## Compare elements of each row in a 2D matrix java

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I was trying to compare the values of each element in a row of a 2D matrix to make sure every element is different from each other.

Here's my function:

private static boolean CompareComponentValue(int[][] m) { int k = 0; for (int i = 0; i < m.length; i = i + 1) { for (int j = 1; j < m[i].length; j = j + 1) { if (m[i][k] == m[i][j]) { return false; } } k = k + 1; } return true; }

I was thinking that I had to campare each element like this: [0][0] is different from [0][1], [0][2] ...[0][n], then: [0][1] is different from [0][2]...[0][n] and also for each row of course, [1][0] is different from [1][1], [1][2]...[1][n].

I can't get it to work properly because it will return false sometimes when it shouldn't.

*"... to make sure every element is different from each other.*"

What you need is a Set. Adding an element to a set, and testing whether an element is in the set already, are easy to do using the `add`

and `contains`

operations. Also, these operations both take O(1) time, so the solution will be efficient; no need to search the matrix again and again.

Also, use the "enhanced `for`

loop" if you just need the values, not the indices.

import java.util.*; public class CheckMatrix { public static boolean hasDuplicate(int[][] matrix) { Set<Integer> seen = new HashSet<>(); for(int[] row : matrix) { // empty the set, to only check for duplicates within a single row seen.clear(); for(int x : row) { if(seen.contains(x)) { return true; } seen.add(x); } } return false; } }

If you want to be fancy, you can use the fact that the add method returns a `boolean`

indicating whether `x`

was not already present: replace the inner loop body with:

if(!seen.add(x)) { return true; }

**Java: Compare elements of 2D array(matrix) in the column And ,** First of all I think you are confused about row and column of an array. Your array has 3 rows and 4 columns. I will assume you want to find the� We use arrayname.length to determine the number of rows in a 2D array because the length of a 2D array is equal to the number of rows it has. The number of columns may vary row to row, which is why the number of rows is used as the length of the 2D array. When calling the length of a column, we pinpoint the row before using .length.

You can do it as follows:

import java.io.IOException; import java.util.Arrays; import java.util.Set; import java.util.stream.Collectors; public class Solution { public static void main(String[] args) throws IOException { Integer n[][] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; Integer x[][] = { { 1, 2, 2 }, { 4, 5, 6 }, { 7, 8, 9 } }; System.out.println(compareComponentValue(n)); System.out.println(compareComponentValue(x)); } private static boolean compareComponentValue(Integer[][] m) { for (int i = 0; i < m.length; i++) { // Add the row i.e. m[i] to a Set Set<Integer> set = Arrays.stream(m[i]).collect(Collectors.toSet()); // If there will be any duplicate values in m[i], they will be rejected by the // set and therefore the size of the set won't be equal to the size of m[i] if (set.size() != m[i].length) return false; } return true; } }

The solution is based on the fact that a set rejects non-unique values.

**Output:**

true false

**Compare rows in 2d array in java using deepequals method ,** Compare rows in 2d array in java using deepequals method. Elements are stored in an array of size [n][4]. Input Format: First line of the input is an integer “n” that denotes the number of rows. Java Program to find the sum of each row and each column of a matrix. In this program, we need to calculate the sum of elements in each row and each column of the given matrix. Above diagram shows the sum of elements of each row and each column of a matrix. Algorithm. STEP 1: START; STEP 2: DEFINE rows, cols, sumRow, sumCol

First of all thank you guys for your answers but they really went above my level, as I've stated at the beggining I'm not using any packages for now because I'm still a beeginer and I want to stay with basic logic and in the end I got an answer by thinking hard.

private static boolean NumerosDistintosFila(int[][] m) { for (int i = 0; i < m.length; i = i + 1) { for (int k = 0; k < m.length - 2; k = k + 1) { for (int j = 1 + k; j < m[i].length; j = j + 1) { if (m[i][k] == m[i][j]) { return false; } } } } return true; }

I added another for loop with a variable k which I noticed didn't ever go above the matrix's lenght - 2, for example if I were to use a 5x5 I'll need it to go as far as 3. This is because the last element I need to compare in the first row for example is [0][3] to [0][4].

My logic was that I needed to compare elements without repeating the comparations between them so what this code does is basically compare the first element with every element in the same row, then compare the second element with every element except the first one then the third element with the fourth and the fifth etc..

Again thank you guys for your solutions I tried them out and all of them worked perfectly but as I said I want to stay with basic logic for now.

**comparing rows in a 2D array,** Hi, I have a 2D array as follows DataSet[10][10] = {{1,1,1,1,1,0,0,1,1,1}, {0,0,0,0,0, 0,1,0,0,0}, CSS � HTML � Java � JavaScript � jQuery � MySQL � Objective-C � PHP � Python � Ruby I want to compare the first row with all the remaining rows to see if they have the (say the first row has a 0 in the last element and the other= Search element in a sorted matrix; Largest Square in a Binary Matrix with at most K 1s for multiple Queries; Median of an unsorted array using Quick Select Algorithm; Check if a string is a scrambled form of another string; Count of smaller elements on right side of each element in an Array using Merge sort

**Find a common element in all rows of a given row-wise sorted matrix ,** Given a matrix where every row is sorted in increasing order. If elements at all last columns are same, then we found the common element. Java. filter_none. edit close. play_arrow. link brightness_4 code This can be checked by comparing HashTable value with row number) Update the value of this key in HashTable� To calculate the sum of elements in each row: Two loops will be used to traverse the array where the outer loop selects a row, and the inner loop represents the columns present in the matrix a. Calculate the sum by adding elements present in a row.

**How to compare 2d array with 1 d array?,** In your case you have array1->lenght 3 and the specific row in lets say records[0 ]-> If you remove element "WNBA" and "5" from your records array the 2 for ( int i = 0; i < records.length; i++) { System.out.println(java.util. A three – dimensional array with 3 array containing 3 rows and 3 columns is shown below: Print 3D array in tabular format: To output all the elements of a Three-Dimensional array, use nested for loops. For this three for loops are required, One to traverse the arrays, second to traverse the rows and another to traverse columns. Example:

**Javanotes 8.1, Section 7.5 -- Two-dimensional Arrays,** The elements of a 2D array are arranged in rows and columns, and the new operator for 2D arrays Java does not actually have two-dimensional arrays. Need for Multidimensional Collections in java? Unlike Arrays we are not bound with the size of any row in Multidimensional collections. Therefore, if we want to use a Multidimensional architecture where we can create any number of objects dynamically in a row, then we should go for Multidimensional collections in Java.

##### Comments

- So your goal is to check whether each row of the matrix consists of unique elements, correct?
- Well, the way you've written it now, sometimes
`k`

will equal`j`

, and then you're checking if something is equal to itself... - Yes, I don't know if this is the right way to answer your comment but that's exactly what I need
- I'd say you could either have an additional space complexity by using a Hashmap to see if an element from the row exists within the map already. Each row gets a new map instantiated.
- As I've said I don't even know what a hashmap is because I'm still a begginer I would rather to keep it simple