Task of Pairing

task_of_pairing.py

def taskOfPairing(freq):
  count = 0
  marker = False
  for i in freq:
    if i != 0:
      count += i // 2
      if i % 2 != 0 and marker:
        count += 1
        marker = False
       elif i % 2 != 0:
        marker = True
    else:
      marker = False
  return count
        

Buddy, 12/15 testcases don't pass with your solution.

Buddy, 12/15 testcases don't pass with your solution.

@therajdeepbiswas Can you please paste some of the test cases here? Let me check and fix it!

Buddy, 12/15 testcases don't pass with your solution.

Change last marker to False
Then all your test cases will run

thanks a lot, bro, but change your code, there is a bug on top where you need to change marker=False

can you give full program

i wrote the same code ... i only passed 3/15 cases ..

def taskOfPairing(freq):
    dap = 0
    test = freq
    for i in range(len(test)):
        if i < len(test)-1:
            if test[i]%2 == 0:
                dap += test[i]//2
                print(i,test[i]//2,dap)
            else:
                dap += test[i]//2
                print(i,test[i]//2,dap)
                test[i+1] += 1
        else:
            dap += test[i]//2
    return dap

This is my code. i think mine and your is same logic. but i only pass 3/15 cases too. so, i copy and paste your code in hacker rank. but your code pass only 3/15 cases too. Is your code is correct?
ps.
oh I found my mistake. i change my code. and it pass all testcase. but i still wonder why when i paste your code it doesnt work...

This is my solution which passes all test cases.

def taskOfPairing(freq):
    # Initialize the number of pairs
    no_of_pairs = 0
    
    # Track any remaining dumbells that can't be paired at each step
    remaining = 0

    # Iterate over the frequency list
    for count in freq:
        if count != 0:
            if remaining != 0:
                # Add pairs that can be formed by combining the current count with the remaining
                no_of_pairs += (count + remaining) // 2
            else:
                # Add pairs that can be formed from the current count alone
                no_of_pairs += count // 2
            
            # Calculate if there's any leftover that can't be paired
            if (count + remaining) % 2 != 0:
                remaining = (count + remaining) % 2
            else:
                remaining = 0
        else:
            # Reset remaining if the count is zero
            remaining = 0

    return no_of_pairs