JavaScript: Get the index of the function in an array of functions which executed the fastest

JavaScript fundamental (ES6 Syntax): Exercise-175 with Solution

Index of Fastest Function

Write a JavaScript program to get the index of the function in an array of functions which executed the fastest.

• Use Array.prototype.map() to generate an array where each value is the total time taken to execute the function after iterations times.
• Use the difference in performance.now() values before and after to get the total time in milliseconds to a high degree of accuracy.
• Use Math.min() to find the minimum execution time, and return the index of that shortest time which corresponds to the index of the most performant function.
• Omit the second argument, iterations, to use a default of 10000 iterations.
• The more iterations, the more reliable the result but the longer it will take.

Sample Solution:

JavaScript Code:

// Define a function 'mostPerformant' that takes an array of functions 'fns' and an optional parameter 'iterations' with a default value of 10000
const mostPerformant = (fns, iterations = 10000) => {
  // Map each function in 'fns' to its execution time in milliseconds
  const times = fns.map(fn => {
    // Record the current timestamp in milliseconds before executing the function
    const before = performance.now();
    // Execute the function 'iterations' number of times
    for (let i = 0; i < iterations; i++) fn();
    // Calculate the elapsed time by subtracting the start time from the current time
    return performance.now() - before;
  });
  // Find the index of the minimum time in the 'times' array and return it
  return times.indexOf(Math.min(...times));
};

// Log the index of the most performant function among the provided functions
console.log(mostPerformant([
  // First function: Loops through the entire array before returning `false`
  () => {
    [1, 2, 3, 4, 5, 6, 7, 8, 9, '10'].every(el => typeof el === 'number');
  },
  // Second function: Only needs to reach index `1` before returning false
  () => {
    [1, '2', 3, 4, 5, 6, 7, 8, 9, 10].every(el => typeof el === 'number');
  }
])); // 1

Output:

1

Flowchart:

Live Demo:

See the Pen javascript-basic-exercise-174-1 by w3resource (@w3resource) on CodePen.

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