Solution to Hackerrank's problem: Sherlock and Subqueries with segment tree

Sherlock and subqueries

#include <iostream>
using namespace std;
const int nmax = 1e5 + 1;
const int MIN = -1;
int arr[nmax];
pair<int, int> tree [nmax * 4];


void build(int id, int l, int r) {
    if (l == r) {
        tree[id] = make_pair(arr[l], 1);
        return;
    }
    int mid = (l + r) / 2;
    build(2 * id, l, mid);
    build(2 * id + 1, mid + 1, r);
    pair<int, int> left = tree[2 * id], right = tree[2 * id + 1];
    if (left.first >= right.first) {
        tree[id] = make_pair(left.first, left.second + (left.first == right.first? right.second : 0));
    } else {
        tree[id] = make_pair(right.first, right.second);
    }
}

void update(int id, int l, int r, int idx, int val) {
    if (l == r) {
        tree[id] = make_pair(val, 1);
        return;
    }
    int mid = (l + r) / 2;
    if (idx <= mid) {
        update(2 * id, l, mid, idx, val);
    } else {
        update(2 * id + 1, mid + 1, r, idx, val);
    }
    pair<int, int> left = tree[2 * id], right = tree[2 * id + 1];
    if (left.first >= right.first) {
        tree[id] = make_pair(left.first, left.second + (left.first == right.first? right.second : 0));
    } else {
        tree[id] = make_pair(right.first, right.second);
    }
}

pair<int,int> query(int id, int left, int right, int a, int b) {
    if (left > b || right < a) {
        return make_pair(MIN, 0);
    }
    if (left >= a && right <= b) {
        return tree[id];
    }
    int mid = (left + right) / 2;
    pair<int,int> l = query(2 * id, left, mid, a, b);
    pair<int,int> r = query(2 * id + 1, mid + 1, right, a, b);
    if (l.first > r.first) {
        return l;
    } else if (l.first == r.first) {
        return make_pair(l.first, l.second + r.second);
    }
    return r;
}

int main() {

    int n, q;

    cin >> n >> q;

    for (int i = 1; i <= n; i++) {
        cin >> arr[i];
    }
    build(1, 1, n);
    for (int i = 0; i < q; i++) {
        
        // query
        int a, b;
        cin >> a >> b;
        cout << query(1, 1, n, a, b).second << "\n";
    }
    return 0;
}