CODEWITHSUNDEEP
javascriptarrays
TheGeekZn
TheGeekZn

Reputation: 3914

Find a group of objects in an array with javascript

Question has been moved to CodeReview: https://codereview.stackexchange.com/questions/154804/find-a-list-of-objects-in-an-array-with-javascript

Having an array of objects - such as numbers - what would be the most optimal (Memory and CPU efficiency) way if finding a sub group of objects? As an example:

demoArray = [1,2,3,4,5,6,7]
Finding [3,4,5] would return 2, while looking for 60 would return -1.
The function must allow for wrapping, so finding [6,7,1,2] would return 5

I have a current working solution, but I'd like to know if it could be optimized in any way.

var arr = [
  1,
  5,2,6,8,2,
  3,4,3,10,9,
  1,5,7,10,3,
  5,6,2,3,8,
  9,1]
var idx = -1
var group = []
var groupSize = 0

function findIndexOfGroup(g){
  group = g
  groupSize = g.length
  var beginIndex = -2
  
  while(beginIndex === -2){
    beginIndex = get()
  }
  
  return beginIndex
}

function get(){
    idx = arr.indexOf(group[0], idx+1);
    
    if(idx === -1 || groupSize === 1){
      return idx;
    }
    var prevIdx = idx
    
    for(var i = 1; i < groupSize; i++){
      idx++
      
      if(arr[getIdx(idx)] !== group[i]){
        idx = prevIdx
        break
      }
      
      if(i === groupSize - 1){
        return idx - groupSize + 1
      }
    }
    return -2
}

function getIdx(idx){
  if(idx >= arr.length){
    return idx - arr.length
  }
  return idx
}

console.log(findIndexOfGroup([4,3,10])) // Normal
console.log(findIndexOfGroup([9,1,1,5])) // Wrapping

Upvotes: 3

Views: 106

Answers (2)

Angelos Chalaris
Angelos Chalaris

Reputation: 6767

My take on the problem is to use slice() and compare each subarray of length equal to the group's length to the actual group array. Might take a bit long, but the code is short enough:

// The array to run the tests on
var arr = [
  1,
  5, 2, 6, 8, 2,
  3, 4, 3, 10, 9,
  1, 5, 7, 10, 3,
  5, 6, 2, 3, 8,
  9, 1
];
// Check arrays for equality, provided that both arrays are of the same length
function arraysEqual(array1, array2) {
  for (var i = array1.length; i--;) {
    if (array1[i] !== array2[i])
      return false;
  }
  return true;
}
// Returns the first index of a subarray matching the given group of objects
function findIndexOfGroup(array, group) {
  // Get the length of both arrays
  var arrayLength = array.length;
  var groupLength = group.length;
  // Extend array to check for wrapping
  array = array.concat(array);
  var i = 0;
  // Loop, slice, test, return if found
  while (i < arrayLength) {
    if (arraysEqual(array.slice(i, i + groupLength), group))
      return i;
    i++;
  }
  // No index found
  return -1;
}
// Tests
console.log(findIndexOfGroup(arr,[4,3,10])); // Normal
console.log(findIndexOfGroup(arr,[9,1,1,5])); // Wrapping
console.log(findIndexOfGroup(arr,[9,2,1,5])); // Not found

If the group is longer than the array, some errors might occur, but I leave it up to you to extend the method to deal with such situations.

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386654

You could use the reminder operator % for keeping the index in the range of the array with a check for each element of the search array with Array#every.

function find(search, array) {
    var index = array.indexOf(search[0]);

    while (index !== -1) {
        if (search.every(function (a, i) { return a === array[(index + i) % array.length]; })) {
            return index;
        }
        index = array.indexOf(search[0], index + 1);
    }
    return -1;
}

console.log(find([3, 4, 5], [1, 2, 3, 4, 5, 6, 7]));          //  2
console.log(find([6, 7, 1, 2], [1, 2, 3, 4, 5, 6, 7]));       //  5
console.log(find([60], [1, 2, 3, 4, 5, 6, 7]));               // -1
console.log(find([3, 4, 5], [1, 2, 3, 4, 6, 7, 3, 4, 5, 9])); //  6
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 4

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