Tips for improving the efficiency of the find_average_number(array) simple function?

Question

Tips for improving the efficiency of the find_average_number(array) simple function?

I successfully completed a CodeWars challenge where I wrote the function find_average to calculate the average of numbers in an array. However, I now have some questions:

1) How can I optimize this code in terms of Big-O notation? Is there a way to reduce loop iterations using array.every for input value testing (excluding TypeError when dealing with non-number elements in the array)?

2) I believe that the complexity is currently Θ(n), but am I correct in my assessment?

function find_average(array) {
  // your code here
  if (Array.isArray(array) && array.every(elem => typeof elem == 'number')) {
    const numberCount = array.length;
    if (numberCount == 1) {
      return array[0];
    }
    let sum = array.reduce((accumulator, currentValue) => {
      return accumulator + currentValue;
    })
    const averageNumber = sum/numberCount;
    return averageNumber;
  }
  throw new TypeError('Array values are not all numbers.');
}

javascript arrays algorithm validation big-o

Answer 1

Answer №1

Instead of iterating twice, once to check if every element is a number and once to add them up, you can optimize by iterating just once. If any element is not a number, you can throw an error immediately:

function calculate_average(array) {
  if (!Array.isArray(array)) {
    throw new TypeError('Values in the array are not numbers.');
  }
  const sum = array.reduce((a, b) => {
    if (typeof b !== 'number') {
      throw new TypeError('Values in the array are not numbers.');
    }
    return a + b;
  }, 0);
  return sum / array.length;
}

function calculate_average(array) {
  if (!Array.isArray(array)) {
    throw new TypeError('Values in the array are not numbers.');
  }
  const sum = array.reduce((a, b) => {
    if (typeof b !== 'number') {
      throw new TypeError('Values in the array are not numbers.');
    }
    return a + b;
  }, 0);
  return sum / array.length;
}

console.log(calculate_average([1, 2, 3]));
console.log(calculate_average([1, 'foo', 3]));

Both this optimized solution and your original implementation have a time complexity of O(n).

Answer 2

Instead of iterating twice, once to check if every element is a number and once to add them up, you can optimize by iterating just once. If any element is not a number, you can throw an error immediately:

function calculate_average(array) {
  if (!Array.isArray(array)) {
    throw new TypeError('Values in the array are not numbers.');
  }
  const sum = array.reduce((a, b) => {
    if (typeof b !== 'number') {
      throw new TypeError('Values in the array are not numbers.');
    }
    return a + b;
  }, 0);
  return sum / array.length;
}

function calculate_average(array) {
  if (!Array.isArray(array)) {
    throw new TypeError('Values in the array are not numbers.');
  }
  const sum = array.reduce((a, b) => {
    if (typeof b !== 'number') {
      throw new TypeError('Values in the array are not numbers.');
    }
    return a + b;
  }, 0);
  return sum / array.length;
}

console.log(calculate_average([1, 2, 3]));
console.log(calculate_average([1, 'foo', 3]));

Both this optimized solution and your original implementation have a time complexity of O(n).

Answer 3

Answer №2

To enhance efficiency, you can implement a single loop to first check if the input is an array.

Not only does this method operate in O(n) time complexity, it also outperforms traditional array iteration techniques.

function find_average(array) {
    // your code here
    var sum = 0,
        count = 0,
        i;
  
    if (!Array.isArray(array) || !array.length) return; // or throw error

    for (i = 0; i < array.length; i++) {
        if (typeof array[i] !== 'number') continue;
        sum += array[i];
        count++;
    }
    if (count === 0) return; // or throw error
    return sum / count;
}

console.log(find_average([null, 3, 5, '10', undefined]));
console.log(find_average());
console.log(find_average([]));

Answer 4

To enhance efficiency, you can implement a single loop to first check if the input is an array.

Not only does this method operate in O(n) time complexity, it also outperforms traditional array iteration techniques.

function find_average(array) {
    // your code here
    var sum = 0,
        count = 0,
        i;
  
    if (!Array.isArray(array) || !array.length) return; // or throw error

    for (i = 0; i < array.length; i++) {
        if (typeof array[i] !== 'number') continue;
        sum += array[i];
        count++;
    }
    if (count === 0) return; // or throw error
    return sum / count;
}

console.log(find_average([null, 3, 5, '10', undefined]));
console.log(find_average());
console.log(find_average([]));

Tips for improving the efficiency of the find_average_number(array) simple function?

Answer №1

Answer №2

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