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I want to write an algorithm to remove duplicates and sort an array with single loop in javascript. I want to do it with algorithms not declarative methods. I can sort it with single loop but can't remove duplicates. I will be glad if someone help me.

Number of elements in array is 10^6, each element is equal to or greater than 0, less than 300

This is my sorting algorithm:

var a = [2, 7, 5, 1, 3, 2, 7];

for (let i = 0; i < a.length - 1; i++) {
  if (a[i] > a[i + 1]) {
    let temp = a[i]
    a[i] = a[i + 1]
    a[i + 1] = temp
    i = -1
  }
}

console.log(a)

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  • Show the sort loop please, and others might help improve Commented Jan 5, 2022 at 12:16
  • 3
    const output = Array.from(new Set(input)).sort((a,b) => a - b) Commented Jan 5, 2022 at 12:20
  • @mplungjan I want to do it with algorithms not declarative methods. Commented Jan 5, 2022 at 12:26
  • 4
    Since the number of keys is limited to 300, you can create a simple array (hash) of length 301 -> iterate over original array and increment the count for each element -> iterate over the hash and if count is greater than 1, then insert the element into result array. It will amount to 2 loops, but the time complexity remains O(n) + O(300) i.e. O(n) Commented Jan 5, 2022 at 12:39
  • 2
    @lukas You can write any algorithm with a single loop by putting a state machine inside that loop. Doesn't make it good code though. Or just write it without a "loop" altogether but use recursion! The interview task just doesn't make sense. Ignore the requirements and discuss the possible solutions with the interviewer, that's what they really care about. Commented Jan 5, 2022 at 13:07

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It's honorable that you was curious enough to try to find a solution, even after the interview. A good sign for a programmer.

  1. I would use an object to keep track of any duplicates.
  2. Also loop from the end, mostly because it feels right, not messing up the index.
  3. Change position in a if current < previous
  4. Else add to the object.
  5. Check if object has previous index, or
  6. Check if current is the same as previous (ex. 8, 8 at the end of the array)
  7. then splice the previous index.
  8. Added numberOfLoops just because it's fun to know.

var a = [2, 2, -1, 8, 5, 1, 3, 3, 13, 2, 8, 8];
var temp = {};  // 1
var result = [];
var current, previous;
var numberOfLoops = 0;

for (let i = a.length - 1; i > 0 ; i--) {  // 2
  current = a[i];
  previous = a[i - 1];
  numberOfLoops++;  // 8

  if (temp.hasOwnProperty(previous) ||   // 5
      current == previous) {             // 6
    a.splice(i - 1, 1);                  // 7
    // i++;
  } else if (current < previous) {  // 3
    a[i - 1] = current;
    a[i] = previous;
    i += 2;
    delete temp[current];
  } else {                          // 4
    temp[current] = current;
  }
}

console.log({numberOfLoops});  // 23
console.log(a);  // [-1, 1, 2, 3, 5, 8, 13]

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