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 &lt; 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 &gt;= arr.length){
    return idx - arr.length
  }
  return idx
}

console.log(findIndexOfGroup([4,3,10])) // Normal
console.log(findIndexOfGroup([9,1,1,5])) // Wrapping
You could first get the index of the first element of the search array.

Then loop while index !== -1 and check all elements in the search array with the elements in the base array. For checking outside of the range of the array use the reminder operator %.

Return index if all elements are equal.

If not get a new index with indexOf and an incremented start value.

Array#every breaks the iteration if the callback returns false.



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; }