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Minimize Marked Fruits Python Trilogy Innovations def minimize_marked_fruits(A, B, C):     marked = [False] * A     count = 0     for i in range(C):         if not marked[B[i]-1]:             marked[B[i]-1] = True             count += 1     min_count = count     for i in range(C, A):         if not marked[B[i]-1]:             marked[B[i]-1] = True             count += 1         if not marked[B[i-C]-1]:             count -= 1             marked[B[i-C]-1] = False         min_count = min(min_count, count)     return min_count Minimize Marked Fruits Python Trilogy

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)

ᴡᴇ ᴅᴏɴ'ᴛ ɴᴇᴇᴅ ᴀɴʏ ᴏᴛʜᴇʀ ᴛʜɪɴɢs ᴊᴜsᴛ ʀᴇᴀᴄᴛ ᴏɴ ᴘᴏsᴛs ᴀɴᴅ ᴍᴀᴋᴇ ᴜs ʜᴀᴘᴘʏ ❤️🤩 Share @Coding_000❤️

XOR Subsequences Python 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'))

ᴡᴇ ᴅᴏɴ'ᴛ ɴᴇᴇᴅ ᴀɴʏ ᴏᴛʜᴇʀ ᴛʜɪɴɢs ᴊᴜsᴛ ʀᴇᴀᴄᴛ ᴏɴ ᴘᴏsᴛs ᴀɴᴅ ᴍᴀᴋᴇ ᴜs ʜᴀᴘᴘʏ ❤️🤩 Share @Coding_000❤️

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

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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