Tel nullen in een rij- en kolomsgewijs gesorteerde matrix
Gegeven een n x n binaire matrix (elementen in de matrix kunnen 1 of 0 zijn) waarbij elke rij en kolom van de matrix in oplopende volgorde is gesorteerd, tel dan het aantal nullen dat daarin aanwezig is.
Voorbeelden:
Invoer:
[0 0 0 0 1]
[0 0 0 1 1]
[0 1 1 1 1]
[1 1 1 1 1]
[1 1 1 1 1]
Uitgang: 8
Invoer:
[0 0]
[0 0]
Uitgang: 4
Invoer:
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
Uitgang:
Het idee is heel eenvoudig. We beginnen vanuit de linkerbenedenhoek van de matrix en herhalen de onderstaande stappen totdat we de boven- of rechterrand van de matrix vinden.
- Verlaag de rij-index totdat we een 0 vinden.
- Voeg het aantal nullen toe in de huidige kolom, dat wil zeggen de huidige rijindex + 1, aan het resultaat en ga naar rechts naar de volgende kolom (verhoog de colindex met 1).
De bovenstaande logica werkt omdat de matrix rijsgewijs en kolomsgewijs is gesorteerd. De logica werkt ook voor elke matrix die niet-negatieve gehele getallen bevat.
Hieronder ziet u de implementatie van bovenstaand idee:
C++ #include #include using namespace std ; // Function to count number of 0s in the given // row-wise and column-wise sorted binary matrix. int countZeroes ( const vector < vector < int >>& mat ) { int n = mat . size (); // start from the bottom-left corner int row = n - 1 col = 0 ; int count = 0 ; while ( col < n ) { // move up until you find a 0 while ( row >= 0 && mat [ row ][ col ]) { row -- ; } // add the number of 0s in the current // column to the result count += ( row + 1 ); // move to the next column col ++ ; } return count ; } int main () { vector < vector < int >> mat = { { 0 0 0 0 1 } { 0 0 0 1 1 } { 0 1 1 1 1 } { 1 1 1 1 1 } { 1 1 1 1 1 } }; cout < < countZeroes ( mat ); return 0 ; }
C // C program to count number of 0s in the given // row-wise and column-wise sorted binary matrix. #include // define size of square matrix #define N 5 // Function to count number of 0s in the given // row-wise and column-wise sorted binary matrix. int countZeroes ( int mat [ N ][ N ]) { // start from bottom-left corner of the matrix int row = N - 1 col = 0 ; // stores number of zeroes in the matrix int count = 0 ; while ( col < N ) { // move up until you find a 0 while ( mat [ row ][ col ]) // if zero is not found in current column // we are done if ( -- row < 0 ) return count ; // add 0s present in current column to result count += ( row + 1 ); // move right to next column col ++ ; } return count ; } // Driver Program to test above functions int main () { int mat [ N ][ N ] = { { 0 0 0 0 1 } { 0 0 0 1 1 } { 0 1 1 1 1 } { 1 1 1 1 1 } { 1 1 1 1 1 } }; printf ( '%d' countZeroes ( mat )); return 0 ; }
Java import java.util.Arrays ; public class GfG { // Function to count number of 0s in the given // row-wise and column-wise sorted binary matrix. public static int countZeroes ( int [][] mat ) { int n = mat . length ; // start from the bottom-left corner int row = n - 1 col = 0 ; int count = 0 ; while ( col < n ) { // move up until you find a 0 while ( row >= 0 && mat [ row ][ col ] == 1 ) { row -- ; } // add the number of 0s in the current // column to the result count += ( row + 1 ); // move to the next column col ++ ; } return count ; } public static void main ( String [] args ) { int [][] mat = { { 0 0 0 0 1 } { 0 0 0 1 1 } { 0 1 1 1 1 } { 1 1 1 1 1 } { 1 1 1 1 1 } }; System . out . println ( countZeroes ( mat )); } }
Python # Function to count number of 0s in the given # row-wise and column-wise sorted binary matrix. def count_zeroes ( mat ): n = len ( mat ) # start from the bottom-left corner row = n - 1 col = 0 count = 0 while col < n : # move up until you find a 0 while row >= 0 and mat [ row ][ col ]: row -= 1 # add the number of 0s in the current # column to the result count += ( row + 1 ) # move to the next column col += 1 return count if __name__ == '__main__' : mat = [ [ 0 0 0 0 1 ] [ 0 0 0 1 1 ] [ 0 1 1 1 1 ] [ 1 1 1 1 1 ] [ 1 1 1 1 1 ] ] print ( count_zeroes ( mat ))
C# // Function to count number of 0s in the given // row-wise and column-wise sorted binary matrix. using System ; using System.Collections.Generic ; class Program { static int CountZeroes ( int [] mat ) { int n = mat . GetLength ( 0 ); // start from the bottom-left corner int row = n - 1 col = 0 ; int count = 0 ; while ( col < n ) { // move up until you find a 0 while ( row >= 0 && mat [ row col ] == 1 ) { row -- ; } // add the number of 0s in the current // column to the result count += ( row + 1 ); // move to the next column col ++ ; } return count ; } static void Main () { int [] mat = { { 0 0 0 0 1 } { 0 0 0 1 1 } { 0 1 1 1 1 } { 1 1 1 1 1 } { 1 1 1 1 1 } }; Console . WriteLine ( CountZeroes ( mat )); } }
JavaScript // Function to count number of 0s in the given // row-wise and column-wise sorted binary matrix. function countZeroes ( mat ) { const n = mat . length ; // start from the bottom-left corner let row = n - 1 col = 0 ; let count = 0 ; while ( col < n ) { // move up until you find a 0 while ( row >= 0 && mat [ row ][ col ]) { row -- ; } // add the number of 0s in the current // column to the result count += ( row + 1 ); // move to the next column col ++ ; } return count ; } const mat = [ [ 0 0 0 0 1 ] [ 0 0 0 1 1 ] [ 0 1 1 1 1 ] [ 1 1 1 1 1 ] [ 1 1 1 1 1 ] ]; console . log ( countZeroes ( mat ));
Uitvoer
8
Tijdcomplexiteit van de bovenstaande oplossing is O(n), aangezien de oplossing een enkel pad volgt van de linkerbenedenhoek naar de boven- of rechterrand van de matrix.
Hulpruimte gebruikt door het programma is O(1). omdat er geen extra ruimte is ingenomen.
