# Minimum difference between maximum and minimum value of Array with given Operations

Given an array **arr[]** and an integer **K**. The following operations can be performed on any array element:

- Multiply the array element with
**K**. - If the element is divisible by
**K**, then divide it by**K**.

The above two operations can be applied any number of times including zero on any array element. The task is to find the minimum difference possible between the maximum and minimum value of the array.

**Examples:**

Input:arr[] = {1, 5000, 9999}, K = 10000

Output:5000

Explanation:

The minimum possible difference between maximum element and minimum element is 5000. When element 1 is multiplied by K, the maximum element of the array becomes 10000 and minimum element is 5000. And, this is the least possible value.

Input:arr[] = {1, 2, 3, 4, 5, 10, 7}, K = 5

Output:5

Explanation:

In the first step, the elements 5 and 10 can be divided with 5 making it 1 and 2 respectively.

In the second step, 1 can be multiplied by 5. This makes 2 the minimum element and 7 the maximum element in the array. Therefore, the difference is 5.

**Approach:** The idea is to use Priority Queue as a Multiset. The following steps can be followed to compute the answer:

- Initially, all the possible values (i.e.) the value obtained when the array element is multiplied by K, divided by K are inserted in the multiset along with the indices.
- Continuously pop-out values from the multiset. At each instance, the popped-out value
**X**with index**i**is the maximum and if at least one value has been popped out for all indices, the current answer will be**X – min of (max of previously popped-out values for an index)**for all indices except i. - Update the answer if the current difference is less then what is calculated till now.
- This is continued till there are elements left in the multiset.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to calculate Minimum ` `// difference between maximum and ` `// minimum value of the array ` `int` `calculateMinDiff(` `int` `arr[], ` `int` `k , ` `int` `n) ` `{ ` ` ` ` ` `// Variable to store minimum difference ` ` ` `int` `ans = INT_MAX; ` ` ` ` ` `// PriorityQueue which is used as a multiset ` ` ` `// to store all possible values ` ` ` `priority_queue< pair<` `int` `,` `int` `> > pq; ` ` ` ` ` `// Iterate through all the values and ` ` ` `// add it to the priority queue ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` ` ` `// If the number is divisible by k ` ` ` `// divide it by K and add to priority queue ` ` ` `if` `(arr[i] % k == 0) ` ` ` `pq.push(make_pair( arr[i] / k, i )); ` ` ` ` ` `// Adding number as it is ` ` ` `pq.push(make_pair( arr[i], i )); ` ` ` ` ` `// Adding number after multiplying it by k ` ` ` `pq.push(make_pair(arr[i] * k, i )); ` ` ` `} ` ` ` ` ` `// HashMap to keep track of current values ` ` ` ` ` `map<` `int` `,` `int` `>mp; ` ` ` ` ` `while` `(!pq.empty()) ` ` ` `{ ` ` ` `pair<` `int` `,` `int` `>temp = pq.top(); ` ` ` `pq.pop(); ` ` ` `mp.insert(temp); ` ` ` ` ` `// if for every index there is at-least ` ` ` `// 1 value we calculate the answer ` ` ` `if` `(mp.size() == n) ` ` ` `{ ` ` ` `int` `min_value = INT_MAX; ` ` ` ` ` `for` `(` `auto` `x:mp) ` ` ` `min_value=min(min_value, x.second); ` ` ` ` ` `ans = min(ans, temp.first - min_value); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Returning the calculated answer ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `// Input Array ` ` ` `int` `arr[7] = { 1, 2, 3, 4, 5, 10, 7 }; ` ` ` `int` `K = 5; ` ` ` ` ` `// Size of the array ` ` ` `int` `N = ` `sizeof` `(arr)/` `sizeof` `(` `int` `); ` ` ` ` ` `// Printing final output ` ` ` `cout << (calculateMinDiff(arr, K, N)); ` `} ` ` ` `// This code is contributed by ishayadav2918 ` |

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## Java

`// Java implementation of the above approach ` `import` `java.io.*; ` `import` `java.util.*; ` ` ` `class` `GfG { ` ` ` ` ` `// Function to calculate Minimum ` ` ` `// difference between maximum and ` ` ` `// minimum value of the array ` ` ` `private` `static` `int` `calculateMinDiff(` `int` `arr[], ` `int` `k) ` ` ` `{ ` ` ` `// Length of the array ` ` ` `int` `n = arr.length; ` ` ` ` ` `// Variable to store minimum difference ` ` ` `int` `ans = Integer.MAX_VALUE; ` ` ` ` ` `// PriorityQueue which is used as a multiset ` ` ` `// to store all possible values ` ` ` `PriorityQueue<` `int` `[]> pq ` ` ` `= ` `new` `PriorityQueue<>((` `int` `x[], ` `int` `y[]) -> x[` `0` `] - y[` `0` `]); ` ` ` ` ` `// Iterate through all the values and ` ` ` `// add it to the priority queue ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` ` ` `// If the number is divisible by k ` ` ` `// divide it by K and add to priority queue ` ` ` `if` `(arr[i] % k == ` `0` `) ` ` ` `pq.add(` `new` `int` `[] { arr[i] / k, i }); ` ` ` ` ` `// Adding number as it is ` ` ` `pq.add(` `new` `int` `[] { arr[i], i }); ` ` ` ` ` `// Adding number after multiplying it by k ` ` ` `pq.add(` `new` `int` `[] { arr[i] * k, i }); ` ` ` `} ` ` ` ` ` `// HashMap to keep track of current values ` ` ` `HashMap<Integer, Integer> map = ` `new` `HashMap<>(); ` ` ` ` ` `while` `(!pq.isEmpty()) { ` ` ` `int` `temp[] = pq.poll(); ` ` ` `map.put(temp[` `1` `], temp[` `0` `]); ` ` ` ` ` `// if for every index there is at-least ` ` ` `// 1 value we calculate the answer ` ` ` `if` `(map.size() == n) { ` ` ` `ans = Math.min(ans, ` ` ` `temp[` `0` `] ` ` ` `- Collections.min(map.values())); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Returning the calculated answer ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `// Input Array ` ` ` `int` `arr[] = { ` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `10` `, ` `7` `}; ` ` ` `int` `K = ` `5` `; ` ` ` ` ` `// Printing final output ` ` ` `System.out.println(calculateMinDiff(arr, K)); ` ` ` `} ` `} ` |

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**Output:**

5

**Time Complexity:** O(NlogN)

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