Udaljenost najbliže ćelije koja ima 1 u binarnoj matrici
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S obzirom na binarni rešetka[][] . Pronađite udaljenost najbližeg 1 u mreži za svaku ćeliju.
Udaljenost se računa kao |i 1 - i 2 | + |j 1 - j 2 | gdje ja 1 j 1 su broj retka i broj stupca trenutne ćelije i i 2 j 2 su broj retka i broj stupca najbliže ćelije koja ima vrijednost 1.
Bilješka: U mreži mora postojati barem jedna ćelija s vrijednošću 1.
Primjeri:
Ulazni: mreža[][] = [[0 1 1 0]
[1 1 0 0]
[0 0 1 1]]
Izlaz: [[1 0 0 1]
[0 0 1 1]
[1 1 0 0]]
Obrazloženje:
ćelija (0 1) ima najbliži 1 u ćeliji (0 0) - udaljenost = |0-0| + |0-1| = 1
ćelija (0 2) ima najbliži 1 u ćeliji (0 3) - udaljenost = |0-0| + |3-2| = 1
ćelija (1 0) ima najbliži 1 u ćeliji (0 0) - udaljenost = |1-0| + |0-0| = 1
ćelija (1 1) ima najbliži 1 u ćeliji (1 2) - udaljenost = |1-1| + |1-2| = 1
ćelija (2 2) ima najbliži 1 u ćeliji (2 1) - udaljenost = |2-2| + |2-1| = 1
ćelija (2 3) ima najbliži 1 u ćeliji (1 3) - udaljenost = |2-1| + |3-3| = 1
Sve ostalo su ćelije koje imaju 1, tako da je njihova udaljenost od najbliže ćelije koja ima 1 0.Ulazni: mreža[][] = [[1 0 1]
[1 1 0]
[1 0 0]]
Izlaz: [[0 1 0]
[0 0 1]
[0 1 2]]
Obrazloženje:
ćelija (0 0) ima najbliži 1 u ćeliji (0 1) - udaljenost = |0-0| + |0-1| = 1
ćelija (0 2) ima najbliži 1 u ćeliji (0 1) - udaljenost = |0-0| + |2-1| = 1
ćelija (1 0) ima najbliži 1 u ćeliji (0 1) - udaljenost = |1-0| + |0-1| = 2
ćelija (1 1) ima najbliži 1 u ćeliji (1 2) - udaljenost = |1-1| + |1-2| = 1
ćelija (2 0) ima najbliži 1 u ćeliji (2 1) - udaljenost = |2-2| + |2-1| = 1
ćelija (2 2) ima najbliži 1 u ćeliji (2 1) - udaljenost = |2-2| + |2-1| = 1
Sve ostalo su ćelije koje imaju 1, tako da je njihova udaljenost od najbliže ćelije koja ima 1 0.
Sadržaj
- [Naivni pristup] - O((n*m)^2) Vrijeme i O(n * m) Prostor
- [Očekivani pristup] - Korištenje pretraživanja prvo u širinu - O(n * m) vremena i O(n * m) prostora
[Naivni pristup] - O((n*m)^2) Vrijeme i O(n * m) Prostor
C++Ideja je prijeći cijelu mrežu i izračunati udaljenost svake ćelije do najbliže 1:
- Ako ćelija sadrži 1, njezina udaljenost je 0.
- Ako ćelija sadrži 0, prelazimo kroz cijelu rešetku kako bismo pronašli najbližu ćeliju koja sadrži 1.
- Za svaku ćeliju 0 izračunajte udaljenost Manhattana do svih ćelija s 1 i uzmite najmanju udaljenost.
Pohranite ovu minimalnu udaljenost u odgovarajuću ćeliju matrice rezultata. Ponovite za sve ćelije u mreži.
//Driver Code Starts #include #include #include using namespace std ; //Driver Code Ends vector < vector < int >> nearest ( vector < vector < int >> & grid ) { int n = grid . size (); int m = grid [ 0 ]. size (); vector < vector < int >> ans ( n vector < int > ( m INT_MAX )); // visit each cell of the grid for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { // if the cell has 1 // then the distance is 0 if ( grid [ i ][ j ] == 1 ) { ans [ i ][ j ] = 0 ; continue ; } // iterate over all the cells // and find the distance of the nearest 1 for ( int k = 0 ; k < n ; k ++ ) { for ( int l = 0 ; l < m ; l ++ ) { if ( grid [ k ][ l ] == 1 ) { ans [ i ][ j ] = min ( ans [ i ][ j ] abs ( i - k ) + abs ( j - l )); } } } } } return ans ; } //Driver Code Starts int main () { vector < vector < int >> grid = {{ 0 1 1 0 } { 1 1 0 0 } { 0 0 1 1 }}; vector < vector < int >> ans = nearest ( grid ); for ( int i = 0 ; i < ans . size (); i ++ ) { for ( int j = 0 ; j < ans [ i ]. size (); j ++ ) { cout < < ans [ i ][ j ] < < ' ' ; } cout < < endl ; } return 0 ; } //Driver Code Ends
Java //Driver Code Starts import java.util.ArrayList ; class GFG { //Driver Code Ends static ArrayList < ArrayList < Integer >> nearest ( int [][] grid ) { int n = grid . length ; int m = grid [ 0 ] . length ; ArrayList < ArrayList < Integer > > ans = new ArrayList <> (); // initialize all cells with maximum value for ( int i = 0 ; i < n ; i ++ ) { ArrayList < Integer > row = new ArrayList <> (); for ( int j = 0 ; j < m ; j ++ ) { row . add ( Integer . MAX_VALUE ); } ans . add ( row ); } // visit each cell of the grid for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { // if the cell has 1 distance is 0 if ( grid [ i ][ j ] == 1 ) { ans . get ( i ). set ( j 0 ); continue ; } // iterate over all cells to find nearest 1 for ( int k = 0 ; k < n ; k ++ ) { for ( int l = 0 ; l < m ; l ++ ) { if ( grid [ k ][ l ] == 1 ) { int distance = Math . abs ( i - k ) + Math . abs ( j - l ); if ( distance < ans . get ( i ). get ( j )) { ans . get ( i ). set ( j distance ); } } } } } } return ans ; } //Driver Code Starts public static void main ( String [] args ) { int [][] grid = { { 0 1 1 0 } { 1 1 0 0 } { 0 0 1 1 } }; ArrayList < ArrayList < Integer > > ans = nearest ( grid ); for ( ArrayList < Integer > row : ans ) { for ( Integer val : row ) { System . out . print ( val + ' ' ); } System . out . println (); } } } //Driver Code Ends
Python def nearest ( grid ): n = len ( grid ) m = len ( grid [ 0 ]) ans = [[ float ( 'inf' )] * m for _ in range ( n )] # visit each cell of the grid for i in range ( n ): for j in range ( m ): # if the cell has 1 # then the distance is 0 if grid [ i ][ j ] == 1 : ans [ i ][ j ] = 0 continue # iterate over all the cells # and find the distance of the nearest 1 for k in range ( n ): for l in range ( m ): if grid [ k ][ l ] == 1 : ans [ i ][ j ] = min ( ans [ i ][ j ] abs ( i - k ) + abs ( j - l )) return ans #Driver Code Starts if __name__ == '__main__' : grid = [[ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ]] ans = nearest ( grid ) for i in range ( len ( ans )): for j in range ( len ( ans [ i ])): print ( ans [ i ][ j ] end = ' ' ) print () #Driver Code Ends
C# //Driver Code Starts using System ; using System.Collections.Generic ; class GfG { //Driver Code Ends static List < List < int > > nearest ( int [ ] grid ) { int n = grid . GetLength ( 0 ); int m = grid . GetLength ( 1 ); List < List < int > > ans = new List < List < int > > (); for ( int i = 0 ; i < n ; i ++ ) { List < int > row = new List < int > (); for ( int j = 0 ; j < m ; j ++ ) { row . Add ( int . MaxValue ); } ans . Add ( row ); } // Visit each cell of the grid for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { // If the cell has 1 distance is 0 if ( grid [ i j ] == 1 ) { ans [ i ][ j ] = 0 ; continue ; } // iterate over all the cells // and find the distance of the nearest 1 for ( int k = 0 ; k < n ; k ++ ) { for ( int l = 0 ; l < m ; l ++ ) { if ( grid [ k l ] == 1 ) { int distance = Math . Abs ( i - k ) + Math . Abs ( j - l ); if ( distance < ans [ i ][ j ]) { ans [ i ][ j ] = distance ; } } } } } } return ans ; } //Driver Code Starts static void Main () { int [ ] grid = { { 0 1 1 0 } { 1 1 0 0 } { 0 0 1 1 } }; List < List < int > > ans = nearest ( grid ); for ( int i = 0 ; i < ans . Count ; i ++ ) { for ( int j = 0 ; j < ans [ i ]. Count ; j ++ ) { Console . Write ( ans [ i ][ j ] + ' ' ); } Console . WriteLine (); } } } //Driver Code Ends
JavaScript function nearest ( grid ) { let n = grid . length ; let m = grid [ 0 ]. length ; let ans = new Array ( n ); for ( let i = 0 ; i < n ; i ++ ) { ans [ i ] = new Array ( m ). fill ( Infinity ); } // visit each cell of the grid for ( let i = 0 ; i < n ; i ++ ) { for ( let j = 0 ; j < m ; j ++ ) { // if the cell has 1 // then the distance is 0 if ( grid [ i ][ j ] === 1 ) { ans [ i ][ j ] = 0 ; continue ; } // iterate over all the cells // and find the distance of the nearest 1 for ( let k = 0 ; k < n ; k ++ ) { for ( let l = 0 ; l < m ; l ++ ) { if ( grid [ k ][ l ] === 1 ) { ans [ i ][ j ] = Math . min ( ans [ i ][ j ] Math . abs ( i - k ) + Math . abs ( j - l )); } } } } } return ans ; } // Driver Code //Driver Code Starts let grid = [ [ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ] ]; let ans = nearest ( grid ); for ( let i = 0 ; i < ans . length ; i ++ ) { console . log ( ans [ i ]. join ( ' ' )); } //Driver Code Ends
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1 0 0 1 0 0 1 1 1 1 0 0
[Očekivani pristup] - Korištenje pretraživanja prvo u širinu - O(n * m) vremena i O(n * m) prostora
C++Problem se može učinkovito riješiti korištenjem BFS pristupa s više izvora. Svaka ćelija u mreži tretira se kao čvor s rubovima koji povezuju susjedne ćelije (gore dolje lijevo desno). Umjesto pokretanja zasebnog pretraživanja za svaku ćeliju 0, stavljamo u red sve ćelije koje sadrže 1 na početku i izvodimo jedan BFS iz ovih višestrukih izvora istovremeno. Kako se BFS širi sloj po sloj, mi ažuriramo udaljenost svake neposjećene nulte ćelije da bude za jednu veću od udaljenosti njenog roditelja. To jamči da svaka stanica prima minimalnu udaljenost do najbliže 1 na optimalan i učinkovit način.
//Driver Code Starts #include #include #include #include using namespace std ; //Driver Code Ends vector < vector < int >> nearest ( vector < vector < int >> & grid ) { int n = grid . size (); int m = grid [ 0 ]. size (); vector < vector < int >> ans ( n vector < int > ( m INT_MAX )); // to store the indices of the cells having 1 queue < pair < int int >> q ; // visit each cell of the grid for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { // if the cell has 1 // then the distance is 0 if ( grid [ i ][ j ] == 1 ) { ans [ i ][ j ] = 0 ; q . push ({ i j }); } } } // iterate over all the cells // and find the distance of the nearest 1 while ( ! q . empty ()) { int len = q . size (); for ( int i = 0 ; i < len ; i ++ ) { int x = q . front (). first ; int y = q . front (). second ; q . pop (); // check all the four directions vector < vector < int >> directions = {{ 0 1 } { 0 -1 } { 1 0 } { -1 0 }}; for ( int j = 0 ; j < directions . size (); j ++ ) { int dx = directions [ j ][ 0 ]; int dy = directions [ j ][ 1 ]; // if the cell is within the grid // and the distance is not calculated yet if ( x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && ans [ x + dx ][ y + dy ] == INT_MAX ) { ans [ x + dx ][ y + dy ] = ans [ x ][ y ] + 1 ; q . push ({ x + dx y + dy }); } } } } return ans ; } //Driver Code Starts int main () { vector < vector < int >> grid = {{ 0 1 1 0 } { 1 1 0 0 } { 0 0 1 1 }}; vector < vector < int >> ans = nearest ( grid ); for ( int i = 0 ; i < ans . size (); i ++ ) { for ( int j = 0 ; j < ans [ i ]. size (); j ++ ) { cout < < ans [ i ][ j ] < < ' ' ; } cout < < endl ; } return 0 ; } //Driver Code Ends
Java //Driver Code Starts import java.util.ArrayList ; import java.util.Queue ; import java.util.LinkedList ; import java.util.Arrays ; class GfG { //Driver Code Ends static ArrayList < ArrayList < Integer >> nearest ( int [][] grid ) { int n = grid . length ; int m = grid [ 0 ] . length ; int [][] ans = new int [ n ][ m ] ; for ( int i = 0 ; i < n ; i ++ ) { Arrays . fill ( ans [ i ] Integer . MAX_VALUE ); } // to store the indices of the cells having 1 Queue < int []> q = new LinkedList <> (); // visit each cell of the grid for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { // if the cell has 1 // then the distance is 0 if ( grid [ i ][ j ] == 1 ) { ans [ i ][ j ] = 0 ; q . add ( new int [] { i j }); } } } // iterate over all the cells // and find the distance of the nearest 1 while ( ! q . isEmpty ()) { int len = q . size (); for ( int i = 0 ; i < len ; i ++ ) { int [] front = q . poll (); int x = front [ 0 ] ; int y = front [ 1 ] ; // check all the four directions int [][] directions = {{ 0 1 } { 0 - 1 } { 1 0 } { - 1 0 }}; for ( int j = 0 ; j < directions . length ; j ++ ) { int dx = directions [ j ][ 0 ] ; int dy = directions [ j ][ 1 ] ; // if the cell is within the grid // and the distance is not calculated yet if ( x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && ans [ x + dx ][ y + dy ] == Integer . MAX_VALUE ) { ans [ x + dx ][ y + dy ] = ans [ x ][ y ] + 1 ; q . add ( new int [] { x + dx y + dy }); } } } } ArrayList < ArrayList < Integer >> result = new ArrayList <> (); for ( int i = 0 ; i < n ; i ++ ) { ArrayList < Integer > row = new ArrayList <> (); for ( int j = 0 ; j < m ; j ++ ) { row . add ( ans [ i ][ j ] ); } result . add ( row ); } return result ; } //Driver Code Starts public static void main ( String [] args ) { int [][] grid = {{ 0 1 1 0 } { 1 1 0 0 } { 0 0 1 1 }}; ArrayList < ArrayList < Integer >> ans = nearest ( grid ); for ( ArrayList < Integer > row : ans ) { for ( int val : row ) { System . out . print ( val + ' ' ); } System . out . println (); } } } //Driver Code Ends
Python #Driver Code Starts from collections import deque import sys #Driver Code Ends def nearest ( grid ): n = len ( grid ) m = len ( grid [ 0 ]) ans = [[ sys . maxsize for _ in range ( m )] for _ in range ( n )] # to store the indices of the cells having 1 q = deque () # visit each cell of the grid for i in range ( n ): for j in range ( m ): # if the cell has 1 # then the distance is 0 if grid [ i ][ j ] == 1 : ans [ i ][ j ] = 0 q . append (( i j )) # iterate over all the cells # and find the distance of the nearest 1 while q : len_q = len ( q ) for _ in range ( len_q ): x y = q . popleft () # check all the four directions directions = [( 0 1 ) ( 0 - 1 ) ( 1 0 ) ( - 1 0 )] for dx dy in directions : # if the cell is within the grid # and the distance is not calculated yet if 0 <= x + dx < n and 0 <= y + dy < m and ans [ x + dx ][ y + dy ] == sys . maxsize : ans [ x + dx ][ y + dy ] = ans [ x ][ y ] + 1 q . append (( x + dx y + dy )) return ans #Driver Code Starts if __name__ == '__main__' : grid = [[ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ]] ans = nearest ( grid ) for row in ans : print ( ' ' . join ( map ( str row ))) #Driver Code Ends
C# //Driver Code Starts using System ; using System.Collections.Generic ; class GFG { //Driver Code Ends static List < List < int >> nearest ( int [] grid ) { int n = grid . GetLength ( 0 ); int m = grid . GetLength ( 1 ); int [] ans = new int [ n m ]; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { ans [ i j ] = int . MaxValue ; } } // to store the indices of the cells having 1 Queue < Tuple < int int >> q = new Queue < Tuple < int int >> (); // visit each cell of the grid for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < m ; j ++ ) { // if the cell has 1 // then the distance is 0 if ( grid [ i j ] == 1 ) { ans [ i j ] = 0 ; q . Enqueue ( new Tuple < int int > ( i j )); } } } // iterate over all the cells // and find the distance of the nearest 1 while ( q . Count > 0 ) { int len = q . Count ; for ( int i = 0 ; i < len ; i ++ ) { var node = q . Dequeue (); int x = node . Item1 ; int y = node . Item2 ; // check all the four directions int [] directions = new int [] { { 0 1 } { 0 - 1 } { 1 0 } { - 1 0 } }; for ( int j = 0 ; j < 4 ; j ++ ) { int dx = directions [ j 0 ]; int dy = directions [ j 1 ]; // if the cell is within the grid // and the distance is not calculated yet if ( x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && ans [ x + dx y + dy ] == int . MaxValue ) { ans [ x + dx y + dy ] = ans [ x y ] + 1 ; q . Enqueue ( new Tuple < int int > ( x + dx y + dy )); } } } } // Convert 2D array to List > before returning
List < List < int >> result = new List < List < int >> (); for ( int i = 0 ; i < n ; i ++ ) { List < int > row = new List < int > (); for ( int j = 0 ; j < m ; j ++ ) { row . Add ( ans [ i j ]); } result . Add ( row ); } return result ; } //Driver Code Starts static void Main () { int [] grid = new int [] { { 0 1 1 0 } { 1 1 0 0 } { 0 0 1 1 } }; List < List < int >> ans = nearest ( grid ); for ( int i = 0 ; i < ans . Count ; i ++ ) { for ( int j = 0 ; j < ans [ i ]. Count ; j ++ ) { Console . Write ( ans [ i ][ j ] + ' ' ); } Console . WriteLine (); } } } //Driver Code Ends
JavaScript //Driver Code Starts const Denque = require ( 'denque' ); //Driver Code Ends function nearest ( grid ) { let n = grid . length ; let m = grid [ 0 ]. length ; // Initialize answer matrix with Infinity let ans = []; for ( let i = 0 ; i < n ; i ++ ) { ans . push ( new Array ( m ). fill ( Infinity )); } // to store the indices of the cells having 1 let q = new Denque (); // visit each cell of the grid for ( let i = 0 ; i < n ; i ++ ) { for ( let j = 0 ; j < m ; j ++ ) { // if the cell has 1 // then the distance is 0 if ( grid [ i ][ j ] === 1 ) { ans [ i ][ j ] = 0 ; q . push ([ i j ]); } } } // iterate over all the cells // and find the distance of the nearest 1 while ( ! q . isEmpty ()) { let [ x y ] = q . shift (); // check all the four directions let directions = [ [ 0 1 ] [ 0 - 1 ] [ 1 0 ] [ - 1 0 ] ]; for ( let dir of directions ) { let dx = dir [ 0 ]; let dy = dir [ 1 ]; // if the cell is within the grid // and the distance is not calculated yet if ( x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && ans [ x + dx ][ y + dy ] === Infinity ) { ans [ x + dx ][ y + dy ] = ans [ x ][ y ] + 1 ; q . push ([ x + dx y + dy ]); } } } return ans ; } //Driver Code Starts // Driver Code let grid = [ [ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ] ]; let ans = nearest ( grid ); for ( let i = 0 ; i < ans . length ; i ++ ) { console . log ( ans [ i ]. join ( ' ' )); } //Driver Code Ends
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1 0 0 1 0 0 1 1 1 1 0 0Napravi kviz
