Comparing 2D arrays: consider rows with same elements in different order being equal

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ardellia

Comparing 2D arrays: consider rows with same elements in different order being equal

in Programming Questions • 1 year ago

Hello everyone!

I need to compare two 2D arrays and append equal ones to a new empty array but this time I would need to two arrays to result equal even if the same elements are in different order, i.e. so {1,2,3,4} and {2,1,3,4} would be considered equal rows... Arrays.equal no longer applicable in this case.

I tried to solve this looping through each element but unfortunately this attempt failed:

int [][] array1 = { //all possible combinations

{1, 2, 3, 4},

{0, 1, 2, 4},

{3, 4, 5, 6},

{2, 3, 9, 3},

{3, 4, 6, 8},

{3, 4, 12, 13}};

int [][] array2 = {//all combinations for this player

{2, 1, 3, 4},

{0, 1, 2, 4},

{3, 4, 5, 6},

{2, 3, 10, 11},

{3, 4, 12, 13}};

int [][]array3 = new int [6][4];

int count1;

for (int i=0; i<array1.length;i++) {

for (int k=0; k<array2.length;k++){

for (int j=0;j<4;j++) {

if (array1[i][j]==array2[k][j]) {

count1=0;

array3[i][j]=array1[i][j];

count1++;

}

Does anyone know if there is any way to solve this (probably also in a more elegant way)?
Thanks a lot in advance!

Replies(5)

baud.r

Re: Comparing 2D arrays: consider rows with same elements in different order being equal

1 year ago

I added an extra bool array - each member is set to true if the members match. If all the bools in the test array are true, the subarray gets copied to array3. Well, it seems to work. I hope this helps!

int[][] array1 = {
{1, 2, 3, 4},
{0, 1, 2, 4},
{3, 4, 5, 6},
{2, 3, 9, 3},
{3, 4, 6, 8},
{3, 4, 12, 13}};
int[][] array2 = {
{4, 3, 2, 1},
{0, 1, 2, 4},
{3, 4, 5, 7},
{2, 3, 3, 9},
{3, 4, 12, 13}};
int[][] array3 = new int[6][4];
boolean[][] array2test = new boolean[array2.length][array2[0].length];
for (int i = 0; i < array2.length; i++) {
for (int j = 0; j < array2[i].length; j++) {
array2test[i][j] = false;
}
}
int count = 0;
for (int i = 0; i < array2.length; i++) {
for (int j = 0; j < array1[i].length; j++) {
for (int k = 0; k < array2[i].length; k++) {
if (array1[i][j] == array2[i][k]) {
array2test[i][k] = true;
}
}
}
boolean isEqual = true;
for (int j = 0; j < array2test[i].length; j++) {
if (!array2test[i][j]) {
isEqual = false;
}
}
if (isEqual) {
arrayCopy(array1[i], array3[count++]);
}
}
for (int i = 0; i < array3.length; i++) {
for (int j = 0; j < array3[i].length; j++) {
print(array3[i][j]);
}
}

baud.r

Re: Comparing 2D arrays: consider rows with same elements in different order being equal

1 year ago

Thought a bit more about this one, and made it a bit safer. Now it checks first to make sure the two subarrays are of equal length. It also checks that an array element hasn't already been 'used' (so that {3, 3, 3, 3} won't 'equal' {3, 1, 1, 1} any more). Also moved the bool array so that it has reduced scope, and commented it a bit.

int[][] array1 = {
{1, 2, 3, 4},
{0, 1, 2, 4},
{3, 4, 5, 6},
{4, 4, 4, 4},
{3, 4, 6, 8},
{3, 4, 12, 13}};
int[][] array2 = {
{4, 3, 2, 1},
{0, 1, 2, 4},
{3, 4, 5, 7},
{2, 4, 3, 9},
{3, 4, 12, 13}};
int[][] array3 = new int[array2.length][array2[0].length];
int count = 0;
for (int i = 0; i < array2.length; i++) {
if (array2[i].length == array1[i].length) {
// set up array used for checking equal members
boolean[] isEqualMember = new boolean[array2[i].length];
for (int j = 0; j < array2[i].length; j++) {
isEqualMember[j] = false;
}
// iterate through subarrays
for (int j = 0; j < array1[i].length; j++) {
for (int k = 0; k < array2[i].length; k++) {
// If two members match, and the array2 value has not already been 'used', set that the array value has an equal
if (array1[i][j] == array2[i][k] && !isEqualMember[k]) {
isEqualMember[k] = true;
}
}
}
// Assume the two subarrays match
boolean isEqual = true;
// Loop through the checking array, and make sure all the values are true
for (int j = 0; j < isEqualMember.length; j++) {
if (!isEqualMember[j]) {
isEqual = false;
}
}
// If all values in checking array are true, then isEqual will be true, and the arrays will be copied to the next available subarray slot
if (isEqual) {
arrayCopy(array1[i], array3[count++]);
}
}
}

ardellia

Re: Re: Comparing 2D arrays: consider rows with same elements in different order being equal

1 year ago

Hello baud.r,

thanks a lot for this solution, it works perfectly! but only one thing - right now for some reason it does not append the last row { 3, 4, 12, 13} although it is equal for both array1 and array2. Why do you think this is happening? I suspect it has something to do with setting the length of boolean array to the length of array2, I am trying to fix it right now but I am not sure...

Chrisir

Re: Comparing 2D arrays: consider rows with same elements in different order being equal

1 year ago

array 1 & 2 they don't have the same length - 6 versus 5 and he's doing a for loop over the shorter (array2)

add a line to array 2 and it should work.

ardellia

Re: Re: Comparing 2D arrays: consider rows with same elements in different order being equal

1 year ago

Hi Chrisir!

Thanks a lot for your comment! Unfortunately, I have already tried adding a new row to array2 to ensure they are of the same length, but the problem persists...

It seems that the code does not recognize rows being equal if they are on different positions, i.e. if array1[0]= {3, 4, 12, 13} and array2[2] = {3, 4, 12, 13} they will not be recognised as equal...

Maybe the algorithm should be like this:

(1) Check if first element in first row of array1 is in first row of array2:

(2) if false, pass to next row of array2 and repeat until match is found...

(3) if true, continue searching for next elements in first row of array2 until all 4 of them match

3.1 if true, repeat (2)

3.2. if false append to array3 and pass to next row of array1 and repeat;

Trying to implement this but so far not successfully... would appreciate any help or suggestions! Thank you!

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