🚨 NUMBER OF WAYS – VERIFIED CODE 🚨

✅ 100% Working Solution
💯 All Test Cases Passed
⚡ Dynamic Programming Approach
🔥 Clean & Optimized Code

MOD = 1000000007

n = int(raw_input())

dp = [[0] * 5 for _ in range(n + 1)]

if n >= 1:
dp[1][1] = 1

if n >= 3:
dp[3][3] = 1

if n >= 4:
dp[4][4] = 1

nums = [1, 3, 4]

for s in range(1, n + 1):
for cur in nums:
if s >= cur:
prev_sum = s - cur

if prev_sum == 0:
continue

total = 0

for prev in nums:
if prev != cur:
total = (total + dp[prev_sum][prev]) % MOD

dp[s][cur] = (dp[s][cur] + total) % MOD

ans = (dp[n][1] + dp[n][3] + dp[n][4]) % MOD

print ans

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import java.io.*;
import java.util.*;

public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

List<Integer> input = new ArrayList<>();
String line;
while ((line = br.readLine()) != null) {
StringTokenizer st = new StringTokenizer(line);
while (st.hasMoreTokens()) {
input.add(Integer.parseInt(st.nextToken()));
}
}

if (input.isEmpty()) return;

int idx = 0;
int N = input.get(idx++);
int K = input.get(idx++);
int L = input.get(idx++);
int M = input.get(idx++);

int[] A = new int[N];
for (int i = 0; i < N; i++) {
A[i] = input.get(idx++);
}

int numSubarrays = N - L + 1;
long[] stdSum = new long[numSubarrays];
long[] altSum = new long[numSubarrays];

for (int i = 0; i < numSubarrays; i++) {
long sVal = 0;
long aVal = 0;
int sign = 1;

for (int j = 0; j < L; j++) {
int val = A[i + j];
sVal += val;
aVal += sign * val;
sign *= -1;
}

stdSum[i] = sVal;
altSum[i] = aVal;
}

long NEG = Long.MIN_VALUE / 4;
long[][][] dp = new long[N + 1][K + 1][M + 1];

for (int i = 0; i <= N; i++) {
for (int k = 0; k <= K; k++) {
Arrays.fill(dp[i][k], NEG);
}
}

dp[0][0][0] = 0;

for (int i = 0; i < N; i++) {
for (int k = 0; k <= K; k++) {
for (int m = 0; m <= M; m++) {
if (dp[i][k][m] == NEG) continue;

dp[i + 1][k][m] = Math.max(dp[i + 1][k][m], dp[i][k][m]);

if (k < K && i + L <= N) {
dp[i + L][k + 1][m] = Math.max(
dp[i + L][k + 1][m],
dp[i][k][m] + stdSum[i]
);

if (m < M) {
dp[i + L][k + 1][m + 1] = Math.max(
dp[i + L][k + 1][m + 1],
dp[i][k][m] + altSum[i]
);
}
}
}
}
}

long ans = NEG;
for (int m = 0; m <= M; m++) {
ans = Math.max(ans, dp[N][K][m]);
}

System.out.println(ans);
}
} maximum sum of k non

🚀 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;
https://whatsapp.com/channel/0029VahiS3p2v1IyoS891Y1g
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;
}
https://whatsapp.com/channel/0029VahiS3p2v1IyoS891Y1g
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;
https://whatsapp.com/channel/0029VahiS3p2v1IyoS891Y1g
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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📌 Problem: Best Time to Buy & Sell Stock II

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