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https://www.hackerearth.com/challenges/new/competitive/amazon-ml-challenge-2023/
Amazon ML Challenge is a two stage competition where students from all engineering campuses across India will get a unique opportunity to work on Amazonβs dataset to bring in fresh ideas and build innovative solutions for a real world problem statement. Top three winning teams will receive cash prizes and certificates.
Registrations will remain open from 3 April 2023 to 20 April 2023 11:59 PM IST. Participation starts from 21 April 2023 12 AM IST to 23 April 2023 11:59 PM IST.
Amazon ML Challenge is a two stage competition where students from all engineering campuses across India will get a unique opportunity to work on Amazonβs dataset to bring in fresh ideas and build innovative solutions for a real world problem statement. Top three winning teams will receive cash prizes and certificates.
Registrations will remain open from 3 April 2023 to 20 April 2023 11:59 PM IST. Participation starts from 21 April 2023 12 AM IST to 23 April 2023 11:59 PM IST.
HackerEarth
Amazon ML Challenge 2023
Registrations Closed!
About Amazon ML Challenge
Amazon ML Challenge is a two stage competition where students from all engineering campuses across India will get a unique opportunity to work on Amazonβs dataset to bring in fresh ideas and build innovativeβ¦
About Amazon ML Challenge
Amazon ML Challenge is a two stage competition where students from all engineering campuses across India will get a unique opportunity to work on Amazonβs dataset to bring in fresh ideas and build innovativeβ¦
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XOR Subsequences
Python
Trilogy Innovations
def bit_count(x):
cnt = 0
while x > 0:
cnt += x & 1
x >>= 1
return cnt
n = int(input())
a = list(map(int, input().split()))
dp = [set() for i in range(n)]
dp[0].add((a[0],))
ans = 0
for i in range(1, n):
dp[i].add((a[i],))
for b in dp[i-1]:
if a[i] > b[-1]:
b2 = b + (a[i],)
dp[i].add(b2)
if len(b) > 1 and bit_count(b[-2] ^ b[-1]) == bit_count(b[-2] ^ a[i]):
b2 = b[:-1] + (a[i],)
dp[i].add(b2)
for b in dp[i]:
x = b[0]
for j in range(1, len(b)):
x ^= b[j]
ans |= x
print(bin(ans).count('1'))
XOR Subsequences
Python
Trilogy Innovations
Python
Trilogy Innovations
def bit_count(x):
cnt = 0
while x > 0:
cnt += x & 1
x >>= 1
return cnt
n = int(input())
a = list(map(int, input().split()))
dp = [set() for i in range(n)]
dp[0].add((a[0],))
ans = 0
for i in range(1, n):
dp[i].add((a[i],))
for b in dp[i-1]:
if a[i] > b[-1]:
b2 = b + (a[i],)
dp[i].add(b2)
if len(b) > 1 and bit_count(b[-2] ^ b[-1]) == bit_count(b[-2] ^ a[i]):
b2 = b[:-1] + (a[i],)
dp[i].add(b2)
for b in dp[i]:
x = b[0]
for j in range(1, len(b)):
x ^= b[j]
ans |= x
print(bin(ans).count('1'))
XOR Subsequences
Python
Trilogy Innovations
π5β€1
Good Arrays Code
C++
Trilogy Innovations
int n = arr.size();
int A = arr[0];
int B = arr[n-1];
int k = 0;
for (int i = 1; i < n-1; i++) {
if (arr[i] > A) {
k++;
}
}
long long ways = 1;
for (int i = 1; i <= k; i++) {
ways = (ways * i) % 1000000007;
}
return ways;
Good Arrays Code
C++
Trilogy Innovations
C++
Trilogy Innovations
int n = arr.size();
int A = arr[0];
int B = arr[n-1];
int k = 0;
for (int i = 1; i < n-1; i++) {
if (arr[i] > A) {
k++;
}
}
long long ways = 1;
for (int i = 1; i <= k; i++) {
ways = (ways * i) % 1000000007;
}
return ways;
Good Arrays Code
C++
Trilogy Innovations
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XOR Subsequences
Pythonβ β
def bit_count(x):
cnt = 0
while x > 0:
cnt += x & 1
x >>= 1
return cnt
n = int(input())
a = list(map(int, input().split()))
dp = [set() for i in range(n)]
dp[0].add((a[0],))
ans = 0
for i in range(1, n):
dp[i].add((a[i],))
for b in dp[i-1]:
if a[i] > b[-1]:
b2 = b + (a[i],)
dp[i].add(b2)
if len(b) > 1 and bit_count(b[-2] ^ b[-1]) == bit_count(b[-2] ^ a[i]):
b2 = b[:-1] + (a[i],)
dp[i].add(b2)
for b in dp[i]:
x = b[0]
for j in range(1, len(b)):
x ^= b[j]
ans |= x
print(bin(ans).count('1'))
Pythonβ β
def bit_count(x):
cnt = 0
while x > 0:
cnt += x & 1
x >>= 1
return cnt
n = int(input())
a = list(map(int, input().split()))
dp = [set() for i in range(n)]
dp[0].add((a[0],))
ans = 0
for i in range(1, n):
dp[i].add((a[i],))
for b in dp[i-1]:
if a[i] > b[-1]:
b2 = b + (a[i],)
dp[i].add(b2)
if len(b) > 1 and bit_count(b[-2] ^ b[-1]) == bit_count(b[-2] ^ a[i]):
b2 = b[:-1] + (a[i],)
dp[i].add(b2)
for b in dp[i]:
x = b[0]
for j in range(1, len(b)):
x ^= b[j]
ans |= x
print(bin(ans).count('1'))
π4π₯2
def count_xor_values(A):
def generate_subsequences(A, i, B):
if i == len(A):
if len(B) > 1 and all(B[j] < B[j+1] for j in range(len(B)-1)):
subsequences.append(B[:])
else:
generate_subsequences(A, i+1, B)
B.append(A[i])
generate_subsequences(A, i+1, B)
B.pop()
def count_set_bits(n):
count = 0
while n > 0:
count += n & 1
n >>= 1
return count
subsequences = []
generate_subsequences(A, 0, [])
xor_values = set()
for B in subsequences:
xor_value = B[0]
for i in range(1, len(B)):
xor_value ^= B[i]
if count_set_bits(xor_value) >= count_set_bits(B[i]):
xor_values.add(xor_value)
return len(xor_values)
def generate_subsequences(A, i, B):
if i == len(A):
if len(B) > 1 and all(B[j] < B[j+1] for j in range(len(B)-1)):
subsequences.append(B[:])
else:
generate_subsequences(A, i+1, B)
B.append(A[i])
generate_subsequences(A, i+1, B)
B.pop()
def count_set_bits(n):
count = 0
while n > 0:
count += n & 1
n >>= 1
return count
subsequences = []
generate_subsequences(A, 0, [])
xor_values = set()
for B in subsequences:
xor_value = B[0]
for i in range(1, len(B)):
xor_value ^= B[i]
if count_set_bits(xor_value) >= count_set_bits(B[i]):
xor_values.add(xor_value)
return len(xor_values)
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