JavaScript: Find the number which appears most in a given array of integers
JavaScript Basic: Exercise-94 with Solution
Write a JavaScript program to find the number appearing most frequently in a given array of integers.
Visual Presentation:
Sample Solution:
JavaScript Code:
// Function to find the mode (most frequently occurring element) in an array
function array_element_mode(arr) {
var ctr = [], // Counter array to store frequencies of elements
ans = 0; // Variable to store the index of the mode
// Initialize the counter array with zeros for each possible element
for (var i = 0; i < 10; i++) {
ctr.push(0);
}
// Iterate through the input array to update the frequencies in the counter array
for (var i = 0; i < arr.length; i++) {
ctr[arr[i] - 1]++; // Increment the frequency of the current element
if (ctr[arr[i] - 1] > ctr[ans]) {
ans = arr[i] - 1; // Update the index of the mode if a higher frequency is found
}
}
return ans + 1; // Return the mode (add 1 to convert from zero-based index to element value)
}
// Example usage
console.log(array_element_mode([1, 2, 3, 2, 2, 8, 1, 9])); // 2
Output:
2
Live Demo:
See the Pen javascript-basic-exercise-94 by w3resource (@w3resource) on CodePen.
Flowchart:
ES6 Version:
// Function to find the mode (most frequent element) in an array
const array_element_mode = (arr) => {
let ctr = []; // Array to store the count of each element
let ans = 0; // Variable to store the index with the highest count
// Initialize the count array with zeros for each possible element (1 to 10)
for (let i = 0; i < 10; i++) {
ctr.push(0);
}
// Iterate through the input array to count occurrences of each element
for (let i = 0; i < arr.length; i++) {
ctr[arr[i] - 1]++; // Increment the count for the corresponding element
if (ctr[arr[i] - 1] > ctr[ans]) {
ans = arr[i] - 1; // Update the index with the highest count
}
}
return ans + 1; // Return the mode (element with the highest count)
};
// Example usage
console.log(array_element_mode([1, 2, 3, 2, 2, 8, 1, 9]));
Improve this sample solution and post your code through Disqus.
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