Given a binary matrix that is, it contains 0s and 1s only, we need to find sum of coverage of all zeros of the matrix where coverage for a particular 0 is defined as total number of ones around a zero in left, right, up and bottom directions. The ones can be anywhere till corner point in a direction.
Examples:
Input : mat[][] = {0 0 0 0
1 0 0 1
0 1 1 0
0 1 0 0}
Output : 20
First four zeros are surrounded by only
one 1. So coverage for zeros in first
row is 1 + 1 + 1 + 1
Zeros in second row are surrounded by
three 1's. Note that there is no 1 above.
There are 1's in all other three directions.
Coverage of zeros in second row = 3 + 3.
Similarly counting for others also, we get
overall count as below.
1 + 1 + 1 + 1 + 3 + 3 + 2 + 2 + 2 + 2 + 2 = 20
Input : mat[][] = {1 1 1 0
1 0 0 1}
Output : 8
Coverage of first zero is 2
Coverages of other two zeros is 3
Total coverage = 2 + 3 + 3 = 8
A simple solution to solve this problem is by counting ones around zeros independently i.e. we run loop four times in each direction for each cell for the given matrix. Whenever we find a 1 in any loop, we break the loop and increment result by 1.
An efficient solution is to do following.
- Traverse all rows from left to right, increment result if a 1 is already seen (in current traversal) and current element is 0.
- Traverse all rows from right to left, increment result if a 1 is already seen (in current traversal) and current element is 0.
- Traverse all columns from top to bottom, increment result if a 1 is already seen (in current traversal) and current element is 0.
- Traverse all columns from bottom to top, increment result if a 1 is already seen (in current traversal) and current element is 0.
In below code a Boolean variable isOne is taken, which is made true as soon as a one is encountered in current traversal, for all zeros after that iteration, result is incremented by one, same procedure is applied in all four directions to get final answer. We reset isOne to false after every traversal.
C++
// C++ program to get total coverage of all zeros in// a binary matrix#include <bits/stdc++.h>using namespace std;#define R 4#define C 4// Returns total coverage of all zeros in mat[][]int getTotalCoverageOfMatrix(int mat[R][C]){ int res = 0; // looping for all rows of matrix for (int i = 0; i < R; i++) { bool isOne = false; // 1 is not seen yet // looping in columns from left to right // direction to get left ones for (int j = 0; j < C; j++) { // If one is found from left if (mat[i][j] == 1) isOne = true; // If 0 is found and we have found // a 1 before. else if (isOne) res++; } // Repeat the above process for right to // left direction. isOne = false; for (int j = C-1; j >= 0; j--) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } } // Traversing across columns for up and down // directions. for (int j = 0; j < C; j++) { bool isOne = false; // 1 is not seen yet for (int i = 0; i < R; i++) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } isOne = false; for (int i = R-1; i >= 0; i--) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } } return res;}// Driver code to test above methodsint main(){ int mat[R][C] = {{0, 0, 0, 0}, {1, 0, 0, 1}, {0, 1, 1, 0}, {0, 1, 0, 0} }; cout << getTotalCoverageOfMatrix(mat); return 0;} |
Java
// Java program to get total // coverage of all zeros in // a binary matriximport java .io.*;class GFG {static int R = 4;static int C = 4;// Returns total coverage// of all zeros in mat[][]static int getTotalCoverageOfMatrix(int [][]mat){ int res = 0; // looping for all // rows of matrix for (int i = 0; i < R; i++) { // 1 is not seen yet boolean isOne = false; // looping in columns from // left to right direction // to get left ones for (int j = 0; j < C; j++) { // If one is found // from left if (mat[i][j] == 1) isOne = true; // If 0 is found and we // have found a 1 before. else if (isOne) res++; } // Repeat the above // process for right // to left direction. isOne = false; for (int j = C - 1; j >= 0; j--) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } } // Traversing across columns // for up and down directions. for (int j = 0; j < C; j++) { // 1 is not seen yet boolean isOne = false; for (int i = 0; i < R; i++) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } isOne = false; for (int i = R - 1; i >= 0; i--) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } } return res;}// Driver code static public void main (String[] args){ int [][]mat = {{0, 0, 0, 0}, {1, 0, 0, 1}, {0, 1, 1, 0}, {0, 1, 0, 0}};System.out.println( getTotalCoverageOfMatrix(mat));}}// This code is contributed by anuj_67. |
Python3
# Python3 program to get total coverage of all zeros in# a binary matrixR = 4C = 4# Returns total coverage of all zeros in mat[][]def getTotalCoverageOfMatrix(mat): res = 0 # looping for all rows of matrix for i in range(R): isOne = False # 1 is not seen yet # looping in columns from left to right # direction to get left ones for j in range(C): # If one is found from left if (mat[i][j] == 1): isOne = True # If 0 is found and we have found # a 1 before. else if (isOne): res += 1 # Repeat the above process for right to # left direction. isOne = False for j in range(C - 1, -1, -1): if (mat[i][j] == 1): isOne = True else if (isOne): res += 1 # Traversing across columns for up and down # directions. for j in range(C): isOne = False # 1 is not seen yet for i in range(R): if (mat[i][j] == 1): isOne = True else if (isOne): res += 1 isOne = False for i in range(R - 1, -1, -1): if (mat[i][j] == 1): isOne = True else if (isOne): res += 1 return res# Driver codemat = [[0, 0, 0, 0],[1, 0, 0, 1],[0, 1, 1, 0],[0, 1, 0, 0]]print(getTotalCoverageOfMatrix(mat))# This code is contributed by shubhamsingh10 |
C#
// C# program to get total coverage // of all zeros in a binary matrixusing System;class GFG { static int R = 4;static int C = 4;// Returns total coverage of all zeros in mat[][]static int getTotalCoverageOfMatrix(int [,]mat){ int res = 0; // looping for all rows of matrix for (int i = 0; i < R; i++) { // 1 is not seen yet bool isOne = false; // looping in columns from left to // right direction to get left ones for (int j = 0; j < C; j++) { // If one is found from left if (mat[i,j] == 1) isOne = true; // If 0 is found and we // have found a 1 before. else if (isOne) res++; } // Repeat the above process for // right to left direction. isOne = false; for (int j = C-1; j >= 0; j--) { if (mat[i,j] == 1) isOne = true; else if (isOne) res++; } } // Traversing across columns // for up and down directions. for (int j = 0; j < C; j++) { // 1 is not seen yet bool isOne = false; for (int i = 0; i < R; i++) { if (mat[i,j] == 1) isOne = true; else if (isOne) res++; } isOne = false; for (int i = R-1; i >= 0; i--) { if (mat[i,j] == 1) isOne = true; else if (isOne) res++; } } return res;}// Driver code to test above methods static public void Main () { int [,]mat = {{0, 0, 0, 0}, {1, 0, 0, 1}, {0, 1, 1, 0}, {0, 1, 0, 0}}; Console.WriteLine(getTotalCoverageOfMatrix(mat)); }}// This code is contributed by vt_m. |
Javascript
<script> // Javascript program to get total // coverage of all zeros in // a binary matrix let R = 4; let C = 4; // Returns total coverage // of all zeros in mat[][] function getTotalCoverageOfMatrix(mat) { let res = 0; // looping for all // rows of matrix for (let i = 0; i < R; i++) { // 1 is not seen yet let isOne = false; // looping in columns from // left to right direction // to get left ones for (let j = 0; j < C; j++) { // If one is found // from left if (mat[i][j] == 1) isOne = true; // If 0 is found and we // have found a 1 before. else if (isOne) res++; } // Repeat the above // process for right // to left direction. isOne = false; for (let j = C - 1; j >= 0; j--) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } } // Traversing across columns // for up and down directions. for (let j = 0; j < C; j++) { // 1 is not seen yet let isOne = false; for (let i = 0; i < R; i++) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } isOne = false; for (let i = R - 1; i >= 0; i--) { if (mat[i][j] == 1) isOne = true; else if (isOne) res++; } } return res; } let mat = [[0, 0, 0, 0], [1, 0, 0, 1], [0, 1, 1, 0], [0, 1, 0, 0]]; document.write(getTotalCoverageOfMatrix(mat));</script> |
Output
20
Time Complexity: O(n2)
Auxiliary Space: O(1)
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