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allcoding1

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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 20 707 obunachidan iborat bo'lib, Taʼlim toifasida 9 717-o'rinni va Hindiston mintaqasida 19 395-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 20 707 obunachiga ega bo‘ldi.

01 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -420 ga, so‘nggi 24 soatda esa -12 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 2.97% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.03% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 615 marta ko‘riladi; birinchi sutkada odatda 214 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 02 Noyabr, 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.

20 707

Obunachilar

-1224 soatlar

-1007 kun

-42030 kun

615

Post ko'rishlar

~ 21424 soatlar

~ 32148 soatlar

2.97%

Muloqot nisbati

Ma'lumot yo'q

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20 708

C++ Telegram:- @allcoding1

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magic Number Code python 3 All test cases passed Telegram:- @allcoding1

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XOR elements C++ language Telegram:- @allcoding1

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C++ Question:- You are given Queries.each query consists of 3 integer X[i],Y[i] and N[i] where i donates the ith query The answer to ith query X[i],Y[i] and N[i] is denoted By (X[i],Y[i])^((X[i]+1),Y[i])^....^((X[i],N[i]*Y[i]) Find the sum of all answers modulo 10^9+7 Telegram:- @allcoding1

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You are given Queries.each query consists of 3 integer X[i],Y[i] and N[i] where i donates the ith query The answer to ith query X[i],Y[i] and N[i] is denoted By (X[i],Y[i])^((X[i]+1),Y[i])^....^((X[i],N[i]*Y[i]) Find the sum of all answers modulo 10^9+7

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#include <bits/stdc++.h> using namespace std; int count(string s) { int N, i, cnt = 0, ans = 0; N = s.length(); for (i = 0; i < N; i++) { if (s[i] == 'R') cnt++; if (s[i] == 'L') ans += cnt; } return ans; } int main() { string s = "RRLL"; cout << count(s) << endl; return 0; } C++ Telegram - @allcoding1

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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

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Python Find Minimum Distance Between two numbers in Array code Telegram:- @allcoding1

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if (n > k && n > a) { printf("N is the Biggest number is %d", n); return n; } if (k > n && k > a) { printf("k is the Biggest number is %d", k); return k; } if (a > n && a > k) { printf("a is the Biggest number is %d", a); return a; } Telegram:- @allcoding1

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from itertools import permutations def count(arr): z=[] perm = permutations(arr) for i in list(perm): z.append(list(i)) q=[] for i in range(len(arr)-1): x,y=arr[i],arr[i+1] for j in range(len(z)): if z[j].index(x)!=len(z[j])-1: if z[j][z[j].index(x)+1]==y: q.append(z[j]) for i in range(len(q)): if q[i] in z: z.remove(q[i]) return len(z) a= int(input()) b=list(map(int,input().strip().split())) print(count(b)) Python Telegram - @allcoding1

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Java language Telegram:- @allcoding1

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Java language Telegram:- @allcoding1

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Python 3 Telegram:- @allcoding1

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Infosys exam Answer's (22/10/22) https://www.joboffersadda.com/2022/01/infosys-exam-answer.html?m=1 https://www.joboffersadda.com/2022/01/infosys-exam-answer.html?m=1 https://www.joboffersadda.com/2022/01/infosys-exam-answer.html?m=1 Telegram:- @Infosys_Ans