Count number of 1's at specific position in a Matrix
Last Updated :
24 Apr, 2023
Given a binary matrix mat[][] of size m*n (0-based indexing), find the number of 1's such that 1 is located at a specific position where the same row and same column don't have any other 1's.
Examples:
Input: mat[][] = [[0, 0, 1], [0, 1, 0], [1, 0, 0]]
Output: 3
Explanation: All 1's in the matrix satisfy the given condition.
Input: mat[][] = [[1, 0, 1], [0, 1, 0], [1, 0, 0]]
Output: 1
Explanation: mat[1][1] satisfy the given condition.
Approach: The approach is based on matrix traversal.
Below are the steps for the above approach:
- Initialize auxiliary array rows[] and cols[] to track how many 1's are in the corresponding row or column.
- Initialize a variable say, ans to store the number of 1's that satisfy the given condition.
- Now traverse through the matrix mat[][] and check if mat[i][j]==1 then increment rows[i]++(for row) and increment cols[j]++(for column).
- Run a loop from i = 0 to i < m and check if row[i] == 1,
- Run a loop from j = 0 to j < n, and check if the element at the current position is 1 in mat[][] if mat[i][j] == 1
- Check if cols[j] == 1, increment ans, else break.
- Return ans.
Below is the code for the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
int findNumberOfOnes(vector<vector<char> >& mat)
{
const int m = mat.size();
const int n = mat[0].size();
int ans = 0;
// Auxiliary row and col for tracking
// how many B at that corresponding
// row and col
vector<int> rows(m);
vector<int> cols(n);
// Traversing through the mat[][]
// matrix. If mat[i][j] == B then
// increment rows[i]++(for row) and
// increment cols[j]++(for column).
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
if (mat[i][j] == 1) {
++rows[i];
++cols[j];
}
// Traversing through the mat[][]
// matrix again and if
// rows[i]==1 || cols[j]==1
// then ans++, else continue.
for (int i = 0; i < m; ++i)
if (rows[i] == 1)
for (int j = 0; j < n; ++j)
// If the current position
// is 1 in this row then
// break and search
// the next row
if (mat[i][j] == 1) {
if (cols[j] == 1)
++ans;
break;
}
return ans;
}
// Drivers code
int main()
{
vector<vector<char> > mat
= { { 0, 0, 1 }, { 0, 1, 0 }, { 1, 0, 0 } };
// Function Call
cout << findNumberOfOnes(mat);
return 0;
}
/*package whatever // do not write package name here */
// A program for finding the lonely color
import java.io.*;
class GFG {
public static int findLonelyColor(char[][] painting)
{
int m = painting.length, n = painting[0].length;
// Auxiliary row and col for tracking how many B at
// that corresponding row and col
int[] rows = new int[m];
int[] cols = new int[n];
// Traversing through the painting matrix. If
// painting[i][j]==B then increment rows[i]++(for
// row) and increment cols[j]++(for column).
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (painting[i][j] == 'B') {
rows[i]++;
cols[j]++;
}
}
}
int ans = 0;
// Traversing through the painting matrix again and
// if rows[i]==1 || cols[j]==1 then ans++, else
// continue.
for (int i = 0; i < m; ++i) {
if (rows[i] == 1) {
for (int j = 0; j < n; ++j) {
// If the current position is 'B' in this
// row then break and search the next
// row
if (painting[i][j] == 'B'
&& cols[j] == 1) {
ans++;
break;
}
}
}
}
return ans;
}
public static void main(String[] args)
{
char[][] painting = { { 'W', 'W', 'B' },
{ 'W', 'B', 'W' },
{ 'B', 'W', 'W' } };
int ans = findLonelyColor(painting);
System.out.println(ans);
}
}
# A program for finding the lonely color
from typing import List
def findLonelyColor(self, painting: List[List[str]]) -> int:
# Auxiliary row and col for tracking how many B at that
# corresponding row and col
m, n = len(painting), len(painting[0])
rows, cols = [0] * m, [0] * n
# Traversing through the painting matrix.
# If painting[i][j]== B then increment rows[i]++(for row)
# and increment cols[j]++(for column).
for i in range(m):
for j in range(n):
if painting[i][j] == 'B':
rows[i] += 1
cols[j] += 1
ans = 0
# Traversing through the painting matrix again and
# if rows[i]== 1 || cols[j]== 1 then ans++, else continue
for i in range(m):
if rows[i] == 1:
for j in range(n):
# If the current position is 'B' in this row
# then break and search the next row
if painting[i][j] == 'B' and cols[j] == 1:
ans += 1
break
return ans
if __name__ == "__main__":
painting = [['W', 'W', 'B'],
['W', 'B', 'W'],
['B', 'W', 'W']]
print(findLonelyColor(__name__, painting))
using System;
public class GFG {
public static int FindLonelyColor(char[, ] painting)
{
int m = painting.GetLength(0);
int n = painting.GetLength(1);
// Auxiliary row and col for tracking how many B at
// that corresponding row and col
int[] rows = new int[m];
int[] cols = new int[n];
// Traversing through the painting matrix. If
// painting[i,j] == 'B' then increment rows[i]++(for
// row) and increment cols[j]++(for column).
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (painting[i, j] == 'B') {
rows[i]++;
cols[j]++;
}
}
}
int ans = 0;
// Traversing through the painting matrix again and
// if rows[i]==1 || cols[j]==1 then ans++, else
// continue.
for (int i = 0; i < m; ++i) {
if (rows[i] == 1) {
for (int j = 0; j < n; ++j) {
// If the current position is 'B' in this
// row then break and search the next
// row
if (painting[i, j] == 'B'
&& cols[j] == 1) {
ans++;
break;
}
}
}
}
return ans;
}
public static void Main()
{
char[, ] painting = { { 'W', 'W', 'B' },
{ 'W', 'B', 'W' },
{ 'B', 'W', 'W' } };
int ans = FindLonelyColor(painting);
Console.WriteLine(ans);
}
}
// This code is contributed by Prajwal Kandekar
// JavaScript code for the above approach:
function findNumberOfOnes(mat) {
const m = mat.length;
const n = mat[0].length;
let ans = 0;
// Auxiliary row and col for tracking
// how many B at that corresponding
// row and col
const rows = new Array(m).fill(0);
const cols = new Array(n).fill(0);
// Traversing through the mat[][]
// matrix. If mat[i][j] == B then
// increment rows[i]++(for row) and
// increment cols[j]++(for column).
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (mat[i][j] === 1) {
++rows[i];
++cols[j];
}
}
}
// Traversing through the mat[][]
// matrix again and if
// rows[i]==1 || cols[j]==1
// then ans++, else continue.
for (let i = 0; i < m; ++i) {
if (rows[i] === 1) {
for (let j = 0; j < n; ++j) {
// If the current position
// is 1 in this row then
// break and search
// the next row
if (mat[i][j] === 1) {
if (cols[j] === 1) {
++ans;
}
break;
}
}
}
}
return ans;
}
// Drivers code
const mat = [
[0, 0, 1],
[0, 1, 0],
[1, 0, 0],
];
// Function Call
console.log(findNumberOfOnes(mat));
Output :
3
Time Complexity: O(m*n + m*n), Where m and n are the sizes of rows and columns
Auxiliary Space: O(m + n), used auxiliary array rows and cols of size m and n respectively
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