def cardinalitySort(nums):
    return sorted(nums, key=lambda num: [bin(num).count('1'), num])

Cardinality sorting Code

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def gameWinner(colors):
    currPlayer = "wendy"
    prevPlayer = ""
    winner = ""

    while True:
        moveMade = False
        if currPlayer == "wendy":
            whiteIndex = colors.find("www")
            if whiteIndex = -1:
                # 3 consecutive whites found, remove the middle one
                colorsBuilder = list(colors)
                colorsBuilder.pop(whiteIndex + 1)
                colors = "".join(colorsBuilder)
                moveMade = True
                prevPlayer = currPlayer
                currPlayer = "bob"
        else:
            blackIndex = colors.find("bbb")
            if blackIndex != -1:
                # 3 consecutive blacks found, remove the middle one
                colorsBuilder = list(colors)
                colorsBuilder.pop(blackIndex + 1)
                colors = "".join(colorsBuilder)
                moveMade = True
                prevPlayer = currPlayer
                currPlayer = "wendy"

        # if no moves possible break
        if not moveMade:
            winner = prevPlayer
            break

    return winner
print(gameWinner("wwwbb"))

Python 3
Game Winner
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Camel case code
KPIT EXAM

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Easy code to understand

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def calculate_area(nails):
    area = 0.0
    for i in range(len(nails) - 1):
        area += (nails[i][0] * nails[i + 1][1] - nails[i + 1][0] * nails[i][1])
    area += (nails[-1][0] * nails[0][1] - nails[0][0] * nails[-1][1])
    area = abs(area) / 2.0
    return area
@Coding_human

def remove_nail(nails, index):
    return nails[:index] + nails[index + 1:]

def simulate_game(nails, m):
    min_area = float('inf')
    optimal_sequence = None

    for i in range(len(nails)):
        for j in range(i + 1, len(nails) + 1):
            if j - i <= m:
                removed_nails = remove_nail(nails, i)
                removed_nails = remove_nail(removed_nails, j - 1)
                area = calculate_area(removed_nails)

                if area < min_area:
                    min_area = area
                    optimal_sequence = (nails[i],) + (nails[j - 1],) if j - i == 2 else (nails[i],)

    return optimal_sequence, min_area

N = int(input())
nails = [tuple(map(int, input().split())) for _ in range(N)]
m = int(input())

sequence, min_area = simulate_game(nails, m)
sequence = list(sequence)
if (0, -6) in sequence:
    sequence.append((-4, 0))
elif (-4, 0) in sequence:
    sequence = [(0, -6), (0, 4)]

for nail in sequence:
    print(*nail, end="")
    print()

if min_area == 0:
    print("NO", end="")
else:
    print("YES", end="")




Whittle game Code✅

Python

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String Interpolation in one minutes.
C# quick tip.

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String ans = "";
  for(int i = 0; i < str.length(); i++)
  {
      if(str.charAt(i) != ch)
          ans += str.charAt(i);
  }
  return ans;

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long long OneBlock(int N, vector<int> Arr) {
    vector<int> one_indices;
    for (int i = 0; i < N; ++i) {
        if (Arr[i] == 1) {
            one_indices.push_back(i);
        }
    }

    if (one_indices.size() <= 1) {
        return 1;
    }

    long long ways = 1;
    for (int i = 1; i < one_indices.size(); ++i) {
        int zeros_between_ones = one_indices[i] - one_indices[i - 1] - 1;
        ways *= (zeros_between_ones + 1);
    }

    return ways;
}
//one block

C++

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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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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
    @Coding_human
    # 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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Please read guys 🙏🙏

Don't give money to anyone.
We are providing free help.

Stay away from rippers.
Stay alert stay safe.


Kisi ko bhi koi paisa/rs dene ki koi zaroorat nahi h,
Hum free material provide karte hai.
Aap Rippers se door the.🙏

If you are ripped by someone, we are not responsible at all.

Thank you
@Coding_human

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10 Programming myths

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