allcoding1_official
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📈 Telegram 频道 allcoding1_official 的分析概览
频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 84 738 名订阅者,在 技术与应用 类别中位列第 1 498,并在 印度 地区排名第 3 519 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 84 738 名订阅者。
根据 08 七月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -1 499,过去 24 小时变化为 -64,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 2.62%。内容发布后 24 小时内通常能获得 0.88% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 2 218 次浏览,首日通常累积 743 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 1。
- 主题关注点: 内容集中在 dsa, stack, namaste, javascript, dev 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新(最新数据采集于 09 七月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 技术与应用 类别中的关键影响点。
84 738
订阅者
-6424 小时
-4007 天
-1 49930 天
帖子存档
84 717
Guys ❤️
✅SALES
🧿telegram ( @allcoding1 channel) =30k rupees
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If anybody want
contact:-@Priya_MD
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84 717
Christmas Eve code in java :
public class Example2 {
// this will be the solution
static int solutionCount = 0;
public static void canCollect(int remainingAmount, int remainingHouses, int previousAmount) {
if (remainingHouses == 0)
return;
if (remainingAmount < 0)
return;
if (remainingAmount == 0) {
solutionCount++;
return;
}
for (int i = 1; i < previousAmount; i++) {
//deduct amount collected from this house-visit (i), deduct 1 from remaining houses, and set previousAmount to what you are collecting here
canCollect(remainingAmount - i, remainingHouses - 1, i);
}
}
// This is your solution function, although you'll need to change allcoding1 it enough to satisfy test-cases.
public static int solutionFunction(int x, int y) {
for (int i = 0; i <= x; i++) {
canCollect(x - i, y-1, i);
}
return solutionCount;
}
//main is just to execute it here. Omit it in your submission
public static void main(String args[]) {
System.out.println(solutionFunction(5,5)); //3
}
}
Java
Christmas Eve
Telegram - @allcoding1
84 717
Repost from allcoding1_official
Playing with Number
static int[] shiftArray(int input1, int[] input2, int input3)
{
int ar[]=new int[input2.length];
int j=0;
for(int i=input3;i<input2.length;i++)
{
ar[j++]=input2[i];
}
for(int i=0; i<input3;i++)
{
ar[j++]=input2[i];
}
return ar;
}
Test cases
input1=7;
input2[]= {1,2,3,4,5,6,7};
input3=2;
output:
3 4 5 6 7 1 2
Array Sort(nick checks)
private static int getCount(int input1, int[] input2)
{
int ar[]=new int[input2.length];
for(int i =0;i<input2.length;i++)
{
ar[i]=input2[i];
}
Arrays.sort(ar);
int num=ar[0];
for(int i=0;i<input2.length;i++)
{
if(ar[i]!=num)
return 0;
num++;
}
return 1;
}
Test Case:
input1:6
input2[]: {3,7,2,5,4,6}
output:
1
Fancy Occurrence
private static String fancyOcc(String input1, String input2)
{
String s="";
char com=input2.charAt(0);
for(int i= 0; i<input1.length();i++)
{
char ch=input1.charAt(i);
if(ch!=com)
s=s+ch;
}
return s;
}
Test Case:
input1:Welcome to metti
input2:i
output:
Welcome to mett
String Within String
private static String isPermutation(String input1,String input2)
{
input1=input1.toLowerCase();
input2=input2.toLowerCase();
int arr1[] = new int[26];
int arr2[] = new int[26];
for(int i=0; i<input1.length();i++)
{
char ch = input1.charAt(i);
arr1[ch-97]++;
}
for(int i=0;i<input2.length();i++)
{
char ch1 = input2.charAt(i);
arr2[ch1-97]++;
if(arr1[ch1-97]<arr2[ch1-97])
return "no";
}
return "yes";
}
Test Case:
input1: abab
input2: ab
output:
yes
Palindrome
private static int isPalindrome(String input1)
{
int j = input1.length();
for(int i=0;i<s.length()/2;i++)
{
char ch= input1.charAt(i);
if(ch!= input1.charAt(j-1))
return 0;
j--;
}
return 1;
}
Test Case:
input1: level
output:
1
Max occurring character
private static String maxOccurance(String input1)
{
char[] c=new char[26];
String s="";
for(int i=0;i<input1.length();i++)
{
char ch= input1.charAt(i);
c[ch-97]++;
}
int max=0;
for(int i=1;i<26;i++)
{
if(c[max]<c[i])
max=i;
}
for(int i=0;i<26;i++)
{
if(c[i]==c[max]&&i!=max)
return "0";
}
return s+(char)(max+97);
}
Test cases:
input1 : abcdd
output : d
Number Sum
private static int sum(int input1, int[] input2)
{
int max=input2[0];
int min=input2[0];
for(int i=1;i<input2.length;i++)
{
if(max<input2[i])
max=input2[i];
if(min>input2[i])
min=input2[i];
}
return max+min;
}
Test cases:
input 1: 7
input 2: { 7,2,6,15,54,10,23}
output : 56
Anagram
private static String isAnagarm (String input1,String input2)
{
int n1 = input1.length();
int n2 = input2.length();
if (n1 != n2)
return "no";
input1=input1.toUpperCase();
input2=input2.toUpperCase();
int a[]= new int[26];
int a2[]= new int[26];
for(int i=0;i<n1;i++)
{
char ch= input1.charAt(i);
a[ch-65]++;
}
for(int i=0;i<n2;i++)
{
char ch= input2.charAt(i);
a2[ch-65]++;
}
for (int i = 0; i < 26; i++)
if (a[i] != a2[i])
return "no";
return "yes";
}
Test cases:
input1:build
input : dubli
output : yes
Coin Counts
private static int getcount(int input1)
{
int coin=0;
while(input1!=0)
{
coin=coin+(input1*input1);
input1--;
}
return coin;
}
Test cases:
input1 : 2
output : 5
Electro Static
private static int electroStatic(int[] input1, String input2, int input3)
{
int sum=0;
for(int i = 0;i<input3;i++)
{
char ch = input2.charAt(i);
switch(ch)
{
case 'P':
sum=sum+input1[i];
break;
case 'N':
sum=sum-input1[i];
break;
}
}
return sum*100;
}
Test cases:
input1 : {4,3,5}
input2 : PNP
input3 : 3
output : 600
Reverse Array
public static int[] reverseArray(int [] input1,int input2) {
int [] ars=new int[input2];
int k=0;
for(int i=input2-1;i>=0;i--)
{
ars[k]=input1[i];
k++;
}
return ars;
}
@allcoding1
84 717
struct Result pushZeroesEnd(int input1[], int input2){
Result* ans = new Result();
int k = 0, count = 0;
for(int i = 0; i < input2; i++){
if(input1[i] == 0)
count += 1;
else
ans->output1[k++] = input1[i];
}
while(count--)
ans->output1[k++] = 0;
return *ans;
};
Python & C++
Push zeros to end
Telegram - @allcoding1
84 717
Repost from allcoding1_official
Missing Braces
Python 3
Telegram👇
@allcoding1 @allcoding1
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@allcoding1 @allcoding1
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84 717
Repost from allcoding1_official
Missing Braces
Python 3
Accenture Answer's
Telegram👇
@allcoding1 @allcoding1
@allcoding1 @allcoding1
@allcoding1 @allcoding1
⚠️ Share post in ur college and telegram Group's
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Both of the given options
Virtusa exam Ans 9:30 AM
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$1,200
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The amount paid...85
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75%
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6,9,5
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Hello
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