allcoding1
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📈 Analytical overview of Telegram channel allcoding1
Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 21 584 subscribers, ranking 9 062 in the Education category and 19 025 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 21 584 subscribers.
According to the latest data from 29 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -360 over the last 30 days and by -8 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 5.78%. Within the first 24 hours after publication, content typically collects 1.10% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 247 views. Within the first day, a publication typically gains 238 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
- Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.
📝 Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 30 August, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.
21 584
Subscribers
-824 hours
-787 days
-36030 days
Posts Archive
21 578
def longest_equal_subarray():
n = int(input())
A = [int(input()) for _ in range(n)]
A = [-1 if x == 0 else 1 for x in A]
prefix_sum_map = {}
prefix_sum = 0
max_length = 0
for i in range(n):
prefix_sum += A[i]
if prefix_sum == 0:
max_length = i + 1
if prefix_sum in prefix_sum_map:
max_length = max(max_length, i - prefix_sum_map[prefix_sum])
else:
prefix_sum_map[prefix_sum] = i
return max_length
print(longest_equal_subarray())
Infosys Longest Subarray code
21 578
class TreeNode:
def init(self, value=0, left=None, right=None):
self.value = value
self.left = left
self.right = right
def count_nodes(node, counts):
if node is None:
return
if node.value in counts:
counts[node.value] += 1
else:
counts[node.value] = 1
count_nodes(node.left, counts)
count_nodes(node.right, counts)
def find_double_roots(root):
counts = {}
count_nodes(root, counts)
double_roots = [value for value, count in counts.items() if count > 1]
return double_roots
def main():
# Example tree:
# 1
# / \
# 2 3
# / \
# 2 4
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)
result = find_double_roots(root)
print("Nodes with double roots:", result)
if name == "main":
main()
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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
21 578
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
21 578
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
21 578
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))
21 578
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)
21 578
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)
21 578
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
21 578
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21 578
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)
