3

first of all i am sorry if this an easy question, but iam having a difficulties to select only unique array in my program (i am just learning javasript btw). i know i can easily do this with built in function like filter or map, but i am forbidden to do that. This is what i"ve got so far : 

function countUnique (numbers) { 
var sorted_numbers = numbers.sort(function(a,b){return a-b})
var result = []
for (var i = 0 ; i<numbers.length; i++){
if (numbers[i]!==numbers[i+1]){
  result.push(numbers[i])
  console.log(result)
}
}
var sum = 0
for (var j = 0; j<result.length; j++) {
sum = sum + result[j]
}
 return sum
}

console.log(countUnique([ 5, 5, 6, 6, 3, 1, 2, 7, 7])) // 6
console.log(countUnique([ 3, 6, 3, 6, 1, 1, 2, 1 ]))  // 2
console.log(countUnique([ 3, 3, 3, 3, 4, 5, 8, 10, 11 ])) // 38

Example: input: [ 5, 5, 6, 6, 3, 1, 2, 7, 7 ] proses: 3 + 1 + 2 result: 6

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3
  • can you use Set operator? Commented Jul 2, 2018 at 2:42
  • well, there is no mention of Set operator, so i guess yeah Commented Jul 2, 2018 at 2:45
  • Set constructor can not filter unique values, it would just not allow to have non-unique values which is not a matter of this problem. Commented Jul 2, 2018 at 2:56

4 Answers 4

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5

Use indexOf and lastIndexOf. If the result of the indexOf is equal to the result of the lastIndexOf it means that the element is unique in the array.

let arr = [ 3, 3, 3, 3, 4, 5, 8, 10, 11 ];
let sum = 0;
arr.forEach(a => {
  sum += (arr.indexOf(a) === arr.lastIndexOf(a)) ? a : 0;
});
console.log(sum);

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1 Comment

sorry, i forget tto tell that the constraint is also indexOf
2

setup a inner loop like this

for (var i = 0 ; i<numbers.length; i++){
  for(var j = 0; j<numbers.length; j++) {
    if (numbers[i] + 1 === numbers[j]){
      result.push(numbers[i])
      console.log(result)
    }
  }
}

or use to do less loops since its sorted

  function countUnique(numbers) { 
    var sorted_numbers = numbers.sort(function(a,b){return a-b})
    var result = []
    
    /*
    for (var i = 0 ; i < numbers.length; i++){
      for(var j = 0; j < sorted_numbers.length; j++) {
        if (numbers[i] + 1 === sorted_numbers[j]){
          result.push(numbers[i]);
        }
      }
    }
    */
      
      var working = false;
      var current_count = 0;
      var counted_values = [];
       
      //create a new array for counting which numbers were used.
      for(i = 0; i < sorted_numbers.length; i++) {
        counted_values.push(0);
      }
      
      var current_value = 0;
      var put = false;
      for(var i = 0; i < sorted_numbers.length; i++) {
        current_value = sorted_numbers[i];
        put = false;
        for(var j = 0; j < sorted_numbers.length; j++) {
          if (current_value + 1 === sorted_numbers[j] && counted_values[j] == 0 && counted_values[i] == 0) {
            if(!put) {
              result.push(current_value);
              put = true; 
            }
            result.push(sorted_numbers[j]);
            counted_values[j] = 1;
            current_value = sorted_numbers[j];
          }

        }
      }

console.log(result);

    var sum = 0;
    for (var j = 0; j < result.length; j++) {
      sum = sum + result[j];
    }
    return sum;
  }

  console.log(countUnique([ 5, 5, 6, 6, 3, 1, 2, 7, 7])); // 6
  console.log(countUnique([ 3, 6, 3, 6, 1, 1, 2, 1 ]));  // 2
  console.log(countUnique([ 3, 3, 3, 3, 4, 5, 8, 10, 11 ])); // 38

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5 Comments

can you elaborate more, my unserstanding of nested loop is a little bit low
the nested loop would scan from the beginning always so it won't miss a number like your loop misses since it wont be the next number it might be like 3 numbers in or more in, you have to loop it or otherwise you would need to hardcode every number in a array to check.
numbers[i]!==numbers[i+1] only checks i and i+1, but it might be i and i+5.. so you loop all the i's on the other side to check every number.
nevermind my solution is probably wrong sorry, try it out, it anyways.
I fixed it a little still don't get your numbers but thats how i understood the problem.
2

You can use objects to determine the unique number. It might be a bit complicated but it can be really useful if your arrays are lengthy.

    function countUnique(numbers){
      var counter={};
      var sum=0;
      //Record all numbers accordingly
      for(var n in numbers){
        //The current number in the array
        var num=numbers[n];
        /*
          This basically checks if the counter object contains the property
          (Which is any number in the array). If it doesn't contain it then it 
          will add it as a property with a property value of 1 
          (meaning there is currently one unique number in the array). If it does 
          contain it then it will simply add 1 to the property value indicating there is 
          more than one of the same number
        */
        (counter[num]==null)?counter[num]=1:counter[num]++;
      }
      /*
        After we are done with going through the array, we will go 
        through the object properties and only add the ones with a value of 1 
        (indicating they are unique) and add 0 if the property's value is more 
        than one
      */
      for(var c in counter){
        sum+=(counter[c]==1)?c:0;
      }
      return sum;
    }

    console.log(countUnique([ 5, 5, 6, 6, 3, 1, 2, 7, 7])); // 6
    console.log(countUnique([ 3, 6, 3, 6, 1, 1, 2, 1 ]));  // 2
    console.log(countUnique([ 3, 3, 3, 3, 4, 5, 8, 10, 11 ])); // 38

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3 Comments

this looks like the best one
thank you for the insight, but i am barely touch objects :)
Oh, I apologize I forgot to take in account that you might have restrictions or that you might not understand. I have added in comments to take you through the whole process if you don't understand.
1 

// Here you have the exact same arrays you used in your question
    
    const numbers1 = [ 5, 5, 6, 6, 3, 1, 2, 7, 7];  // 6
    const numbers2 = [ 3, 6, 3, 6, 1, 1, 2, 1 ];  // 2
    const numbers3 = [ 3, 3, 3, 3, 4, 5, 8, 10, 11 ]; // 38

// Here we have a function that finds non-repeating numbers
// We don't only remove duplicates if a number appears more than once, we don't include it at all

    function notRepeatedNum(arr) {
     let result = [];
     let ctr = 0; 
      for (let x = 0; x < arr.length; x++) {
       ctr = 0;
      for (let y = 0; y < arr.length; y++) {
      if (arr[x] === arr[y]) {   // If we have a match increment the counter by 1
        ctr+= 1;
        }
       }

 // If the counter is >= 2 then the number is repeating, otherwise push it to the array

    if (ctr < 2) {
      result.push(arr[x]);
      }
    }
    return result;
    }

// Here we use the function to find those non-repeating and store it in a variable 
    
    const nonReapiting1 = notRepeatedNum(numbers1);
    const nonReapiting2 = notRepeatedNum(numbers2);
    const nonReapiting3 = notRepeatedNum(numbers3);
    
// Here we are summing the non-repeating numbers into a single value
    
    function sumIt(numbers) {
      let sum = 0;
      for (let b = 0; b < numbers.length; b++) {
        sum += numbers[b];
      }
      return sum;
    }

// And finally we log it to the console :)

console.log(sumIt(nonReapiting1));  // 6
console.log(sumIt(nonReapiting2));  // 2
console.log(sumIt(nonReapiting3));  // 38

Note: This could have been done a lot faster, cleaner, and in a much more efficient way, with array helpers, but I'm pretty sure this is what you asked for. No Babel/ ES6, not even objects ware used :)

Run the code snippet and see if this workes for you

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