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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 78 794 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 627-o'rinni va Hindiston mintaqasida 3 772-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 78 794 obunachiga ega bo‘ldi.

29 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 610 ga, so‘nggi 24 soatda esa -51 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.18% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.61% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 718 marta ko‘riladi; birinchi sutkada odatda 478 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.

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Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 30 Oktabr, 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.

78 794
Obunachilar
-5124 soatlar
-3677 kun
-1 61030 kun
Postlar arxiv
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8) H

7) 3

6) both I and 3

1) BUSMUF 2) KXVTL 3) D 4) CANNOT BE DETERMINED 5) E WA SITTING FRONT OF I

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All this 300 rupees only Contact:- @meterials_available
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All this 300 rupees only Contact:- @meterials_available

if not intervals:         return 0         intervals.sort(ke
if not intervals:
        return 0
    
    intervals.sort(key=lambda x: x[0])  # Sort intervals based on start points
    merged_intervals = [intervals[0]]  # Initialize merged intervals with the first interval
    
    for interval in intervals[1:]:
        prev_interval = merged_intervals[-1]
        
        if interval[0] <= prev_interval[1]:  # If overlapping with previous interval
            prev_interval[1] = max(prev_interval[1], interval[1])  # Merge intervals
        else:
            merged_intervals.append(interval)  # Add non-overlapping interval
    
    total_length = sum(end - start for start, end in merged_intervals)
    return total_length

# Example usage:
intervals = [(1, 3), (2, 6), (8, 10), (15, 18)]
total_length_covered = merge_intervals(intervals)
print("Total length covered after merging overlapping intervals:", total_length_covered)