You are given an integer array nums and an integer k. You may partition nums into one or more subsequences such that each element in nums appears in exactly one of the subsequences. Return the minimum number of subsequences needed such that the diff

partitionArray.js

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var partitionArray = function(nums, k) {
    if (nums.length === 0) return 0;

    // Step 1: Sort the array so that we can group close numbers together
    nums.sort((a, b) => a - b);

    let count = 1; // At least one subsequence to start
    let start = nums[0]; // Start of the current subsequence

    // Step 2: Iterate through the sorted array
    for (let i = 1; i < nums.length; i++) {
        // If the current number is too far from the start of the subsequence,
        // we start a new group
        if (nums[i] - start > k) {
            count++;
            start = nums[i]; // Reset the start to the new group's first number
        }
    }

    return count;
};