🚀 Minimum Cost for Tickets – Python Solution Executed 💯✅

from bisect import bisect_left

m = int(raw_input())
days = list(map(int, raw_input().split()))
cost = list(map(int, raw_input().split()))

dp = {}

def solve(i, t):
if i == m:
return 0

if (i, t) in dp:
return dp[(i, t)]

ans = cost[0] + solve(i + 1, t)

if t:
ans = min(ans, solve(i + 1, t - 1))

j = bisect_left(days, days[i] + 7)
ans = min(ans, cost[1] + solve(j, t + 1))

j = bisect_left(days, days[i] + 30)
ans = min(ans, cost[2] + solve(j, t + 2))

dp[(i, t)] = ans
return ans
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print(solve(0, 0))

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🚨 100% TEST CASES PASSED ✅🔥

🏆 Minimum Number of Operations to Form Target Array – C++ Solution Available 💯

⚡ Optimized Approach
✅ All Test Cases Passed
🚀 OA Ready Code
💻 Clean & Efficient Implementation

#include <bits/stdc++.h>
using namespace std;
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long long solve(int N, vector<int>& target) {
long long total = 0;
for (int i = 0; i < N; i++) total += target[i];

long long sumX = 0;
long long carry = 0;

for (int i = 0; i + 1 < N; i++) {
long long cap = min((long long)target[i] - carry,
(long long)target[i + 1]);
if (cap < 0) cap = 0;
sumX += cap;
carry = cap;
}
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return total - sumX;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);

int N;
cin >> N;

vector<int> target(N);
for (int i = 0; i < N; i++)
cin >> target[i];

cout << solve(N, target) << '\n';
return 0;
}

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🚀 Thermostat Schedule – C++ Solution Executed 💯✅

🔥 100% Test Cases Passed
⚡ Optimized O(N) Solution
💻 Clean & OA Ready Code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
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int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);

int N;
if (!(cin >> N)) return 0;

vector<long long> t(N);
for (int i = 0; i < N; ++i)
cin >> t[i];

bool is_sorted = true;
for (int i = 0; i < N - 1; ++i) {
if (t[i] > t[i + 1]) {
is_sorted = false;
break;
}
}

if (is_sorted) {
long long total_sum = 0;
for (long long x : t)
total_sum += x;
cout << 0 << "\n" << total_sum << "\n";
return 0;
}
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vector<long long> suffix_min(N);
suffix_min[N - 1] = t[N - 1];

for (int i = N - 2; i >= 0; --i)
suffix_min[i] = min(t[i], suffix_min[i + 1]);

long long max_sum = 0;
long long prefix_sum = 0;

for (int i = 0; i < N; ++i) {
long long prev = (i == 0 ? 0 : t[i - 1]);
long long val = suffix_min[i];

if (val >= prev) {
long long cur = prefix_sum + 1LL * (N - i) * val;
max_sum = max(max_sum, cur);
}

if (i < N - 1 && t[i] > t[i + 1])
break;

prefix_sum += t[i];
}

cout << 1 << "\n" << max_sum << "\n";
return 0;
}

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N = int(raw_input())
K = int(raw_input())
prices = map(int, raw_input().split())

dp = [0] * N
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for j in range(N):
if j > 0:
dp[j] = dp[j - 1]
if j >= K:
dp[j] = max(dp[j], dp[j - K] + prices[j] - prices[j - K])

print(max(dp) if N else 0)

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🔥 FREE OA CODE DROP 🔥
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📌 Problem: Best Time to Buy & Sell Stock II

✅ Plagiarism-Free Solution
✅ Optimized & Submission Ready
✅ Suitable for OA & Coding Assessments

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💻 Problem: Length of Longest Sequences

✅ Optimized Dynamic Programming Solution
✅ Clean & Easy to Understand
✅ Placement OA Ready
✅ Python 3

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import sys

input = sys.stdin.readline

def solve(N, target, nums):
dp = [[-1] * (target + 1) for _ in range(N + 5)]

for i in range(N + 5):
dp[i][0] = 0

for i in range(N - 1, -1, -1):
val = nums[i]
nxt = min(N, i + (val % 4) + 1)

for s in range(target + 1):
dp[i][s] = dp[i + 1][s]

if val <= s and dp[nxt][s - val] != -1:
dp[i][s] = max(dp[i][s], dp[nxt][s - val] + 1)

return dp[0][target]


if name == "main":
try:
N = int(input())
target = int(input())
nums = list(map(int, input().split()))
print(solve(N, target, nums))
except (EOFError, ValueError):
pass

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