allcoding1_official
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π Analytical overview of Telegram channel allcoding1_official
Channel allcoding1_official (@allcoding1_official) in the English language segment is an active participant. Currently, the community unites 84 420 subscribers, ranking 1 497 in the Technologies & Applications category and 3 532 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 84 420 subscribers.
According to the latest data from 13 July, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 602 over the last 30 days and by -72 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 1.89%. Within the first 24 hours after publication, content typically collects 0.82% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 598 views. Within the first day, a publication typically gains 696 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 1.
- Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, dev.
π Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 14 July, 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 Technologies & Applications category.
84 420
Subscribers
-7224 hours
-4097 days
-1 60230 days
Posts Archive
84 420
Repost from allcoding1_official
Accenture Exam
Factorial Number code
python
Telegram π
@allcoding1
@allcoding1
@allcoding1
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@allcoding1
84 420
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ACCENTURE Exam
n=int(input())
fact=1
for i in range(1,n+1):
fact=fact*i;
print(str(fact))
Factorial code
python
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84 420
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ACCENTURE Exam
n=input()
s=list(map(int, input().split(',')))
maxi=max(s) # finding the maximum number
mini=min(s) # finding the minimum number
print((maxi)+(mini))
Number Sum
Python
Telegram π
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84 420
Repost from allcoding1_official
def se(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def check(f,p1,p2,prime):
for i in range(p1,p2+1):
for j in range(i+1,p2+1):
if prime[i] or prime[j]:
if f<=0:
print(i,j)
return
else:
if (i*j)/((j-i)**2)>=f:
print(i,j)
return
print("None")
f,p1,p2=map(int,input().split())
prime=se(p2)
check(f,p1,p2,prime)
POINT FORCE Code
python
Telegram π
@allcoding1
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84 420
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Accenture exam
n=int(input())
sum=0
for i in range(1,n+1):
sum=sum+(i**2)
print(sum)
Adam's Charity
Python 3
Telegram π
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84 420
Repost from allcoding1_official
public class SmithNumbers {
public static boolean isPrime(int number) {
int loop;
int prime = 1;
for(loop = 2; loop < number; loop++) {
if((number % loop) == 0) {
prime = 0;
}
}
if (prime == 1) {
return true;
} else {
return false;
}
}
public static ArrayList<Integer> primeFactors(int number) {
int sum = 0;
ArrayList al = new ArrayList();
for(int i = 2; i< number; i++) {
while(number % i == 0) {
System.out.println(i+" ");
al.add(i);
number = number/i;
}
}
if(number >2) {
System.out.println(number);
al.add(number);
}
return al;
}
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter a number :");
int num = sc.nextInt();
int sum = 0;
if(isPrime(num)) {
System.out.println("Given number is not smith number ");
} else {
ArrayList<Integer> arrayList = primeFactors(num);
System.out.println("Prime factors of the given number are"+arrayList);
Iterator it = arrayList.iterator();
while(it.hasNext()) {
int n = (int) it.next();
System.out.println(n);
while(n!=0) {
sum = sum+n%10;
n = n/10;
}
}
int temp = 0;
while(num != 0) {
temp = temp + num%10;
num = num/10;
}
System.out.println("sum of the digits of the numbers of the prime factorization: "+sum);
System.out.println("Sum of digits in the given number: "+temp);
if(temp==sum) {
System.out.println("Given number is a smith number");
} else {
System.out.println("Given number is not smith number ");
}
}
}
}
Smith number Code
java
Telegram π
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84 420
Repost from allcoding1_official
+1Good sequence code
Python
Accenture Exam Answers
Telegram π
@allcoding1
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84 420
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Cluster code
python
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int replace(int matrix,int n)
{
int i,j;
for(i=0;i<n;i++)
{
for (j=i;j==i;j++)
{
if(i==0)
{
matrix[i][j]=matrix[i][j+1]+matrix[i+1][j]+matrix[i+1][j+1];
}
else if(i==n-1)
{
matrix[i][j]=matrix[i-1][j-1]+matrix[i-1][j]+matrix[i][j-1];
}
else
{
matrix[i][j]=matrix[i-1][j-1] + matrix[i-1][j] + matrix[i-1][j+1]+ matrix[i][j-1] + matrix[i][j+1]+ matrix[i+1][j-1] + matrix[i+1][j] + matrix[i+1][j+1] ;
}
}
}
return matrix;
}
Replace diagonal Code
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84 420
Repost from allcoding1_official
ACCENTURE Exam
n=int(input())
fact=1
for i in range(1,n+1):
fact=fact*i;
print(str(fact))
Factorial code
python
allcoding1 @allcoding1
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84 420
84 420
Accenture Exam
l=int(input())
n=0
i=0
a=[]
while n<100:
i+=1
count=1
for j in range(2,i):
if i%j==0:
count=0
break
//allcoding1
if count==1:
a.append(i)
n+=1
#print(a) #to check the series
print(a[l])
Nth Prime Number
Python
84 420
Accenture Exam
l=int(input())
arr=list(map(int, input().split()))
arr=list(set(arr))
length_list=len(arr)
arr.sort()
#print("set = ",arr)
#print("length of set: ",length_list)
print(arr[length_list-2])
Second Largest Number
Python
84 420
Accenture exam 3/4/22
All codes are there once check it πππβββ
Telegram π
@allcoding1
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84 420
ACCENTURE Exam
b=int(input())
e=int(input())
m=int(input())
print((b**e)%m) //main operation
Modular Exponentiation
Python
