Forwarded from CodingHuman #Coding_Help , All exam codes, Coding solutions, accenture TCS Wipro Nagarro Persistent Cisco
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We are providing free help.
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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
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
⚡Many companies like Wipro, TCS, Accenture are sending Onboarding dates , so check your mail
✅ Telegram:
https://t.me/Coding_human
✅ Telegram:
https://t.me/Coding_human
def cardinalitySort(nums):
return sorted(nums, key=lambda num: [bin(num).count('1'), num])
Cardinality sorting Code
Telegram : https://t.me/Coding_human
return sorted(nums, key=lambda num: [bin(num).count('1'), num])
Cardinality sorting Code
Telegram : https://t.me/Coding_human
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
JP Morgan
Telegram:- https://t.me/Coding_human
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
JP Morgan
Telegram:- https://t.me/Coding_human
Camel case code
KPIT EXAM
C++ language
Easy code to understand
Share with your family and friends : https://t.me/Coding_human
KPIT EXAM
C++ language
Easy code to understand
Share with your family and friends : https://t.me/Coding_human
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Forwarded from CodingHuman #Coding_Help , All exam codes, Coding solutions, accenture TCS Wipro Nagarro Persistent Cisco
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
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
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
Telegram:- @Coding_human
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
Telegram:- @Coding_human
String ans = "";
for(int i = 0; i < str.length(); i++)
{
if(str.charAt(i) != ch)
ans += str.charAt(i);
}
return ans;
Telegram:- @Coding_human
for(int i = 0; i < str.length(); i++)
{
if(str.charAt(i) != ch)
ans += str.charAt(i);
}
return ans;
Telegram:- @Coding_human
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Designation : Associate Software Engineer
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Designation : Associate Software Engineer
Eligibility : Engineering / MCA
Salary CTC : Rs.3.25 LPA & 5.5LPA (for super coders)
Apply Now:- https://registration.techmahindra.com/Candidate/RegDefault.aspx
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++
Telegram:- @Coding_human
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++
Telegram:- @Coding_human
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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)
@Coding_human
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)
@Coding_human
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
# 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
Forwarded from CodingHuman #Coding_Help , All exam codes, Coding solutions, accenture TCS Wipro Nagarro Persistent Cisco
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
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