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def Love(n, k, arr):
    count = 0
    for num in arr:
        if num == k:
            count += 1
        else:
            val = 1
            while True:
                ans = val & k
                if ans == num:
                    count += 1
                    break
                if ans > num or val > k:
                    break
                val += 1
    print(count)

n = int(input())
k = int(input())
arr = list(map(int, input().split()))

Love(n, k, arr)
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What is the difference between

I like you & I love you

Beautifully answered by Buddha :

When you like a flower, you just pluck it But when you love a flower, you water it daily !

One who understand this, understands life.


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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 Amazon’s 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.
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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
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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
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