GRÓF NULY V RADE MÚDRE A STĹPCOVEJ MÚDREJ MATICE

Vzhľadom na binárnu maticu n x n (prvky v matrici môžu byť buď 1 alebo 0), kde je každý riadok a stĺpec matrice zoradený podľa počtu počtu počtu príkazov 0 s prítomným v ňom.

Príklady:

konektivita java

Vstup:
[0 0 0 0 1]
[0 0 0 1 1]
[0 1 1 1]
[1 1 1 1 1]
[1 1 1 1 1]
Výstup: 8

Vstup:
[0 0]
[0 0]
Výstup: 4

Vstup:
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
Výstup:

Myšlienka je veľmi jednoduchá. Začneme z ľavého dolného rohu matrice a opakujeme nižšie kroky, až kým nenájdeme horný alebo pravý okraj matrice.

Index radov, kým nenájdeme 0.
Pridajte číslo 0S v aktuálnom stĺpci, t. J. Aktuálny index riadkov + 1 k výsledku a presuňte sa doprava na ďalší stĺpec (index prírastku COL o 1).

Vyššie uvedená logika bude fungovať, pretože matica je zoradená v riadku a stĺpcový. Logika bude tiež pracovať pre akúkoľvek maticu obsahujúcu nezáporné celé čísla.

Nižšie je uvedená implementácia vyššie uvedenej myšlienky:

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));

Výstup

Zložitosť z vyššie uvedeného roztoku je O (n), pretože roztok sleduje jednu cestu od ľavého dolného rohu po horný alebo pravý okraj matrice.
Pomocný priestor Používa sa v programe O (1). Pretože sa nezaoberal žiadny ďalší priestor.

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Gróf nuly v rade múdre a stĺpcovej múdrej matice