how to calculate price using slider [algorithmic problem]

Solution 1:

Create a table with the thresholds and then in JS iterate over it to accumulate the price.

Here is a runnable snippet:

const table = [
    [10000, 10.00],
    [ 7200,  5.40],
    [ 4800,  3.75],
    [ 2400,  2.50],
    [ 1200,  2.20],
    [    0,  2.10],
];

function convert(points) {
    let total = 0;
    for (let [limit, price] of table) {
        if (points > limit) {
             total += Math.floor((points - limit) / 100) * price;
             points = limit;
        }
    }
    return total;
}

// IO handling
let [rngStart, rngEnd] = document.querySelectorAll("input");
let output = document.querySelector("span");
rngStart.addEventListener("input", refresh);
rngEnd.addEventListener("input", refresh);


function refresh() {
    let start = +rngStart.value;
    let end = +rngEnd.value
    output.textContent = (convert(end) - convert(start)).toFixed(2);
}
refresh();

input[type=range] { width: 80% }

From: <input type="number" min="0" max="20000" step="100" value="3000">
To: <input type="number" min="0" max="20000" step="100" value="12000">

Price: <span></span>

Solution 2:

Here's an algorithm that you can convert to code:

1. Let's have a nested array, or an array of price objects (it's important to be indexable and sorted ascending), for prices called prices sorted ascending as follows:

[[1200, 2400, 2.2],...]

2. Let's have the lower and upper bounds of user input in variables lower and upper.
3. Let's have a variable finalPrice initialized to 0.
4. Find the index of the first element in prices where its upper bound is > lower. Let's call this index i.
5. Now run while lower < upper:

// calc the cost of this range
finalPrice +=  (min(prices[i][1], upper) - lower)/100 * prices[i]
// sets `lower` to the upper bound of the previous price,
// since we already calculated that.
lower = min(prices[i][1], upper) 
// increment i to calc with the next price range
i += 1

Since the slider is constraint to 20000 on the upper side, you can be sure that i never goes out of bounds.