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 21 494 subscribers, ranking 9 167 in the Education category and 19 105 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 21 494 subscribers.
According to the latest data from 04 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -351 over the last 30 days and by -4 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 7.55%. 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 1 625 views. Within the first day, a publication typically gains 270 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 05 September, 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.
21 494
Subscribers
-424 hours
-827 days
-35130 days
Posts Archive
21 494
🎯Accenture Hiring for HR Service Delivery New Associate-HR Operations
Company Name Accenture
Position HR Service Delivery New Associate-HR Operations
Qualifications Any Graduation
Experience 0 – 1 (Years)
Job Location Bengaluru, India
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1_official
21 494
🎯Virtusa Hiring Data Engineer
Qualifications: Bachelor's/ Master's Degree
Salary: Up to 8 LPA (Expected)
Batch: 2019/ 20/ 21/ 22/ 23
Location: Chennai/ Work From Home
Apply Now:- WWW.allcoding1.com
Telegram:- @allcoding1
21 494
Infosys exam coding Ans
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Allcoding Ans are available
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21 494
INFOSYS ALL EXAM ANS ARE AVAILABLE FREE OFF COST
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21 494
You are given a rooted tree of N vertices and an array A of N integers A[i] is the parent of the vertex (i+1 or 0 if the vertex (i+1) is the root.
For each vertex V from 1 to N, the answer K is the total number of subsets of vertices with the LCA (lowest common ancestor) equal to V Since the of K can be large calculate it modulo 10 ^ 9 + 7
Let there be an integer array Res where Res[i] contains the answer for (I + 1)th vertex, Find the array Res
Notes:
• The lowest common ancestor (LCA) is defined for X nodes A1, A2,.... Ax as the lowest node in the tree that has all A1 A2..... Ax as descendants (where we allow a node to be a descendant of itself)
Input Formate
The first line contains an integer N denoting the number of elements in A
Each line 1 of the N subsequent lines (where 0<=i<N) contains an integer
describing All It is given that A[i] denotes the parent of the vertex (i+1) or 0
If the vertex (i+1) is the root
//@allcoding1
const int N = 100005;
const int MOD = 1e9 + 7;
int a[N], dp[N], res[N];
vector<int> g[N];
void dfs(int u) {
dp[u] = 1;
for (int v : g[u]) {
dfs(v);
dp[u] = 1ll * dp[u] * (dp[v] + 1) % MOD;
}
res[u] = (1ll * dp[u] * (1 << g[u].size()) - 1 + MOD) % MOD;
}
vector<int> functionName(int n,vector<int>A)
{
g=A;
dfs(1);
return res;
}
Language c++
21 494
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Telegram:- @allcoding1
@allcoding1
discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
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21 494
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Telegram:- @allcoding1
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discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
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21 494
+1INFOSYS EXAM ANS 10AM
ALL Slots are available
Telegram:- @allcoding1
@allcoding1
discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
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21 494
+1Java
INFOSYS EXAM ANS 10AM
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Telegram:- @allcoding1
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discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
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21 494
// Infosys
// N flowers on a Recatangular pana
int ans = 100000000;
void solve(vector<int> a, int n, int k, int index, int sum,
int maxsum)
{
if (k == 1)
{
maxsum = max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++)
{
sum += a[i];
}
maxsum = max(maxsum, sum);
ans = min(ans, maxsum);
return;
}
sum = 0;
for (int i = index; i < n; i++)
{
sum += a[i];
maxsum = max(maxsum, sum);
solve(a, n, k - 1, i + 1, sum, maxsum);
}
}
int GetMaxBeauty(int N, int K, vector<int> A)
{
solve(A, N, K, 0, 0, 0);
return ans;
}
C++
INFOSYS EXAM ANS 10AM
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Telegram:- @allcoding1
@allcoding1
discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
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21 494
def getLargestString(s, k):
frequency_array = [0] * 26
for i in range(len(s)):
frequency_array[ord(s[i]) -
ord('a')] += 1
ans = ""
i = 25
while i >= 0:
if (frequency_array[i] > k):
temp = k
st = chr( i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
frequency_array[i] -= k
j = i - 1
while (frequency_array[j] <= 0 and
j >= 0):
j -= 1
if (frequency_array[j] > 0 and
j >= 0):
str1 = chr(j + ord( 'a'))
ans += str1
frequency_array[j] -= 1
else:
break
elif (frequency_array[i] > 0):
temp = frequency_array[i]
frequency_array[i] -= temp
st = chr(i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
else:
i -= 1
return ans
if name == "main":
S = input()
k = 3
print (getLargestString(S, k))
Python
Bob code
Telegram:- @allcoding1
21 494
Python
Once check it out put
INFOSYS EXAM ANS 10AM
ALL Slots are available
Telegram:- @allcoding1
@allcoding1
discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
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