Hi everyone, I'm getting TLE in this problem, I'm trying and offline approach (I was getting RE doing it online), solving only for values of v that are given in the queries, I find the adjacents groups of numbers >= v in O(n) using 2 pointers.

The complexity is O(N x (#different values of v) + QlgQ).

Could someone suggest me an optimization or different approach? Thanks :)

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int a[100000];
long long cont[100001];
long long ans[100000];

struct query{
    int v,a,b,id;
    
    query(){}
    
    bool operator < (query X)const{
        return v < X.v;
    }
}q[100000];

int main(){
    int N,Q;
    
    scanf("%d %d",&N,&Q);
    
    for(int i = 0;i < N;++i)
        scanf("%d",&a[i]);
    
    for(int i = 0;i < Q;++i){
        scanf("%d %d %d",&q[i].v,&q[i].a,&q[i].b);
        q[i].b = min(q[i].b,N);
        q[i].id = i;
    }
    
    sort(q,q + Q);
    
    for(int i = 0;i < Q;){
        int e = i;
        
        while(e < Q && q[e].v == q[i].v) ++e;
        
        int v = q[i].v;
        memset(cont,0,sizeof cont);
        
        for(int j = 0;j < N;){
            if(a[j] < v) ++j;
            else{
                int e2 = j;
                
                while(e2 < N && a[e2] >= v) ++e2;
                
                int m = e2 - j;
                
                for(int k = 1;k <= m;++k)
                    cont[k] += m + 1 - k;
                
                j = e2;
            }
        }
        
        for(int j = 1;j <= N;++j)
            cont[j] += cont[j - 1];
        
        for(int j = i;j < e;++j)
            ans[ q[j].id ] = (q[j].a <= q[j].b? cont[ q[j].b ]- cont[ q[j].a - 1 ] : 0);
        
        i = e;
    }
    
    for(int i = 0;i < Q;++i)
        printf("%lld\n",ans[i]);
    
    return 0;
}