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Counting possible APs within an array in JavaScript
Arithmetic Progression
Arithmetic Progression (AP) is a sequence of numbers such that the difference of any two consecutive numbers is a constant value (aka common difference).
For instance, 1, 2, 3, 4, 5, 6,… is an AP, which has a common difference equal to 1 (2 -1).
Problem
We are required to write a JavaScript function that takes in an array of integers, arr, as the first and the only argument.
The task of our function is to return the number of arithmetic progressions of size 3 that are possible from that list. In each progression, the differences between the elements must be the same. We are guaranteed that the input array will be sorted in increasing order. For example, if the input to the function is
For example, if the input to the function is −
Input
const arr = [1, 2, 3, 5, 7, 9];
Output
const output = 5;
Output Explanation
Because the desired APs are −
[1, 2, 3], [1, 3, 5], [1, 5, 9], [3, 5, 7] and [5, 7, 9]
Example
Following is the code −
const arr = [1, 2, 3, 5, 7, 9];
const countAP = (arr = []) => {
let i, j, k;
let { length: len } = arr;
let count = 0;
for (i = 0; i < len - 2; i++){
for (k = i + 2; k < len; k++){
let temp = arr[i] + arr[k];
let div = temp / 2;
if ((div * 2) == temp){
for (j = i + 1; j < k; j++){
if (arr[j] == div){
count += 1;
}
}
}
}
}
return count;
};
console.log(countAP(arr));Output
5
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