LeetCode 2620 Counter Problem Solution

Given an integer n, return a counter function. This counter function initially returns n and then returns 1 more than the previous value every subsequent time it is called (n, n + 1, n + 2, etc).

Example 1:

Input: 
n = 10 
["call","call","call"]
Output: [10,11,12]
Explanation: 
counter() = 10 // The first time counter() is called, it returns n.
counter() = 11 // Returns 1 more than the previous time.
counter() = 12 // Returns 1 more than the previous time.

Example 2:

Input: 
n = -2
["call","call","call","call","call"]
Output: [-2,-1,0,1,2]
Explanation: counter() initially returns -2. Then increases after each sebsequent call.

Constraints:

• -1000 <= n <= 1000
• 0 <= calls.length <= 1000
• calls[i] === "call"

Here is the Solution for the above question :-

/**
 * @param {number} n
 * @return {Function} counter
 */
var createCounter = function(n) {
    return function() {
        return n++;
    };
};

let n = 10;
const counter = createCounter(n);

console.log(counter()); // Output: 10
console.log(counter()); // Output: 11
console.log(counter()); // Output: 12