ŠTEJE NIČLE V VRSTI MODRE IN MODRO RAZVRŠČENE MATRIKE

Glede na n x n binarna matrika (elementi v matriki so lahko 1 ali 0), kjer se vsaka vrstica in stolpec matrice razvrsti v število povečanih številk 0, ki je prisotna v njej.

Primeri:

Vnos:
[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]
Izhod: 8

Vnos:
[0 0]
[0 0]
Izhod: 4

Vnos:
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
Izhod:

Ideja je zelo preprosta. Začnemo od spodnjega levega kota matrice in ponovimo pod koraki, dokler ne najdemo zgornjega ali desnega roba matrice.

Indeks vrst Decrement, dokler ne najdemo 0.
Dodajte število 0 v trenutnem stolpcu, tj. Indeks trenutne vrstice + 1 v rezultat in premaknite desno na naslednji stolpec (indeks povečevanja col za 1).

Zgornja logika bo delovala, saj je matrica razvrščena v vrsto in v stolpcu. Logika bo delovala tudi za katero koli matrico, ki vsebuje ne-negativna cela števila.

Spodaj je izvedba zgornje ideje:

vzorec načrtovanja tovarniške metode

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 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))

// 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));

Izhod

Časovna zapletenost zgornje raztopine je o (n), saj raztopina sledi eni poti od spodaj levega vogala do zgornjega ali desnega roba matrice.
Pomožni prostor Uporablja program O (1). Ker ni bil odvzet dodatni prostor.

TechCodeview

Šteje ničle v vrsti modre in modro razvrščene matrike