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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 84 838 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 497-o'rinni va Hindiston mintaqasida 3 505-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 84 838 obunachiga ega bo‘ldi.
06 Iyul, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 554 ga, so‘nggi 24 soatda esa -61 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 2.83% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.90% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 2 405 marta ko‘riladi; birinchi sutkada odatda 762 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 1 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.
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Yuqori yangilanish chastotasi (oxirgi ma’lumot 07 Iyul, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
84 838
Obunachilar
-6124 soatlar
-3767 kunlar
-1 55430 kunlar
Postlar arxiv
84 838
![array AR of size N perfect_sum(arr, s, result) : x = [0]*len(arr) j = len(arr) - 1 while (s > 0) : x[j] = s % 2 s = s // 2 j](https://cdn.tlmtr.io/_next/static/media/empty-channel.35ed232f.svg)
array AR of size N
perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) - 1
while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print("{",end="");
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= ", ");
print("}, ",end="")
def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)
# Driver code
arr = [ ]
n = int(input("Enter length of array : "))
s=int(input("Enter sum : "))
for i in range(n):
ele=int(input("Enter element : "))
arr.append(ele)
print_subset(arr, s)
Python
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84 838
class CountOccurrences {
public static void main(String[] args) {
String str = "aabbccd";
char a[]=str.toCharArray();
int cnt = 0;
for (int i = 0; i < a.length; i++) {
for (int j = i + 1; j < a.length; j++)
{
if ((a[i] == a[j])) {
cnt++;
}
}
}
System.out.println(cnt);
}
}
Java
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84 838
// 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;
}
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84 838
Number Of Subsequences Code For Infosys
int A[] = {10,13,7,8,14,11};
int n = 6;
int memo[6];//initialized with -1s;
int count(int currIndex){
if (currIndex == n-1) return 1;
if (memo[currIndex] != -1) return memo[currIndex];
int res = 0;
for (int i=currIndex+1 ; i<n ; i++){
if (abss(A[currIndex] - A[i]) <= 3){
res += count(i);
}
}
memo[currIndex] = res;
return res;
}
c++
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84 838
n = int(input())
A = int(input())
A1 = []
A2= []
A3= []
ans = 0
for i in range(n):
A1.append(int(input()))
for j in range(n):
A2.append(int(input()))
for k in range(n):
A2.append([p[k], b[k]])
a.sort()
for z in a:
if a[0] > N:
break
N += a[1]
ans += 1
print(ans)
Try this code for this question... Python 3
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84 838
Infosys exam Ans 29/7/22 at 10AM
http://www.joboffersadda.com/2022/01/infosys-exam-answer.html
http://www.joboffersadda.com/2022/01/infosys-exam-answer.html
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Once check it after msg in group
Allcoding Ans are available
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Infosys sending test mails for SP and DSE Roles. The exam is on 29th July 2022. Please check your mails.
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Ask your 10 friends to follow Instagram And subscribe🛑 & LIKE👍 youtube channel and send screenshot to
He will add you to private channel for tomorrow Ans
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https://youtu.be/ofDvcsP3OWY
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Qualification: BE/B-Tech
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