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Cells with Odd Values in a Matrix - Array - Easy - LeetCode - MiniTV

Cells with Odd Values in a Matrix - Array - Easy - LeetCode - मिनी टीवी

Given n and m which are the dimensions of a matrix initialized by zeros and given an array indices where indices[i] = [ri, ci]. For each pair of [ri, ci] you have to increment all cells in row ri and column ci by 1.

Return the number of cells with odd values in the matrix after applying the increment to all indices.

 

Example 1:

 

Input: n = 2, m = 3, indices = [[0,1],[1,1]]

Output: 6

Explanation: Initial matrix = [[0,0,0],[0,0,0]].

After applying first increment it becomes [[1,2,1],[0,1,0]].

The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.

 

Example 2:

Input: n = 2, m = 2, indices = [[1,1],[0,0]]

Output: 0

Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.

 

Constraints:

1 <= n <= 50

1 <= m <= 50

1 <= indices.length <= 100

0 <= indices[i][0] < n

0 <= indices[i][1] < m

 

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.Array
{
    public class OddCellsSoln
    {
        public int OddCells(int n, int m, int[][] indices)
        {
            var matrix = new int[n, m];

            //initialize matrix to zero.
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < m; j++)
                {
                    matrix[i, j] = 0;
                }
            }

            for (int i = 0; i < indices.Length; i++)
            {
                int x = indices[i][0];
                int y = indices[i][1];

                //increment x axis
                for (int a = 0; a < m; a++)
                {
                    matrix[x,a]++;
                }

                //increment y axis
                for (int b = 0; b < n; b++)
                {
                    matrix[b, y]++;
                }
            }

            int odd = 0;
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < m; j++)
                {
                    if (matrix[i, j] % 2 != 0)
                    {
                        odd++;
                    }
                }
            }

            return odd;
        }
    }
}

 

Time Complexity: O(n^2)

Space Complexity: O(m*n) For storing the result

 

Unit Tests:

using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.Array
{
    [TestClass]
    public class OddCellsSolnTests
    {
        [TestMethod]
        public void OddCellsSoln_First()
        {
            var indices = new int[,] {
                    { 0, 1 },
                    { 1, 1 }
                };
            var n = 2;
            var m = 3;
            var output = 6;

            var res = new OddCellsSoln().OddCells(n, m, ArrayMapper(indices));

            Assert.AreEqual(output, res);
        }

        [TestMethod]
        public void OddCellsSoln_Second()
        {
            var indices = new int[,] {
                    { 1, 1 },
                    { 0, 0 }
                };
            var n = 2;
            var m = 2;
            var output = 0;

            var res = new OddCellsSoln().OddCells(n, m, ArrayMapper(indices));

            Assert.AreEqual(output, res);
        }

        private int[][] ArrayMapper(int[,] matrix)
        {
            var arr = new int[matrix.GetLength(0)][];
            for (int i = 0; i < matrix.GetLength(0); i++)
            {
                arr[i] = new int[matrix.GetLength(1)];
                for (int j = 0; j < matrix.GetLength(1); j++)
                {
                    arr[i][j] = matrix[i, j];
                }
            }

            return arr;
        }
    }
}