Javascript Program for Least frequent element in an array
Given an array, find the least frequent element in it. If there are multiple elements that appear least number of times, print any one of them.
Examples :
Input : arr[] = {1, 3, 2, 1, 2, 2, 3, 1}
Output : 3
3 appears minimum number of times in given
array.
Input : arr[] = {10, 20, 30}
Output : 10 or 20 or 30
A simple solution is to run two loops. The outer loop picks all elements one by one. The inner loop finds frequency of the picked element and compares with the minimum so far. Time complexity of this solution is O(n2)
A better solution is to do sorting. We first sort the array, then linearly traverse the array.
// JavaScript program to find the least frequent element
// in an array.
function leastFrequent(arr, n) {
// Sort the array
arr.sort();
// find the min frequency using
// linear traversal
let min_count = n + 1, res = -1;
let curr_count = 1;
for (let i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
curr_count++;
else {
if (curr_count < min_count) {
min_count = curr_count;
res = arr[i - 1];
}
curr_count = 1;
}
}
// If last element is least frequent
if (curr_count < min_count) {
min_count = curr_count;
res = arr[n - 1];
}
return res;
}
// Driver code
let arr = [1, 3, 2, 1, 2, 2, 3, 1];
let n = arr.length;
console.log(leastFrequent(arr, n));
// JavaScript program to find the least frequent element// in an array.function leastFrequent(arr, n) { // Sort the array arr.sort(); // find the min frequency using // linear traversal let min_count = n + 1, res = -1; let curr_count = 1; for (let i = 1; i < n; i++) { if (arr[i] == arr[i - 1]) curr_count++; else { if (curr_count < min_count) { min_count = curr_count; res = arr[i - 1]; } curr_count = 1; } } // If last element is least frequent if (curr_count < min_count) { min_count = curr_count; res = arr[n - 1]; } return res;}// Driver codelet arr = [1, 3, 2, 1, 2, 2, 3, 1];let n = arr.length;console.log(leastFrequent(arr, n));Output
3
Complexity Analysis:
- Time Complexity : O(n Log n)
- Auxiliary Space : O(1)
An efficient solution is to use hashing. We create a hash table and store elements and their frequency counts as key value pairs. Finally we traverse the hash table and print the key with minimum value.
// JavaScript program to find the least frequent element
// in an array.
function leastFrequent(arr, n) {
// Insert all elements in hash.
let hash = new Map();
for (let i = 0; i < n; i++) {
if (hash.has(arr[i]))
hash.set(arr[i], hash.get(arr[i]) + 1)
else
hash.set(arr[i], 1);
}
// find the min frequency
let min_count = n + 1, res = -1;
hash.forEach((value, key) => {
if (min_count >= value) {
res = key;
min_count = value;
}
});
return res;
}
// driver program
let arr = [1, 3, 2, 1, 2, 2, 3, 1];
let n = arr.length;
console.log(leastFrequent(arr, n));
// JavaScript program to find the least frequent element// in an array.function leastFrequent(arr, n) { // Insert all elements in hash. let hash = new Map(); for (let i = 0; i < n; i++) { if (hash.has(arr[i])) hash.set(arr[i], hash.get(arr[i]) + 1) else hash.set(arr[i], 1); } // find the min frequency let min_count = n + 1, res = -1; hash.forEach((value, key) => { if (min_count >= value) { res = key; min_count = value; } }); return res;}// driver programlet arr = [1, 3, 2, 1, 2, 2, 3, 1];let n = arr.length;console.log(leastFrequent(arr, n));Output
3
Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(n)
Please refer complete article on Least frequent element in an array for more details!
