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 22 590 subscribers, ranking 8 822 in the Education category and 19 518 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 22 590 subscribers.

According to the latest data from 12 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -437 over the last 30 days and by -6 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 5.99%. Within the first 24 hours after publication, content typically collects 1.25% reactions from the total number of subscribers.
Post reach: On average, each post receives 1 353 views. Within the first day, a publication typically gains 283 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
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 13 June, 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.

22 590

Subscribers

-624 hours

-967 days

-43730 days

1 353

Post views

~ 28324 hours

~ 39848 hours

5.99%

Engagement rate

No data

Posts per day

Ads index

beta

Posts Archive

22 580

// coins game import java.util.*; class HelloWorld { static int helper(int a, int b, int i ,int[]x,int n){ if(i==x.length) return 0; int left=(x[i]>n)?0:Math.abs(x[i]-a)+helper(x[i],b,i+1,x,n); int right=(x[i]>n)?0:Math.abs(x[i]-b)+helper(a,x[i],i+1,x,n); return Math.min(left,right); } public static void main(String[] args) { Scanner sc=new Scanner(System.in); int N=sc.nextInt(); int Q=sc.nextInt(); int A=sc.nextInt(); int B=sc.nextInt(); int[]x=new int[Q]; for(int i=0;i<Q;i++){ x[i]=sc.nextInt(); } System.out.println(helper(A,B,0,x,N)); } }

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All codes are available Once check it 👇👇 https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No

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Short String

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Unique subarray sum solution

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Python

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Tree lis Query

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import sys def get_ans(N, K, A): def main(): N = int(sys.stdin.readline().strip()) K = int(sys.stdin.readline().strip()) A = [] for _ in range(N): A.append(int(sys.stdin.readline().strip())) result = get_ans(N, K, A) print(result) if name == "main": main() Subset with LCM code in python

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Python 3 Music melodies

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Music melodies

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This is the code Everyone write neatly All test cases passed Python 3 Infosys

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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

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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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Python

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Sub set with LCM

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Minimum difference pairs in python

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Equilibrium point Infosys all test cases passed

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Minimum unique sum

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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