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🚨 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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