Chocolates With Magical Number

1. Using this magical number, Ninja can either increase or decrease the sweetness of each chocolate. 2. After increasing or decreasing the sweetness, all the new sweetness values must be non-negative. 3. Ninja must use this magic number on each chocolate exactly once.

For ‘PACKETS’ = [1, 2, 3, 4, 5] and 'K' = 1, the absolute difference between two chocolates with maximum (5) and minimum (1) sweetness is 4. Now in order to minimize this value, Ninja increases [1, 2, 3] and decreases [4, 5] by 1 (‘K’ = 1). So, ‘PACKET’ becomes [2,3,4,3,4]. Now, the absolute difference between the two chocolates with maximum (4) and minimum (2) sweetness is 2 which is the minimum possible.

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow. The first line of each test case contains two single space-separated integers 'N' and 'K' denoting the number of chocolates in the 'PACKET' and the magic number, respectively. The second line of each test case contains 'N' single space-separated integers, denoting the sweetness of each chocolate in 'PACKET'.

For each test case, print the minimum possible difference between the maximum and minimum sweetness value of chocolates in the array/list ‘PACKET’. Print the output of each test case in a separate line.

1 <= T <= 100 1 <= N <= 10^5 1 <= K <= 10^5 1 <= PACKET[i] <= 10^5 Where 'T' is the number of test cases, 'N' denotes the number of chocolates in the given array/list 'PACKET' and 'K' denotes the given magic number, respectively. 'PACKET[i]' denotes the sweetness of the i'th chocolate. Time Limit : 1 sec

For the first test case: Given ‘K’ = 1 so, [6, 3], [6, 1], [4, 1] and [4, 3] are all the possible ways in which sweetness of chocolates can be increased or decreased among which [4,3] has the least difference of 1 between the minimum and maximum sweetness value. For the second test case : Given ‘K’ = 2 so [5] and [1] are the only possible ways in which the sweetness of chocolates can be increased or decreased and the minimum difference between the minimum and maximum sweetness value obtained in both ways is 0. For the third test case: Given ‘K’ = 3 and all the elements of ‘PACKET’ are the same. Among all possible ways of increasing or decreasing sweetness values of chocolates by ‘K’, [8, 8, 8, 8] and [2, 2, 2, 2] are the only possible ways which yield a minimum difference of 0 in the maximum and minimum sweetness values.

For the first test case: Given ‘K’ = 10 so we can only increase the sweetness of all chocolates in ‘PACKET’ by ‘K’ because the sweetness of chocolates must be positive. We get [11,12] after modifying the ‘PACKET’. So, the least difference in the maximum and minimum sweetness values is 1. For the second test case: Given ‘K’ = 8 so we can only increase the sweetness of all chocolates in ‘PACKET’ by ‘K’ because the sweetness of chocolates must be positive. We get [9,10,11,12] after modifying the ‘PACKET’. So, the least difference in the maximum and minimum sweetness values is 3. For the third test case : Among all possible ways of increasing or decreasing the sweetness of chocolates in the given ‘PACKET’ by ‘K’, [4,2,4] and [4,6,4] are the optimal chocolate sweetness obtained which have the least difference of 2 between the maximum and the minimum sweetness values among all possibilities.

Sorting

The main observation while solving this problem is that after every update to ‘PACKET’, only the difference between the largest and smallest sweetness level of chocolate matters to us.

The main idea behind this approach is to sort ‘PACKETS’ in increasing order. Then for all the chocolates in the ‘PACKET’, check if subtracting the sweetness level of ‘i’th chocolate by ‘K’ (i.e. PACKET[i]’ - ’K’) or increasing its value by ‘K’ (i.e. ‘PACKET[i]’ + ’K’) makes any changes in our result or not.

Here is the complete algorithm :

Sort the ‘PACKET’ in increasing order and initialize three variables ‘BIG’ = ‘PACKET[N - 1]’ i.e. the largest value in ‘PACKET’, ‘SMALL’ = ‘PACKET[0]’ i.e. the smallest value in ‘PACKET’ and ‘MIN_DIFF’ to store the difference between the largest and the smallest sweetness value.
Now iterate the ‘PACKET’ and for each chocolate, do the following:
- If subtracting and adding ‘K’ to the chocolate does not change ‘SMALL’ and ‘BIG’ respectively then do nothing. Following are those two cases:
  - ‘PACKET[i] - ‘K’’ >= ‘SMALL’
  - ‘PACKET[i] + ‘K’’ <= ‘BIG’
- Else, either the subtraction (‘PACKET[i]’ - ‘K’) will yield the smaller number or the addition (‘PACKET[i]’ - ‘K’) will result in a greater number. Update our ‘SMALL’ or ‘BIG’ variables accordingly:
  - If ‘BIG’ - ‘PACKET[i]’ - ‘K’ results in a smaller difference then update ‘SMALL’.
  - Else update ‘BIG’.
Finally, return the minimum of ‘MIN_DIFF’ and ‘BIG’ - ‘SMALL’.

Time Complexity

O(NlogN), where ‘N’ is the number of chocolates in the ‘PACKET’.

We are sorting the ‘PACKET’ and then iterating it once. So the overall time complexity will be O(NlogN).

Space Complexity

O(1)

Because we are not using any extra space so the overall space complexity will be O(1).

Chocolates With Magical Number

Problem statement

Sample Input 1 :

Sample Output 1 :

Explanation of Sample Input 1 :

Sample Input 2 :

Sample Output 2 :

Explanation of Sample Input 2 :