1

I need to remove similar duplicates as well as real duplicates from 2D array in JavaScript.

let a = [
  [5, 6],
  [1,1],
  [6,5],
  [1,1],
  [3,2],
  [2,3]
]

function makeUnique(arr) {
  var uniques = [];
  var itemsFound = {};
  for(var i = 0, l = arr.length; i < l; i++) {
      var stringified = JSON.stringify(arr[i]);
      if(itemsFound[stringified])  continue; 
      uniques.push(arr[i]);
      itemsFound[stringified] = true;
  }
  return uniques;
}

a=makeUnique(a)
console.log(a);

I have got this output:

[ [ 5, 6 ], [ 1, 1 ], [ 6, 5 ], [ 3, 2 ], [ 2, 3 ] ]

Correct should be:

[ [ 5, 6 ], [ 1, 1 ], [ 2, 3 ] ]

My code removes correctly duplicates, but I need to remove similar duplicates also.

For example if I have [3,2] and [2,3] I should remove [3,2] (the one which has bigger starting index value.)

Could you help me to fix this?

Improve this question

3 Answers 3

Reset to default
2

Here is an example of how you can do it:


function makeUnique(arr) {
    var uniques = [];
    var itemsFound = {};
    arr.sort((a, b) => a[0] + a[1] - (b[0] + b[1]))
    for (var i = 0, l = arr.length; i < l; i++) {
        if (!itemsFound[arr[i]] && !itemsFound[[arr[i][1], arr[i][1]]]) {
            uniques.push(arr[i]);
            itemsFound[arr[i]] = true;
            itemsFound[[arr[i][1], arr[i][0]]] = true;
        }
    }
    return uniques;
}

I hope it helps.

Improve this answer
Sign up to request clarification or add additional context in comments.

3 Comments

Sorry. I didn't see "the one which has bigger starting index value.". Hold on plz.
Does the order of uniques array matter?
nice, thank you, could you please make it that it keeps lower number of two similar duplicates. for example if we have [2,3] and [3,2] it should keep [2,3] because it is lower, your code currently keeps the first element of two similar duplicates, it doesn't check which is lower.
1

There are two parts

  1. similar should be considered
  2. among similar, one with smaller first key should stay

1. Similar should be considered

Here you can just make the key for hashmap in such a way that similar items produce same key.

One way to do that is sort the items in the tuple and then form the key, as there are two items only, first one will be min and second one will be max

let a = [
  [5, 6],
  [1,1],
  [6,5],
  [1,1],
  [3,2],
  [2,3]
]

function makeUnique(arr) {
  var uniques = [];
  var itemsFound = {};
  for(var i = 0, l = arr.length; i < l; i++) {
      let [a,b] = arr[i];
      const hashKey = [ Math.min(a,b), Math.max(a,b)];
      var stringified = JSON.stringify(hashKey);
      if(itemsFound[stringified])  continue; 
      uniques.push(arr[i]);
      itemsFound[stringified] = true;
  }
  return uniques;
}

let ans1=makeUnique(a)
console.log(ans1);

2. Among similar, the one with smaller first key should stay

Now you can remember in the hashmap what the value for a key was and keep updating it based on the correct candidate

let a = [
  [5, 6],
  [1,1],
  [6,5],
  [1,1],
  [3,2],
  [2,3]
]

function makeUniqueSmallerFirst(arr) {
  var items = {};
  for(var i = 0, l = arr.length; i < l; i++) {
      let [a,b] = arr[i];
      const hashKey = [ Math.min(a,b), Math.max(a,b)];
      var stringified = JSON.stringify(hashKey);
      
      if (stringified in items) {
          let previous = items[stringified];
          if (previous[0] > arr[i][0]) {
              items[stringified] = arr[i];
          }
      } else {
          items[stringified] =  arr[i]  // I am just storing  the array because if I see a similar item next time, I can compare if that has first item smaller or not 
      }
  }

  return Object.values(items); // this doesn't guarantee output order though
  // if you want order as well . you can iterate over input array once more and arrange the items in the preferred order.
}

let ans2=makeUniqueSmallerFirst(a)
console.log(ans2);
Improve this answer

Comments

1

UPDATED (More simple and faster example for ES5+):

function makeUnique(arr) {
    return new Set(a.map(
        arr => JSON.stringify(arr.sort((a, b) => a - b)))
    )
}

const m = makeUnique(a)
console.log(m) //

OLD:

This is an example of code that makes a two-dimensional array with arrays of any length unique.

let a = [
    [5, 6],
    [1, 1],
    [6, 5],
    [1, 5],
    [3, 2],
    [2, 3],
    [6, 5, 3],
    [3, 5, 6]
]

function isUnique(uniqueArray, checkedArray) {
    let checked = [...checkedArray];
    let unique = [...uniqueArray];
    let uniqueValue = 0;
    unique.forEach(value => {
        if (checked.includes(value)) {
            checked.splice(checked.indexOf(value), 1)
        } else uniqueValue++;
    })
    return uniqueValue > 0;
}

function makeUnique(array2d) {
    let unique = [array2d[0]]
    array2d.forEach(checkedArray => {
        if (unique.some(uniqueArray => {
                if (checkedArray.length !== uniqueArray.length) return false;
                return !isUnique(uniqueArray, checkedArray)
            }
        )) return 0;
        else unique.push(checkedArray)
    })
    return unique
}

console.log(makeUnique(a)) // [ [ 5, 6 ], [ 1, 1 ], [ 1, 5 ], [ 3, 2 ], [ 6, 5, 3 ] ]

isUnique() this function checks if the numbers in both arrays are unique, and if they are, it outputs true. We use the copy through spread operator, so that when you delete a number from an array, the array from outside is not affected.

makeUnique() function makes the array unique, in the following way: It checks if our unique two-dimensional array has at least one array that is identical to checkedArray

The first check if the arrays are of different lengths - they are unique, skip and check for uniqueness, if !isUnique gives out true, then the array is skipped by return 0

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.