STL's EASY WAY TO FIND maximum & minimum IN AN ARRAY

- 
 *max_element(a,a+n)-*min_element(a,a+n)












































Comments











Post a Comment






























Popular posts from this blog









Divide Intervals Into Minimum Number of Groups






-















  Divide Intervals Into Minimum Number of Groups  Leetcode Problem  Divide Intervals Into Minimum Number of Groups   is very much similar to the number-of-the-smallest-unoccupied-chair  Here we have to divide the given intervals into minimum number of groups such that each interval is in exactly one group, and no two intervals that are in the same group intersect each other. Solution: We will think each interval as combination of two events ,  startTime  and endTime. 1) We create a list of events and then sort that list.  2) After that, we Initialise a min heap. which can be thinked like different baskets or groups.  3) We iterate through the events one by one, and assign each events to groups (picked from minheap).      While iterating starting event we pick a group from min heap and keep track in an array.       While iterating ending event through our tracking array, we release the group again in min...








Read more










B. Dreamoon and WiFi  :calculate no. of ways : recursive solution (branch and bound )






-















  http://codeforces.com/contest/476/problem/B    #include < bits / stdc ++. h >  using  namespace  std ;  string  s1 , s2 ;  int  val = 0 , n ;  int  dp [ 15 ][ 15 ];  int  solve ( int  s , int  l , int  f )  {      int  & ret = dp [ s ][ l ];      if ( s == l )      {          if ( f == val )              return  1 ;          else              return  0 ;      }          int  ct = 0 ;                                   ///number of ways in starting is 0      if ( s2 [ s ]== '+' )          ct += solve ( s + 1 , l , f + 1 );      if ( s2 [ s ]== '-' )          ct += solve ( s + 1 , l , f - 1 );      if ( s2 [ s ]== '?' )          ct += solve ( s + 1 , l , f + 1 )+ solve ( s + 1 , l , f - 1 );      return  ret = ct ;                              /// In the last return this value   }  int  main ()  {      int  i , j , k , l ;      double  ans ;      cin >> s1 >> s2 ;      n = s1 . size ();      for ( i = 0 ; i < n ; i ++)      {          if ...








Read more










A. Dreamoon and Stairs : minimum steps to reach : recursion solution.






-















   http://codeforces.com/contest/476/problem/A    #include < bits / stdc ++. h >   using  namespace  std ;  int  dp [ 10001 ][ 5055 ];   int   fun ( int  cu , int  n , int  m , int  st )  {     if ( st % m == 0  &&  cu == n )    {      //  cout<<st<<endl;        return  st ;    }      if ( cu > n ||  st > n ||  st > 5050  )     return  999999 ;  int  & ret = dp [ cu ][ st ];  if ( ret )      return  ret ;   int  k1 = fun ( cu + 1 , n , m , st + 1 );    int  k2 = fun ( cu + 2 , n , m , st + 1 );   return  ret = min ( k1 , k2 );   }      int  main ()  {   int  n , m , i , j , k , l ;   cin >> n >> m ;    //if(n>m*2)     // cout<<-1<<endl;    k = fun ( 0 , n , m , 0 );     if ( k == 999999 )      cout <<- 1 << endl ;    else      cout << k << endl ;   return  0 ;  }   








Read more