Oglinzi maxime care pot transfera lumina de jos la dreapta
Este dată o matrice pătrată în care fiecare celulă reprezintă fie un gol, fie un obstacol. Putem plasa oglinzi în poziție goală. Toate oglinzile vor fi situate la 45 de grade, adică pot transfera lumina de jos în dreapta dacă nu există niciun obstacol în calea lor.
În această întrebare trebuie să numărăm câte astfel de oglinzi pot fi plasate în matrice pătrată care poate transfera lumina de jos în dreapta.
Exemple:
Output for above example is 2. In above diagram mirror at (3 1) and (5 5) are able to send light from bottom to right so total possible mirror count is 2.
Putem rezolva această problemă prin verificarea poziției unor astfel de oglinzi în matrice, oglinda care poate transfera lumina de jos în dreapta nu va avea niciun obstacol în calea lor, de exemplu.
dacă o oglindă este acolo la indicele (i j) atunci
nu va exista nici un obstacol la indicele (k j) pentru toate k i < k <= N
nu va exista nici un obstacol la indicele (i k) pentru toate k j < k <= N
Ținând cont de două ecuații de deasupra, putem găsi obstacolul cel mai din dreapta la fiecare rând într-o iterație a matricei date și putem găsi obstacolul cel mai de jos la fiecare coloană dintr-o altă iterație a matricei date. După stocarea acestor indici într-o matrice separată, putem verifica la fiecare index dacă nu satisface nicio condiție de obstacol sau nu și apoi crește numărul corespunzător.
Mai jos este implementată soluția pentru conceptul de mai sus, care necesită timp O(N^2) și spațiu suplimentar O(N).
C++ // C++ program to find how many mirror can transfer // light from bottom to right #include using namespace std ; // method returns number of mirror which can transfer // light from bottom to right int maximumMirrorInMatrix ( string mat [] int N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) int horizontal [ N ] vertical [ N ]; // initialize both array as -1 signifying no obstacle memset ( horizontal -1 sizeof ( horizontal )); memset ( vertical -1 sizeof ( vertical )); // looping matrix to mark column for obstacles for ( int i = 0 ; i < N ; i ++ ) { for ( int j = N -1 ; j >= 0 ; j -- ) { if ( mat [ i ][ j ] == 'B' ) continue ; // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( int j = 0 ; j < N ; j ++ ) { for ( int i = N -1 ; i >= 0 ; i -- ) { if ( mat [ i ][ j ] == 'B' ) continue ; // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } int res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ]) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code to test above method int main () { int N = 5 ; // B - Blank O - Obstacle string mat [ N ] = { 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' }; cout < < maximumMirrorInMatrix ( mat N ) < < endl ; return 0 ; }
Java // Java program to find how many mirror can transfer // light from bottom to right import java.util.* ; class GFG { // method returns number of mirror which can transfer // light from bottom to right static int maximumMirrorInMatrix ( String mat [] int N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) int [] horizontal = new int [ N ] ; int [] vertical = new int [ N ] ; // initialize both array as -1 signifying no obstacle Arrays . fill ( horizontal - 1 ); Arrays . fill ( vertical - 1 ); // looping matrix to mark column for obstacles for ( int i = 0 ; i < N ; i ++ ) { for ( int j = N - 1 ; j >= 0 ; j -- ) { if ( mat [ i ] . charAt ( j ) == 'B' ) { continue ; } // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( int j = 0 ; j < N ; j ++ ) { for ( int i = N - 1 ; i >= 0 ; i -- ) { if ( mat [ i ] . charAt ( j ) == 'B' ) { continue ; } // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } int res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ] ) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code public static void main ( String [] args ) { int N = 5 ; // B - Blank O - Obstacle String mat [] = { 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' }; System . out . println ( maximumMirrorInMatrix ( mat N )); } } /* This code is contributed by PrinciRaj1992 */
Python3 # Python3 program to find how many mirror can transfer # light from bottom to right # method returns number of mirror which can transfer # light from bottom to right def maximumMirrorInMatrix ( mat N ): # To store first obstacles horizontally (from right) # and vertically (from bottom) horizontal = [ - 1 for i in range ( N )] vertical = [ - 1 for i in range ( N )]; # looping matrix to mark column for obstacles for i in range ( N ): for j in range ( N - 1 - 1 - 1 ): if ( mat [ i ][ j ] == 'B' ): continue ; # mark rightmost column with obstacle horizontal [ i ] = j ; break ; # looping matrix to mark rows for obstacles for j in range ( N ): for i in range ( N - 1 - 1 - 1 ): if ( mat [ i ][ j ] == 'B' ): continue ; # mark leftmost row with obstacle vertical [ j ] = i ; break ; res = 0 ; # Initialize result # if there is not obstacle on right or below # then mirror can be placed to transfer light for i in range ( N ): for j in range ( N ): ''' if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right ''' if ( i > vertical [ j ] and j > horizontal [ i ]): ''' uncomment this code to print actual mirror position also''' res += 1 ; return res ; # Driver code to test above method N = 5 ; # B - Blank O - Obstacle mat = [ 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' ]; print ( maximumMirrorInMatrix ( mat N )); # This code is contributed by rutvik_56.
C# // C# program to find how many mirror can transfer // light from bottom to right using System ; class GFG { // method returns number of mirror which can transfer // light from bottom to right static int maximumMirrorInMatrix ( String [] mat int N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) int [] horizontal = new int [ N ]; int [] vertical = new int [ N ]; // initialize both array as -1 signifying no obstacle for ( int i = 0 ; i < N ; i ++ ) { horizontal [ i ] =- 1 ; vertical [ i ] =- 1 ; } // looping matrix to mark column for obstacles for ( int i = 0 ; i < N ; i ++ ) { for ( int j = N - 1 ; j >= 0 ; j -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( int j = 0 ; j < N ; j ++ ) { for ( int i = N - 1 ; i >= 0 ; i -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } int res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ]) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code public static void Main ( String [] args ) { int N = 5 ; // B - Blank O - Obstacle String [] mat = { 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' }; Console . WriteLine ( maximumMirrorInMatrix ( mat N )); } } // This code is contributed by Princi Singh
JavaScript < script > // JavaScript program to find how many mirror can transfer // light from bottom to right // method returns number of mirror which can transfer // light from bottom to right function maximumMirrorInMatrix ( mat N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) var horizontal = Array ( N ). fill ( - 1 ); var vertical = Array ( N ). fill ( - 1 ); // looping matrix to mark column for obstacles for ( var i = 0 ; i < N ; i ++ ) { for ( var j = N - 1 ; j >= 0 ; j -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( var j = 0 ; j < N ; j ++ ) { for ( var i = N - 1 ; i >= 0 ; i -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } var res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( var i = 0 ; i < N ; i ++ ) { for ( var j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ]) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code var N = 5 ; // B - Blank O - Obstacle var mat = [ 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' ]; document . write ( maximumMirrorInMatrix ( mat N )); < /script>
Ieșire
2
Complexitatea timpului: O(n 2 ).
Spațiu auxiliar: O(n)
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