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Most Efficient Method To Check For Range Of Numbers Within Number Without Duplicates

- 1 answer

Given a number n , a minimum number min , a maximum number max , what is the most efficient method to determine

  1. Number n is or is not within range , inclusive of , min - max

  2. Number n does or does not contain duplicate numbers

  3. Efficiency meaning here that the method or set of methods requires the least amount of computational resources and returns either true or false in the least amount of time

  4. Context: Condition at if within a for loop which could require from thousands to hundreds of thousands of iterations to return a result; where milliseconds required to return true or false as to Number check could affect performance

At Profiles panel at DevTools on a collection of 71,3307 items iterated, RegExp below was listed as using 27.2ms of total 1097.3ms to complete loop . At a collection of 836,7628 items iterated RegExp below used 193.5ms within total of 11285.3ms .

Requirement: Most efficient method to return Booleantrue or false given above parameters , within the least amount of time.

Note: Solution does not have to be limited to RegExp ; used below as the pattern returned expected results.


Current js utilizing RegExpre , RegExp.protype.test()

var min = 2
, max = 7
, re = new RegExp("[" + min + "-" + max + "](.)(?!=\1)", "g")
, arr = [81, 35, 22, 45, 49];

for (var i = 0; i < arr.length; i++) {
  console.log(re.test(arr[i]), i, arr[i])
    /*
      false 0 81 
      true 1 35
      false 2 22
      true 3 45
      false 4 49 
    */
}

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Answer

Associative arrays approach:

This has the advantage of being easily understandable.

function checkDigits(min, max, n) {
    var digits = Array(10);                   // Declare the length of the array (the 10 digits) to avoid any further memory allocation
    while (n) {
        d = (n % 10);                         // Get last digit
        n = n / 10 >>0;                       // Remove it from our number (the >>0 bit is equivalent to compose(Math.floor, Math.abs))
        if (d < min || d > max || digits[d])  // Test if "d" is outside the range or if it has been checked in the "digits" array
            return false;
        else
            digits[d] = true;                 // Mark the digit as existing
    }
}

var min = 2
, max = 7
, arr = [81, 35, 22, 45, 49];

function checkDigits(min, max, n) {
    var digits = Array(10);                   // Declare the length of the array (the 10 digits) to avoid any further memory allocation
    while (n) {
        d = (n % 10);                         // Get last digit
        n = n / 10 >>0;                       // Remove it from our number (the >>0 bit is equivalent to compose(Math.floor, Math.abs))
        if (d < min || d > max || digits[d])  // Test if "d" is outside the range or if it has been checked in the "digits" array
            return false;
        else
            digits[d] = true;                 // Mark the digit as existing
    }
    return true;
}

for (var i = 0; i <

for (var i = 0; i < arr.length; i++) {
  console.log(checkDigits(min, max, arr[i]), i, arr[i])
}

Binary mask approach:

This replaces the Array with an integer that is in effect used as an array of bits. It should be faster.

function checkDigits(min, max, n) {
    var digits = 0;                   
    while (n) {
        d = (n % 10);                         
        n = n / 10 >>0;
        if (d < min || d > max || (digits & (1 << d)))
            return false;
        else
            digits |= 1 << d;
    }
    return true;
}

function checkDigits(min, max, n) {
    var digits = 0;                   
    while (n) {
        d = (n % 10);                         
        n = n / 10 >>0;
        if (d < min || d > max || (digits & (1 << d)))
            return false;
        else
			digits |= 1 << d;
    }
    return true;
}

Explanation for binary mask approach:

1 << d creates a bit mask, an integer with the d bit set and all other bits set to 0.
digits |= 1 << d sets the bit marked by our bit mask on the integer digits.
digits & (1 << d) compares the bit marked by our bit mask with digits, the collection of previously marked bits.
See the docs on bitwise operators if you want to understand this in detail.

So, if we were to check 626, our numbers would go like this:

________n_____626_______________
           |
        d  |  6
     mask  |  0001000000
   digits  |  0000000000
           |
________n_____62________________
           |
        d  |  2
     mask  |  0000000100
   digits  |  0001000000
           |
________n_____6_________________
           |
        d  |  6
     mask  |  0001000000
   digits  |  0001000100
                 ^
               bit was already set, return false
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source: stackoverflow.com
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