CodeChef Contest 184 – All Questions Solved FREE 💻🔥 | 30 April 2025 | 3⭐ Rated
CodeChef Contest 184 – All Questions Solved FREE 💻🔥 | 30 April 2025 | 3⭐ Rated
👇Scroll down to get all solutions!👇
Are you ready for another exciting CodeChef-rated contest? Starters 184 is here to test your coding skills and problem-solving abilities! Whether you're a beginner or an intermediate coder (rated up to 5 stars), this is your chance to compete, improve, and rank up.
🗓️ Contest Date: April 30, 2025
⏰ Time: 08:00 PM – 10:00 PM IST | 🕒 Duration: 2 Hours
🏆 Contest Name: CodeChef Starters 184
🚨 Note: The Code of Conduct has been updated — make sure to follow the latest rules!!
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NOTE :- Don’t copy the full code — use the logic.
We're not responsible for plagiarism issues.
Thank you!
Q1 Solutions -
n = int(input())
rem = n % 3
if rem == 0:
print(n)
elif rem == 1:
print(n - 1)
else:
print(n + 1)
Q2 Solutions -
test_cases = int(input())
for _ in range(test_cases):
length = int(input())
if length == 1:
print(-1)
elif length % 2 == 0:
result = [1, -1] * (length // 2)
print(*result)
else:
result = [1, 2, -3]
for _ in range((length - 3) // 2):
result += [1, -1]
print(*result)
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Q4 Solutions -
import sys
def mod_pow(base, exp, mod):
result = 1
base %= mod
while exp:
if exp & 1:
result = (result * base) % mod
base = (base * base) % mod
exp >>= 1
return result
def solve():
n = int(sys.stdin.readline())
arr = list(map(int, sys.stdin.readline().split()))
MOD = 998244353
# Count occurrences of each value in [0..n-1]
counts = [0]*n
for x in arr:
if 0 <= x < n:
counts[x] += 1
else:
print(0)
return
# Check the “middle” element if n is odd
if n & 1:
mid = n//2
if counts[mid] != 1:
print(0)
return
# Check pairing counts[k] + counts[n-1-k] == 2
for k in range(n//2):
if counts[k] + counts[n-1-k] != 2:
print(0)
return
# All good → output 2^(n/2) mod
print(mod_pow(2, n//2, MOD))
# Driver
t = int(sys.stdin.readline())
for _ in range(t):
solve()
Q5 Solutions
import sys
input = sys.stdin.readline
def process_ops(n, ops):
# state[i] = whether value i has been “used”
used = [False] * (n + 5)
used[0] = True
next_free = 1 # next integer not yet in state
current_size = 1 # how many values are in state
inv_count = 0 # inversion-like counter
cross_count = 0 # cross-term accumulator
result = []
for op in ops:
if op == 1:
# Add a single new element
inv_count += current_size
cross_count += current_size
used[next_free] = True
current_size += 1
# Advance next_free to the next unused integer
while used[next_free]:
next_free += 1
else:
# Double the structure
inv_count = inv_count * 2 + cross_count
cross_count *= 4
current_size *= 2
result.append(inv_count)
return result
def main():
t = int(input())
for _ in range(t):
m = int(input())
ops = list(map(int, input().split()))
print(*process_ops(m, ops))
if __name__ == "__main__":
main()
Q6 Solutions- (if You want Solution , then Subscribe)
import sys
input = sys.stdin.readline
MOD = 998244353
class SMEXCalculator:
def __init__(self, A):
self.A = A
self.N = len(A)
self.MOD = MOD
def modinv(self, x):
# Fermat’s little theorem
return pow(x, self.MOD - 2, self.MOD)
def build_counts(self):
cnt = [0] * (self.N + 2)
for v in self.A:
cnt[v] += 1
return cnt
def build_prefixes(self, cnt):
Zcnt = [0] * (self.N + 3)
P_all = [1] * (self.N + 3)
P_nz = [1] * (self.N + 3)
S_inv = [0] * (self.N + 3)
for k in range(1, self.N + 2):
zero_here = 1 if cnt[k-1] == 0 else 0
Zcnt[k] = Zcnt[k-1] + zero_here
Q = (pow(2, cnt[k-1], self.MOD) - 1) % self.MOD
P_all[k] = P_all[k-1] * Q % self.MOD
if cnt[k-1] > 0:
P_nz[k] = P_nz[k-1] * Q % self.MOD
S_inv[k] = (S_inv[k-1] + self.modinv(Q)) % self.MOD
else:
P_nz[k] = P_nz[k-1]
S_inv[k] = S_inv[k-1]
return Zcnt, P_all, P_nz, S_inv
def build_suffixes(self, cnt):
p2 = [pow(2, c, self.MOD) for c in cnt]
suf = [1] * (self.N + 4)
for v in range(self.N + 1, -1, -1):
suf[v] = p2[v] * suf[v+1] % self.MOD
return suf
def compute(self):
cnt = self.build_counts()
Zcnt, P_all, P_nz, S_inv = self.build_prefixes(cnt)
suf = self.build_suffixes(cnt)
ans = 0
for k in range(1, self.N + 2):
z = Zcnt[k]
if z >= 2:
sumR = 0
elif z == 1:
sumR = P_nz[k]
else:
sumR = P_all[k] * S_inv[k] % self.MOD
Ck = sumR * suf[k+1] % self.MOD
ans = (ans + k * Ck) % self.MOD
# subtract empty-subsequence contribution
return (ans - 1) % self.MOD
def solve_one(A):
return SMEXCalculator(A).compute()
def main():
T = int(input())
out = []
for _ in range(T):
n = int(input())
A = list(map(int, input().split()))
out.append(str(solve_one(A)))
print("\n".join(out))
if __name__ == "__main__":
main()
Q7 Solving Now........
T = int(input())
for _ in range(T):
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
# Check if boundaries match
if A[0] != B[0] or A[N-1] != B[N-1]:
print("No")
continue
# Check blocks of 1s in B
block_ok = True
i = 0
while i < N and block_ok:
if B[i] == 0:
i += 1
continue
l = i
while i < N and B[i] == 1:
i += 1
r = i - 1
# Check if there is at least one 1 in A[l:r+1]
has_one = any(A[k] == 1 for k in range(l, r+1))
if not has_one:
block_ok = False
if not block_ok:
print("No")
continue
# Check positions where B[i] = 0 and A[i] = 1
set_ok = True
for j in range(1, N-1):
if B[j] == 0 and A[j] == 1:
if A[j-1] == 1 or A[j+1] == 1:
set_ok = False
break
if set_ok:
print("Yes")
else:
print("No")
6 error coming
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