CODING SOLUTION - Placement Jobs & Materials
前往频道在 Telegram
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. “ Contact Admin: Mail me for collab offcampusjobsindiaadmin@gmail.com instagram.com/offcampusjobsindia_it Our Website : www.offcampusjobsindia.com
显示更多📈 Telegram 频道 CODING SOLUTION - Placement Jobs & Materials 的分析概览
频道 CODING SOLUTION - Placement Jobs & Materials (@codingsolution_it) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 128 253 名订阅者，在 职业 类别中位列第 157，并在 印度 地区排名第 1 680 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 128 253 名订阅者。
根据 08 十一月, 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 -1 995，过去 24 小时变化为 -103，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 1.50%。内容发布后 24 小时内通常能获得 0.43% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 920 次浏览，首日通常累积 548 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 0。
- 主题关注点： 内容集中在 t.me/codingsolution_it, vector, codevita, string, graduate 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台：
“🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. “
Contact Admin:
Mail me for collab
offcampusjobsindiaadmin@gmail.com
instagram.com/offcampusjobsindia_it
Our Website :
www.offcampusjobsindia.com”
凭借高频更新（最新数据采集于 09 十一月, 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 职业 类别中的关键影响点。
128 253
订阅者
-10324 小时
-5187 天
-1 99530 天
帖子存档
int n=books.size();
vector<vector<int>> cost(n+1,vector<int>(n+1));
for(int i=1;i<=n;i++)
{
int height=books[i-1][1];
int width=books[i-1][0];
cost[i][i]=height;
for(int j=i+1;j<=n;j++)
{
height=max(height,books[j-1][1]);
width+=books[j-1][0];
if(width<=shelfWidth) cost[i][j]=height;
else cost[i][j]=-1;
}
}
vector<int> ans(n+1);
ans[0]=0;
for(int i=1;i<=n;i++)
{
ans[i]=INT_MAX;
for(int j=1;j<=i;j++)
{
if(cost[j][i]==-1) continue;
if(ans[j-1]!=INT_MAX) ans[i]=min(ans[i],ans[j-1]+cost[j][i]);
}
}
return ans[n]
Python
Murder on book shelf Code
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Python
Distant Palindrome code
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Python
Minimum swap Code
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def isPalindrome(s):
return s == s[::-1]
# Driver code
s = "malayalam"
ans = isPalindrome(s)
if ans:
print("Yes")
else:
print("No")
Python
Palindrome Code
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C++
Build road code
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#include <bits/stdc++.h>
using namespace std;
// Utility method to check whether a character
// is operator or not
bool isOperator(char op)
{
return (op == '+' || op == '*');
}
// method prints minimum and maximum value
// obtainable from an expression
void printMinAndMaxValueOfExp(string exp)
{
vector<int> num;
vector<char> opr;
string tmp = "";
// store operator and numbers in different vectors
for (int i = 0; i < exp.length(); i++)
{
if (isOperator(exp[i]))
{
opr.push_back(exp[i]);
num.push_back(atoi(tmp.c_str()));
tmp = "";
}
else
{
tmp += exp[i];
}
}
// storing last number in vector
num.push_back(atoi(tmp.c_str()));
int len = num.size();
int minVal[len][len];
int maxVal[len][len];
// initializing minval and maxval 2D array
for (int i = 0; i < len; i++)
{
for (int j = 0; j < len; j++)
{
minVal[i][j] = INT_MAX;
maxVal[i][j] = 0;
// initializing main diagonal by num values
if (i == j)
minVal[i][j] = maxVal[i][j] = num[i];
}
}
// looping similar to matrix chain multiplication
// and updating both 2D arrays
for (int L = 2; L <= len; L++)
{
for (int i = 0; i < len - L + 1; i++)
{
int j = i + L - 1;
for (int k = i; k < j; k++)
{
int minTmp = 0, maxTmp = 0;
// if current operator is '+', updating tmp
// variable by addition
if(opr[k] == '+')
{
minTmp = minVal[i][k] + minVal[k + 1][j];
maxTmp = maxVal[i][k] + maxVal[k + 1][j];
}
// if current operator is '*', updating tmp
// variable by multiplication
else if(opr[k] == '*')
{
minTmp = minVal[i][k] * minVal[k + 1][j];
maxTmp = maxVal[i][k] * maxVal[k + 1][j];
}
// updating array values by tmp variables
if (minTmp < minVal[i][j])
minVal[i][j] = minTmp;
if (maxTmp > maxVal[i][j])
maxVal[i][j] = maxTmp;
}
}
}
// last element of first row will store the result
cout << "Minimum value : " << minVal[0][len - 1]
<< ", Maximum value : " << maxVal[0][len - 1];
}
// Driver code to test above methods
int main()
{
string expression = "1+2*3+4*5";
printMinAndMaxValueOfExp(expression);
return 0;
}
C++
Minimum Value Expression
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#include<bits/stdc++.h>
int max(int a, int b);
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
if (m == 0 || n == 0)
return 0;
if (X[m-1] == Y[n-1])
return 1 + lcs(X, Y, m-1, n-1);
else
return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Driver program to test above function */
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
int m = strlen(X);
int n = strlen(Y);
printf("Length of LCS is %d", lcs( X, Y, m, n ) );
return 0;
}
C++
Longest Special Subsequences
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