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def solve(N, A): unique_sums = set() for start in range(N): current_sum = 0 for end in range(start, N): current_sum += A[end] unique_sums.add(current_sum) print(len(unique_sums))

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You are given a string S consisting of lowercase latin letters, i.e. {a, b,c, ..z} You can perform the following operation on S any number times Remove two consecutive characters. Find the total number of distinct strings that you can generate. Note: All operations are mutually exclusive. This means that all operations are going to be performed independently on the initial string S. Input Format The first line contains a string, S. denoting the given string. Constraints 1 <= len(S) <= 10^5 Sample Test Cases Case 1 Input aaabcc Output 4 Explanation: S = "aaabcc" we can get the strings "abcc", "aacc", "aaac", and "aaab". Hence, the answer for this case is equal to 4. Case 2 Input aaaaaaaaaa Output 1 Explanation: S = "аaаааааааа" We can get only one string which is "aaaaaaaaaa". Hence, the answer for this case is equal to 1. Case 3 Input: abcdef Output: 5 Explanation: S = "abcdef" We can get the strings "cdef", "adef", "abef", "abcf", and "abcd". Hence, the answer for this case is equal to 5.

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Minimum substring ..

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def count_distinct_strings(S): distinct_strings = set() for i in range(len(S) - 1): new_string = S[:i] + S[i+2:] distinct_strings.add(new_string) return len(distinct_strings) # Read input string S = input().strip() # Get the number of distinct strings that can be generated result = count_distinct_strings(S) print(result)

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def minimum_unique_sum(A): N = len(A) A.sort() total = A[0] for i in range(1, N): if A[i] <= A[i-1]: A[i] = A[i-1] + 1 total += A[i] return total # Input format N = int(input()) A = [] for i in range(N): A.append(int(input())) # Output result = minimum_unique_sum(A) print(result)

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def split_string_cost(S): # Length of the string S len_S = len(S) # To store the cost of the split parts max_cost = 0 # Set to keep track of distinct characters in the first part distinct_chars_A = set() # List to keep track of the cost for the second part from each split position cost_B = [0] * len_S # Set to keep track of distinct characters in the second part distinct_chars_B = set() # Calculate cost for second part from the end for i in range(len_S - 1, -1, -1): distinct_chars_B.add(S[i]) cost_B[i] = len(distinct_chars_B) # Calculate maximum sum of cost for parts A and B for i in range(len_S - 1): distinct_chars_A.add(S[i]) cost_A = len(distinct_chars_A) cost = cost_A + cost_B[i + 1] max_cost = max(max_cost, cost) # Calculate the result as |S| - X result = len_S - max_cost return result # Example usage S = "aaabbb" print(split_string_cost(S)) # Output: 3

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Accenture HackDiva Test Pattern - 2 Codes Round 1-2 Codes (90 min) Round 2-2 Codes (120 min) Round 3-2 Codes (120 min) Note: Each Round is Elimination Round 1 (DATE: 07 JULY 2024) Round 2 (DATE: 14 JULY 2024) Round 3 (DATE: 21 JULY 2024)

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