Related Articles

• Write an Interview Experience
• Matrix Data Structure

Count zeros in a row wise and column wise sorted matrix

• Difficulty Level : Easy
• Last Updated : 19 Aug, 2022

Given a N x N binary matrix (elements in matrix can be either 1 or 0) where each row and column of the matrix is sorted in ascending order, count number of 0s present in it.
Expected time complexity is O(N).

Examples:

```Input:
[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]

Output: 8

Input:
[0, 0]
[0, 0]

Output: 4

Input:
[1, 1, 1, 1]
[1, 1, 1, 1]
[1, 1, 1, 1]
[1, 1, 1, 1]

Output: 0```

The idea is very simple. We start from the bottom-left corner of the matrix and repeat below steps until we find the top or right edge of the matrix.

1. Decrement row index until we find a 0.
2. Add number of 0s in current column i.e. current row index + 1 to the result and move right to next column (Increment col index by 1).

The above logic will work since the matrix is row-wise and column-wise sorted. The logic will also work for any matrix containing non-negative integers.

Below is the implementation of above idea :

C++

 `// C++ program to count number of 0s in the given``// row-wise and column-wise sorted binary matrix.``#include ``using` `namespace` `std;``// 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 }``    ``};` `    ``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;``}` `// This code is contributed by kothavvsaakash.`

Java

 `// Java program to count number of 0s in the given``// row-wise and column-wise sorted binary matrix``import` `java.io.*;` `class` `GFG``{``    ``public` `static` `int` `N = ``5``;``    ` `    ``// Function to count number of 0s in the given``    ``// row-wise and column-wise sorted binary matrix.``    ``static` `int` `countZeroes(``int` `mat[][])``    ``{``        ``// 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] > ``0``)`` ` `                ``// 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``    ``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));``    ``}``}` `// This code is contributed by Pramod Kumar`

Python

 `# Python program to count number``# of 0s in the given row-wise``# and column-wise sorted``# binary matrix.` `# Function to count number``# of 0s in the given``# row-wise and column-wise``# sorted binary matrix.``def` `countZeroes(mat):``    ` `    ``# start from bottom-left``    ``# corner of the matrix``    ``N ``=` `5``;``    ``row ``=` `N ``-` `1``;``    ``col ``=` `0``;` `    ``# stores number of``    ``# zeroes in the matrix``    ``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;``            ``row ``=` `row ``-` `1``;` `        ``# add 0s present in``        ``# current column to result``        ``count ``=` `count ``+` `(row ``+` `1``);` `        ``# move right to``        ``# next column``        ``col ``=` `col ``+` `1``;` `    ``return` `count;``    ` `# Driver Code``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``( countZeroes(mat));` `# This code is contributed``# by chandan_jnu`

C#

 `// C# program to count number of``// 0s in the given row-wise and``// column-wise sorted binary matrix``using` `System;` `class` `GFG``{``    ``public` `static` `int` `N = 5;``    ` `    ``// Function to count number of``    ``// 0s in the given row-wise and``    ``// column-wise sorted binary matrix.``    ``static` `int` `countZeroes(``int` `[,] mat)``    ``{``        ``// 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] > 0)` `                ``// 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 Code``    ``public` `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));``    ``}``}` `// This code is contributed by KRV.`

PHP

 ``

Javascript

 ``

Output

`8`

Time complexity of above solution is O(n) since the solution follows single path from bottom-left corner to top or right edge of the matrix.
Auxiliary space used by the program is O(1). since no extra space has been taken.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up