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Hello everyone  Welcome to our another blog. In this blog i code for some Standard Templete Library(STL) which will help you to coding practice. All the best !! Keep Practicing!! #include <iostream> #include<algorithm> #include<vector> using namespace std; int main() {      vector<int> v; //Defining vector v      v.push_back(1);     v.push_back(5); //Inserting element in vector     v.push_back(8);     v.push_back(9);     cout<<"Finding 5  -> "<< binary_search(v.begin(), v.end(), 5) <<endl; //Applyin binary search for 5     cout<<"lower bound -> "<< lower_bound(v.begin(),v.end(),5)-v.begin() <<endl;      //finding lower bond for 5 cout<<"uper bound -> "<< upper_bound(v.begin(), v.end(),5)-v.begin() <<endl; //finding upper bond for 5     int a =4;     int b = 8;     cout<<&q

ALLOCATE BOOK PROBLEM SOLUTION PROBLEM STATEMENT:- Recommendation: First you have to try then go for solution. For practice  Click here  . Given an array ‘arr’ of integer numbers . where ‘arr[i]’ represents the number of pages in the ‘i-th’ book. There are ‘m’ number of students and the task is to allocate all the books to their students. Allocate books in such a way that: 1. Each student gets at least one book. 2. Each book should be allocated to a student. 3. Book allocation should be in a contiguous manner. You have to allocate the book to ‘m’ students such that the maximum number of pages assigned to a student is minimum. Example: Let’s consider ‘n=4’ (number of books ) and ‘m=2’ (number of students). ‘arr = { 10, 20, 30, 40 }’. All possible way to allocate the ‘4’ books in ‘2’ number of students is - 10 | 20, 30, 40 - sum of all the pages of books which allocated to student-1 is ‘10’, and student-2 is ‘20+ 30+ 40 = 90’ so maximum is ‘max(10, 90)= 90’. 10, 20 | 30, 40 - sum of all

 || Peak Index in a Mountain Array  || Binary Search practice example  || Leet Code poblem #852 Given an integer array arr we have to return the index of peak element.   Array first increase and then decrease.  Format  a[1] < a[2] <  a[3] <........a[n].......> a[2]  > a[1]  We have to return the index of a[n] Input: arr = [0,1,0] Output: 1 Input: arr = [0,2,1,0] Output: 1 Input: arr = [0,10,5,2] Output: 1 ***************CODE**************** //TIME COMPLEXITY IS O(LOG N)  int peakIndexInMountainArray(vector<int>& arr) {         int s=0;         int e=arr.size()-1;         int mid= s + (e-s)/2;         while (s<e){                          if(arr[mid] < arr[mid+1]){                 s = mid +1;             }             else                 e= mid;                          mid= s + (e-s)/2;         }         return s;     } 

 Here I am writting only function solution not main part... ||It is based on Binary Search Example Problem is there is given an array and you have to find a particular key of first and last index of that arry. Example:- Input format 2     // number of test Cases 6  3        // size of array      &&   the key element which have to find there first and last index in array 0  5  5  6  6  6 8  2 0  0  1  1  2  2  2  2   Output -1   -1     ///output is -1 because 3 is not present in the array 4    7    /// output is 4(first occurance) and 7 ( the last Occurance) ***************************************CODE************************************** int firstocc(vector<int>& arr, int n, int key){     int s=0;     int e=n-1;     int mid = s+(e-s)/2;     int ans =-1;     while(s<=e){         if(arr[mid] == key){             ans = mid;             e = mid -1;                     }         else if(key> arr[mid]){   //right me jana hai             s = mid +1;         }         e

  #include   < stdio.h > #include <string.h> #include <math.h> #include <stdlib.h> //Complete the following function. void calculate_the_maximum(int n, int k) {     int maxand =0;     int maxor =0;     int maxxor =0;     for(int i=1; i<=n; i++){         for (int j= i+1; j<=n; j++){             if(((i & j)> maxand) && ((i&j)<k)){                 maxand= i&j;             }                         if(((i | j)> maxor) && ((i|j)<k)){                 maxor= i|j;             }                         if(((i ^ j)> maxxor) && ((i^j)<k)){                 maxxor= i^j;             }         }     }     printf("%d\n%d\n%d", maxand,maxor,maxxor); } int main() {     int n, k;        scanf("%d %d", &n, &k);     calculate_the_maximum(n, k);       return 0; }

  int   main ()   { int   a , b ;      int   n = 0 ;      string   intMap [ 9 ]=   { "one" ,   "two" , "three" , "four" , "five" , "six" , "seven" , "eight" , "nine" };           cin >> a >> b ;           if   (( a <= 9 )&&( b <= 9 )){           for ( n = a ; n <= b ; n ++){              cout   <<   intMap [ n - 1 ]<< endl ;            }   }      else   if   (( a <= 9 )&&( b > 9 )){               for ( n = a ; n <= 9 ; n ++){              cout << intMap [ n - 1 ]<< endl ;          }               for ( n = 10 ; n <= b ; n ++){                   if   ( n % 2 == 0 ){                  cout << "even" << endl ;              }               else   {                   cout << "odd" << endl ;               }           }    }           else   {          for ( n = a ; n <= b ; n ++){