cout<<"No"<

define yes cout<<"Yes"<

define mod 1000000007

void solve() { vectorv; sets; mapmp; ll n,m,j,i,y,z,l,r,d,x,k; cin>>n>>k; ll a[n+1]; ll c=0; for(i=0;i>a[i]; if(a[i]>k)c++; } x=0; ll mx=-1,ans=0; for(i=0;ik) { ans+=(i-mx)*(n-i); mx=i; } } cout<

} int main() { fast; int T=1; cin>>T; while(T--)solve(); return 0; }

Hint: prefix array

Just a humble request to all others who have solved questions like these, please refrain from showing the solutions or giving links to it here in the discussion as it destroys the purpose of solving such problems. Please try to give only intuitions to guide those who need assistance rather than giving them the code.

Thank you, Sincerely, akshay_123

ruby:

gets.to_i.times do
    n,k = gets.strip.split.map &:to_i
    a = gets.strip.split.map &:to_i
    max, prev = (1..n).reduce(:+), 0
    for i in 0...n do
        if a[i] > k
            max -= (1..prev).reduce(:+) || 0
            prev = 0
        else
            prev +=1 end
        i += 1 end
    max -= (1..prev).reduce(:+) || 0
    puts max end