HackerEarth The maximum distance problem solution YASH PAL, 31 July 202414 February 2026 In this HackerEarth The maximum distance problem solution You are given an array a1, a2, …, aN consisting of N elements. You are given Q queries. Each query is of the following types:The format of the first type of query is 1 L R x. You must increase the values of all elements in range L to R (both inclusive) by integer x.The format of the second type of query is 2 L R y. You must multiply the values of all elements in the range L to R (both inclusive) by the integer y.The format of the third type of query is 3 z. You are required to find the maximum distance between elements that are equal to z in the array. If the element is not present in the array, print -1. HackerEarth The maximum distance problem solution.#include <bits/stdc++.h>using namespace std;#define ll long long#define ull unsigned long long#define f(a, b) for (ll i = a; i < b; i++)#define mod 1000000007#define pb push_back#define vll vector<ll>#define pll vector<pair<ll, ll>>#define ld long double#define fr(a, b) for (ll j = a; j >= b; j--)#define fi(a, b) for (ll j = a; j < b; j++)#define fii(a, b) for (ll k = a; k < b; k++)template <class T>ostream &operator<<(ostream &os, vector<T> V){ os << "[ "; for (auto v : V) os << v << " "; os << "]"; return os;}template <class T>ostream &operator<<(ostream &os, multiset<T> S){ os << "{ "; for (auto s : S) os << s << " "; return os << "}";}template <class L, class R>ostream &operator<<(ostream &os, pair<L, R> P){ return os << "(" << P.first << "," << P.second << ")";}template <class L, class R>ostream &operator<<(ostream &os, map<L, R> M){ os << "{ "; for (auto m : M) os << "(" << m.first << ":" << m.second << ") "; return os << "}";}ll n, blockSize, a[100005], add[350], mul[350];multiset<pair<ll, ll>> s[350];int main(){#ifndef ONLINE_JUDGE freopen("input.txt", "rt", stdin); freopen("output.txt", "wt", stdout);#endif ios_base::sync_with_stdio(false); cin.tie(NULL); cin >> n; blockSize = sqrt(n); f(0, n) cin >> a[i], s[i / blockSize].insert({a[i], i}); f(0, 350) mul[i] = 1; ll q; cin >> q; while (q--) { ll c; cin >> c; if (c == 1 || c == 2) { ll l, r, x; cin >> l >> r >> x; if (l > r) swap(l, r); l--, r--; ll lBlock = l / blockSize, rBlock = r / blockSize; if (lBlock == rBlock) { f(l, n) { if (i / blockSize != lBlock) break; s[i / blockSize].erase(s[i / blockSize].find({a[i], i})); a[i] *= mul[i / blockSize], a[i] += add[i / blockSize]; s[i / blockSize].insert({a[i], i}); } fr(l - 1, 0) { if (j / blockSize != lBlock) break; s[j / blockSize].erase(s[j / blockSize].find({a[j], j})); a[j] *= mul[j / blockSize], a[j] += add[j / blockSize]; s[j / blockSize].insert({a[j], j}); } add[lBlock] = 0, mul[lBlock] = 1; f(l, r + 1) { if (c == 1) s[lBlock].erase(s[lBlock].find({a[i], i})), a[i] += x, s[lBlock].insert({a[i], i}); else s[lBlock].erase(s[lBlock].find({a[i], i})), a[i] *= x, s[lBlock].insert({a[i], i}); } } else { f(l, n) { if (i / blockSize != lBlock) break; s[i / blockSize].erase(s[i / blockSize].find({a[i], i})); a[i] *= mul[i / blockSize], a[i] += add[i / blockSize]; if (c == 1) a[i] += x; else a[i] *= x; s[i / blockSize].insert({a[i], i}); } fr(l - 1, 0) { if (j / blockSize != lBlock) break; s[j / blockSize].erase(s[j / blockSize].find({a[j], j})); a[j] *= mul[j / blockSize], a[j] += add[j / blockSize]; s[j / blockSize].insert({a[j], j}); } add[lBlock] = 0, mul[lBlock] = 1; fr(r, 0) { if (j / blockSize != rBlock) break; s[j / blockSize].erase(s[j / blockSize].find({a[j], j})); a[j] *= mul[j / blockSize], a[j] += add[j / blockSize]; if (c == 1) a[j] += x; else a[j] *= x; s[j / blockSize].insert({a[j], j}); } f(r + 1, n) { if (i / blockSize != rBlock) break; s[i / blockSize].erase(s[i / blockSize].find({a[i], i})); a[i] *= mul[i / blockSize], a[i] += add[i / blockSize]; s[i / blockSize].insert({a[i], i}); } add[rBlock] = 0, mul[rBlock] = 1; lBlock++; while (lBlock != rBlock) { if (c == 1) add[lBlock] += x; else mul[lBlock] *= x, add[lBlock] *= x; lBlock++; } } } else { ll x; cin >> x; ll l = -1, r = -1; ll mx = (n - 1) / blockSize, cur = 0; while (cur <= mx) { ll y = x; if (add[cur]) y -= add[cur]; if (y < 0) { cur++; continue; } if (mul[cur]) { if (y % mul[cur] == 0) y /= mul[cur]; else { cur++; continue; } } auto it = s[cur].lower_bound({y, -1e18}); if (it != s[cur].end() && it->first == y) { l = it->second; break; } cur++; } if (l == -1) { cout << "-1n"; continue; } cur = mx; while (cur >= 0) { ll y = x; if (add[cur]) y -= add[cur]; if (y < 0) { cur--; continue; } if (mul[cur]) { if (y % mul[cur] == 0) y /= mul[cur]; else { cur--; continue; } } auto it = s[cur].lower_bound({y + 1, -1e18}); if (it != s[cur].begin()) { it--; if (it->first == y) { r = it->second; break; } } cur--; } cout << r - l + 1 << "n"; } }#ifndef ONLINE_JUDGE cout << "nTime Elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " sn";#endif return 0;} Second solutionINF = 1e9n = int(input())a = list(map(int, input().split(' ')))q = int(input())while q > 0: query = list(map(int, input().split(' '))) t = query[0] if t == 1: [l, r, x] = query[1:] for i in range(l - 1, r): if(a[i] <= INF): a[i] += x elif t == 2: [l, r, x] = query[1:] for i in range(l - 1, r): if(a[i] <= INF): a[i] *= x else: [z] = query[1:] if not (z in a): print(-1) else: print(len(a) - a[::-1].index(z) - a.index(z)) q -= 1 coding problems solutions HackerEarth HackerEarth