HackerEarth Different queries problem solution

In this HackerEarth Different queries problem solution You are given an array A of size N. You have to perform Q queries on the array. Each of these queries belongs to the following types of queries:

1. L R X : Add X to all elements in the range [L,R]
2. L R X : Set the value of all elements in the range [L,R] to X

However, it is not mandatory to perform the queries in order. Your task is to determine the lexicographically largest array that can be obtained by performing all the Q queries.

HackerEarth Different queries problem solution.

#include <bits/stdc++.h>
using namespace std;

long long A[505];
vector< pair<int, pair<int, int> > > negative_add, positive_add, all_set;

void addQueries(vector< pair<int, pair<int, int> > > &X)
{
   for(auto it : X)
   {
      int x = it.first;
      int L = it.second.first, R = it.second.second;
      
      for(int i=L; i<=R; i++)
         A[i] += x;
   }
}

void setQueries(vector< pair<int, pair<int, int> > > &X)
{
   for(auto it : X)
   {
      int x = it.first;
      int L = it.second.first, R = it.second.second;
      
      for(int i=L; i<=R; i++)
         A[i] = x;
   }
}

int main()
{
   int N, Q;
   cin >> N >> Q;
   
   for(int i=1; i<=N; i++)
      cin >> A[i];
   
   while(Q--)
   {
      int type, L, R, x;
      cin >> type >> L >> R >> x;
      
      if(type == 1)
      {
         if(x < 0)
            negative_add.push_back({x, {L, R}});
         else
            positive_add.push_back({x, {L, R}});
      }
      else
         all_set.push_back({x, {L, R}});
   }
   
   sort(all_set.begin(), all_set.end());
   
   addQueries(negative_add);
   setQueries(all_set);
   addQueries(positive_add);
   
   for(int i=1; i<=N; i++)
      cout << A[i] << " ";
   cout << "\n";
   
   return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
const int MAX = 505;
long long a[MAX];
vector<pair<int, pair<int, int>>> p1, p2, p3;

int main(int argc, char* argv[]) {
    if(argc == 2 or argc == 3) freopen(argv[1], "r", stdin);
    if(argc == 3) freopen(argv[2], "w", stdout);
    int n, q, type, l, r, x;
    assert(cin >> n >> q);
    assert(1 <= n and n <= 500);
    assert(1 <= q and q <= 1e5);
    for(int i = 1; i <= n; i++) {
        assert(cin >> a[i]);
        assert(-1e5 <= a[i] and a[i] <= 1e5);
    }
    for(int i = 0; i < q; i++) {
        assert(cin >> type >> l >> r >> x);
        assert(1 <= type and type <= 2);
        assert(1 <= l and l <= r);
        assert(l <= r and r <= n);
        assert(-1e5 <= x and x <= 1e5);
        if (type == 1) {
            if (x < 0) p1.push_back({x, {l, r}});
            else p3.push_back({x, {l, r}});
        }
        else p2.push_back({x, {l, r}});
    }
    sort(p2.begin(), p2.end());
    for(auto p : p1) {
        x = p.first;
        l = p.second.first;
        r = p.second.second;
        for(int j = l; j <= r; j++) {
            a[j] += x;
        }
    }
    for(auto p : p2) {
        x = p.first;
        l = p.second.first;
        r = p.second.second;
        for(int j = l; j <= r; j++) {
            a[j] = x;
        }
    }
    for(auto p : p3) {
        x = p.first;
        l = p.second.first;
        r = p.second.second;
        for(int j = l; j <= r; j++) {
            a[j] += x;
        }
    }
    for(int i = 1; i <= n; i++) {
        cout << a[i] << " \n"[i == n];
    }
    assert(!(cin >> n));
    return 0;
}