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;}`