Endi mavjud! Telegram Tadqiqoti 2025 — yilning asosiy insaytlari 
allcoding1
Kanalga Telegram’da o‘tish
Ko'proq ko'rsatish
📈 Telegram kanali allcoding1 analitikasi
allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 509 obunachidan iborat bo'lib, Taʼlim toifasida 8 844-o'rinni va Hindiston mintaqasida 19 350-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 22 509 obunachiga ega bo‘ldi.
18 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -405 ga, so‘nggi 24 soatda esa -11 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 6.44% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.27% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 450 marta ko‘riladi; birinchi sutkada odatda 287 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, learning kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Kanal uchun tavsif kiritilmagan.
Yuqori yangilanish chastotasi (oxirgi ma’lumot 19 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
22 509
Obunachilar
-1124 soatlar
-887 kunlar
-40530 kunlar
Postlar arxiv
22 509
INFOSYS ALL EXAM ANS ARE AVAILABLE FREE OFF COST
@Infosys_Ans @Infosys_Ans
@Infosys_Ans @Infosys_Ans
NOT:-. ONCE CHECK IT
22 509
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++
22 509
INFOSYS EXAM ANS 3PM
ALL Slots are available
Telegram:- @allcoding1
@allcoding1
discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
Share with your friends
22 509
+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=
Share with your friends
22 509
+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=
Share with your friends
22 509
+1Java
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=
Share with your friends
22 509
// 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
ALL Slots are available
Telegram:- @allcoding1
@allcoding1
discussion group:- @Infosys_exam_Ans
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
Share with your friends
22 509
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
22 509
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=
Share with your friends
22 509
+1Python
INFOSYS EXAM ANS 10AM
ALL Slots are available
Telegram:- @allcoding1_official
@allcoding1_official
discussion group:- @Infosys_examAns
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
Share with your friends
22 509
+1Java
INFOSYS EXAM ANS 10AM
ALL Slots are available
Telegram:- @allcoding1_official
@allcoding1_official
discussion group:- @Infosys_examAns
Coding Ans :- https://instagram.com/allcoding1_official?igshid=OGQ2MjdiOTE=
Share with your friends
22 509
def next_stepping_number(N):
N = str(N)
for i in range(len(N) - 1):
if abs(int(N[i]) - int(N[i+1])) != 1:
return int(N[:i+1] + str(int(N[i]) + 1 if int(N[i]) < int(N[i+1]) else int(N[i]) - 1) + '9'*(len(N)-i-1))
return int(N) + 1
print(next_stepping_number(4)) # should return 5
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=
Share with your friends
22 509
+2Python
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=
Share with your friends
22 509
def shortest_subarray(colors, C):
n = len(colors)
color_count = [0] * (C+1)
left, right = 0, 0
min_length = float('inf')
count = 0
while right < n:
color_count[colors[right]] += 1
if color_count[colors[right]] == 1:
count += 1
while count == C:
min_length = min(min_length, right - left + 1)
color_count[colors[left]] -= 1
if color_count[colors[left]] == 0:
count -= 1
left += 1
right += 1
if min_length == float('inf'):
return -1
else:
return min_length
Python3
