allcoding1_official

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Minimum unique sum

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def max_sum_of_distinct_characters(S): n = len(S) left_chars = set() right_chars = set() left_count = [0] * n right_count = [0] * n for i in range(n): left_chars.add(S[i]) left_count[i] = len(left_chars) for i in range(n-1, -1, -1): right_chars.add(S[i]) right_count[i] = len(right_char) max_sum = 0 for i in range(n-1): max_sum = max(max_sum, left_count[i] + right_count[i+1]) return n- max_sum Split String code Python 3 All passed

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def max_cost_split(s): n = len(s) max_cost = 0 for i in range(1, n): a = s[:i] b = s[i:] cost_a = len(set(a)) cost_b = len(set(b)) max_cost = max(max_cost, cost_a + cost_b) return n - max_cost Infosys max cost split code in python

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import sys det solve(N, A) for i in range(N): if A[i]=0; A[1] ps=0 M-1 pm={0:-1) for i in range(N): ps+=A[i] if ps in pm: m=max(m,i-pm[ps]) else: pm[ps]=i return m def main(): Nint(sys.stdin.readline().strip()) A-[] for_ in range(N): A.append(int(sys.stdin.readline().strip())) result = solve(N, A) print(result) Largest Subarray with equal number Infosys

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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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SAP Off Campus Hiring Fresher For IT Technology Services Associate Location: Bangalore Qualification: Bachelor's Degree Work Experience: Fresher Salary: 7 - 8 LPA Apply Now:- https://jobs.sap.com/job/Bangalore-IT-Technology-Services-Associate-560066/1090655601/?feedId=244601&utm_campaign=limitedlistings&utm_source=LinkedinJobPostings

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📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT 📌Linux 📌Networking 📌Design patterns 📌Donet 📌Docker 📌Entity framework 📌Node.js 📌ASP. Net 📌Aps. Net cro 📌java 📌JavaScript 📌full stack developer Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 𝐂𝐘𝐁𝐄𝐑 𝐒𝐄𝐂𝐔𝐑𝐈𝐓𝐘 𝐀𝐋𝐋 𝐂𝐎𝐔𝐑𝐒𝐄 ⚡️ Basics ⚡️ Reconnaissance and Footprinting ⚡️ Network Scanning ⚡️ Enumeration ⚡️ Firewalls HIDs Honeypot ⚡️ Malware and Threats ⚡️ Mobile Platform ⚡️ Pentesting ⚡️ Sql Injection ⚡️ System Hacking ⚡️ Web Application ⚡️ Wireless Network ⚡️ Cloud Computing ⚡️ Web Server ⚡️ Social Engineering ⚡️ Session Hijacking ⚡️ Sniffing ⚡️ BufferOverflow ⚡️ Cryptography ⚡️ Denial Of Service All courses (100 rupees) Contact:- @meterials_available

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"Amazon Software System Development Engineer Jobs - Apply Now for On-Demand Interviews!" Position: System Development Engineer Job type: FULL TIME Salary: 15 LPA Apply Now:- https://www.amazon.jobs/en/jobs/2598111/system-development-engineer-i

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Infosys Job ID/Reference Code INFSYS-EXTERNAL-184946 Work Experience 2-5 Years Job Title SQL Database https://career.infosys.com/jobdesc?sourceId=4003&jobReferenceCode=INFSYS-EXTERNAL-184946

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