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Your code is wrong. Used Java passed all the test cases: public static int countTeams(int teamSize_1, int teamSize_2, int p) { // Write your code here int maxTeams = Math.min(p / teamSize_1, p / teamSize_2); int maxTeamSize = Math.max(teamSize_1,teamSize_2); int minTeamSize = Math.min(teamSize_1,teamSize_2); for (int i = maxTeams; i >= 0; i --) { if ( (p - i * maxTeamSize) % minTeamSize == 0) { return i + (p - i * maxTeamSize) / minTeamSize; } } return -1; }
I mean the overlap isn't an issue necessarily if I'm first going through each column from 1 to 2024 and coloring the topmost 1000 squares in each column then the area I have left to work with is 1024 completely uncolored rows, from which I can effectively paint 1000 squares in any way I want since it won't matter. ultimately end up with 2024*1000 + 1024*1000 = 3,048,000 no?
hello sir, how do we solve for a p+ polysilicon device? expected to find Vtp and Φms flat band voltage is also not given. Determine the metal - semiconductor work function difference and the threshold voltage for a silicon MOS device at 𝑇 = 3 0 0 𝐾 for the following parameters: 𝑝 + polysilicon gate, 𝑁 𝑎 = 2 × 1 0 1 6 𝑐 𝑚 - 3 , 𝑡 𝑜 𝑥 = 8 𝑛 𝑚 = 8 0 Å , and 𝑄 𝑠 𝑠 ' = 2 × 1 0 1 0 𝑐 𝑚 - 2 .
def countTeams(teamSize_1, teamSize_2, p): # Calculate the maximum number of teams using the smaller team size max_teams = min(p // teamSize_1, p // teamSize_2) # Check if the total number of participants is divisible by both team sizes if p % teamSize_1 == 0 and p % teamSize_2 == 0: return max_teams else: return -1 # Example usage teamSize_1 = 3 teamSize_2 = 4 p = 7 print(countTeams(teamSize_1, teamSize_2, p)) # Output: 2
def repeated_digit(n): # Check if a number has repeated digits s = set() while n != 0: d = n % 10 if d in s: return False s.add(d) n = n // 10 return True def calculate(L, R): answer = 0 for i in range(L, R + 1): if repeated_digit(i): answer += 1 return answer # Example usage L, R = 80, 120 print(calculate(L, R)) # Output: 27
def getMinMachines(start, end): # Combine start and end times into a single list of tuples tasks = [(s, e) for s, e in zip(start, end)] # Sort tasks based on end times tasks.sort(key=lambda x: x[1]) # Initialize a list of machines machines = [] for task in tasks: # Check if there's an available machine assigned = False for machine in machines: if machine[-1][1] <= task[0]: machine.append(task) assigned = True break # If no available machine, create a new one if not assigned: machines.append([task]) # Number of machines created is the answer return len(machines) # Example usage start = [2, 1, 5, 5, 8] end = [5, 3, 8, 6, 12] print(getMinMachines(start, end)) # Output: 3
Here's my attempt at solving this: (Python) ``` def get_freq(word): return {c: word.count(c) for c in set(word)} def get_minimal_moves(word): freq = get_freq(word) moves = 0 new_word = word while any(freq[c] > 1 for c in freq): sorted_freq = sorted(freq.items(), key=lambda x:x[1], reverse=True) c = sorted_freq[0][0] indices = [i for i, x in enumerate(new_word) if x == c] i = freq[c] // 2 index = indices[i] left, right = new_word[:index], new_word[index+1:] if c in left or c in right: new_l = left new_r = right if c in left: j = indices[i-1] new_l = left[:j] + left[j+1:] if c in right: j = indices[i+1] - index - 1 new_r = right[:j] + right[j+1:] new_word = new_l + c + new_r moves += 1 freq = get_freq(new_word) return moves ```
The English lecture at HackerElementary School focuses on teaching students the letters of the alphabet. Here’s how the students can achieve a minimal word length: For each character c in the string: Find the first occurrence of c to the left of the current index. Find the first occurrence of c to the right of the current index. Delete both occurrences (if they exist). Repeat step 1 for all characters in the string. For example, consider the word adabacaea: Choosing index 4 (0-based) with character ‘a’: Delete the first occurrence of ‘a’ to the left (index 2). Delete the first occurrence of ‘a’ to the right (index 6). The resulting word is adbacea. Repeat this process for all characters. The minimum number of moves required to obtain a word of minimal length is the total number of deletions performed.
How to make a screenshot of the text screen and to reload the screen using two batch files and some internal Debug commands: n, r w, q and n, r, l, q and segment address, file name and file size. Test in 4 steps from DosBox command prompt: dir ScreS.bat Screen1.SAV cls ScreL.bat Screen1.SAV @echo off REM ScreS.bat filename REM Save text screen to file. echo n %1>tmp.deb echo rcx>>tmp.deb echo FA0>>tmp.deb echo rds>>tmp.deb echo B800>>tmp.deb echo wds:0>>tmp.deb echo q>>tmp.deb debug<tmp.deb>nul del tmp.deb @echo off REM ScreL.bat filename REM Load text screen from file. echo n %1>tmp.deb echo rcx>>tmp.deb echo FA0>>tmp.deb echo rds>>tmp.deb echo B800>>tmp.deb echo lds:0>>tmp.deb echo q>>tmp.deb debug<tmp.deb>nul del tmp.deb
These are internal Debug commands. All values are hexadecimal. I like to use a Debug version in DosBox that provide 16 bit and 32 bit instructions. DosBox emulates a PC with intel CPU 80386/80387.
