JavaScript: Determine the index of matching elements in two arrays

Question

JavaScript: Determine the index of matching elements in two arrays

Here is a code snippet I am working with:

function findMatch(array_1_small, array2_large) {
var ary = new Array();
for(i = 0;i < array2_large.length; i++)
{
  for(z = 0; z < array_1_small.length; z++)
  {
    if(array2_large[i] == array_1_small[z])
    {
     var idx = array2_large.indexOf(array2_large[i]);
      ary.push(idx);
    }
  }

}
return ary;
}

The code operates on the following arrays:

var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
var all_FORM_NUMBERS = ["", "", "", "", "", "", "0871355066",""];

Its main function is to identify matches and return their indices in the 'all_form_numbers' array.

When I run the code using this setup:

var a = findMatch(all_SMS_TO, all_FORM_NUMBERS);
console.log("Match Found " + a);

I receive the output:

Match Found: 6

This result is accurate. However, when I modify the 'all_form_Numbers' array as follows:

var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];

The resulting output becomes:

Match Found: 1,1

I am seeking assistance to make the output display as:

Match Found 1, 6.

Any help would be greatly appreciated. Thank you.

javascript arrays

Answer 1

Answer №1

Give this a shot:

function checkMatch(arraySmall, arrayLarge) {
var resultArray = new Array();
for(indexLarge = 0; indexLarge < arrayLarge.length; indexLarge++)
{
  for(indexSmall = 0; indexSmall < arraySmall.length; indexSmall++)
  {
    if(arrayLarge[indexLarge] == arraySmall[indexSmall])
    {
      resultArray.push(indexLarge);
    }
  }

}
return resultArray;
}

Answer 2

Give this a shot:

function checkMatch(arraySmall, arrayLarge) {
var resultArray = new Array();
for(indexLarge = 0; indexLarge < arrayLarge.length; indexLarge++)
{
  for(indexSmall = 0; indexSmall < arraySmall.length; indexSmall++)
  {
    if(arrayLarge[indexLarge] == arraySmall[indexSmall])
    {
      resultArray.push(indexLarge);
    }
  }

}
return resultArray;
}

Answer 3

Answer №2

When you execute

var idx = array2_large.indexOf(array2_large[i]);

you are actually searching for the index of the value 0871355066 in the array named array2_large twice. The indexOf method returns the position of the first occurrence of a specified value.

This explains why you see the index value 1 appearing twice since it represents the index of the first occurrence.

To solve this issue, simply push the variable i's value into the ary array. This value already corresponds to the index value of array2_large within the loop.

function findMatch(array_1_small, array2_large) {
var ary = new Array();
for(i = 0;i < array2_large.length; i++)
{
  for(z = 0; z < array_1_small.length; z++)
  {
    if(array2_large[i] == array_1_small[z])
    {
      ary.push(i);
    }
  }

}
return ary;
}


var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
//var all_FORM_NUMBERS = ["", "", "", "", "", "", "0871355066",""];

var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];


var a = findMatch(all_SMS_TO, all_FORM_NUMBERS);
console.log("Match Found " + a);

Answer 4

When you execute

var idx = array2_large.indexOf(array2_large[i]);

you are actually searching for the index of the value 0871355066 in the array named array2_large twice. The indexOf method returns the position of the first occurrence of a specified value.

This explains why you see the index value 1 appearing twice since it represents the index of the first occurrence.

To solve this issue, simply push the variable i's value into the ary array. This value already corresponds to the index value of array2_large within the loop.

function findMatch(array_1_small, array2_large) {
var ary = new Array();
for(i = 0;i < array2_large.length; i++)
{
  for(z = 0; z < array_1_small.length; z++)
  {
    if(array2_large[i] == array_1_small[z])
    {
      ary.push(i);
    }
  }

}
return ary;
}


var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
//var all_FORM_NUMBERS = ["", "", "", "", "", "", "0871355066",""];

var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];


var a = findMatch(all_SMS_TO, all_FORM_NUMBERS);
console.log("Match Found " + a);

Answer 5

Answer №3

To solve the problem, simply ensure that you are pushing the index i in the code below. You may also choose to create an array by declaring it as var res = [];

function findMatch(arraySmall, arrayLarge) {
  var res = []
  for (var i = 0; i < arrayLarge.length; i++) {
    for (var j = 0; j < arraySmall.length; j++) {
      if (arrayLarge[i] === arraySmall[j]) {
        res.push(i);
      }
    }
  }
  return res;
}

Answer 6

To solve the problem, simply ensure that you are pushing the index i in the code below. You may also choose to create an array by declaring it as var res = [];

function findMatch(arraySmall, arrayLarge) {
  var res = []
  for (var i = 0; i < arrayLarge.length; i++) {
    for (var j = 0; j < arraySmall.length; j++) {
      if (arrayLarge[i] === arraySmall[j]) {
        res.push(i);
      }
    }
  }
  return res;
}

Answer 7

Answer №4

To achieve a runtime complexity of O(n+m), one can solve this issue by initially creating a lookup table of positions. Then, each element from the first array is mapped to all positions and these indices are collected in a Set to retain only unique values.

Give this solution a try:

var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];

function findMatch(array_1_small, array2_large) {
    var positions = Array.from(array2_large.entries()).reduce((acc, t) => {
        var index = t[0]
        var element = t[1]
        if (!acc.hasOwnProperty(element)) {
            acc[element] = []
        }
        acc[element].push(index)
        return acc
    }, {})
    var result = new Set()
    array_1_small.forEach(x => {
        if (positions[x] === undefined) {
            return
        }
        positions[x].forEach(index => result.add(index))
    })
    return Array.from(result)
}

console.log("Found a match: " + findMatch(all_SMS_TO, all_FORM_NUMBERS))

Answer 8

To achieve a runtime complexity of O(n+m), one can solve this issue by initially creating a lookup table of positions. Then, each element from the first array is mapped to all positions and these indices are collected in a Set to retain only unique values.

Give this solution a try:

var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];

function findMatch(array_1_small, array2_large) {
    var positions = Array.from(array2_large.entries()).reduce((acc, t) => {
        var index = t[0]
        var element = t[1]
        if (!acc.hasOwnProperty(element)) {
            acc[element] = []
        }
        acc[element].push(index)
        return acc
    }, {})
    var result = new Set()
    array_1_small.forEach(x => {
        if (positions[x] === undefined) {
            return
        }
        positions[x].forEach(index => result.add(index))
    })
    return Array.from(result)
}

console.log("Found a match: " + findMatch(all_SMS_TO, all_FORM_NUMBERS))

JavaScript: Determine the index of matching elements in two arrays

Answer №1

Answer №2

Answer №3

Answer №4

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