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: 
 

1. Multiply the array element with K.
2. 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: 
 

1. 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.
2. 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.
3. Update the answer if the current difference is less then what is calculated till now.
4. This is continued till there are elements left in the multiset.

Below is the implementation of the above approach: 
 

// 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

// 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));
    }
}

# Python3 implementation of the above approach  
import sys

# Function to calculate Minimum  
# difference between maximum and  
# minimum value of the array  
def calculateMinDiff(arr, k , n) :  
      
    # Variable to store minimum difference  
    ans = sys.maxsize
  
    # PriorityQueue which is used as a multiset  
    # to store all possible values  
    pq = [] 
          
    # Iterate through all the values and  
    # add it to the priority queue  
    for i in range(n) :
  
        # If the number is divisible by k  
        # divide it by K and add to priority queue  
        if (arr[i] % k == 0) :
            pq.append((arr[i] // k, i ))  
  
        # Adding number as it is  
        pq.append((arr[i], i))  
  
        # Adding number after multiplying it by k  
        pq.append((arr[i] * k, i))
        
    pq.sort()
    pq.reverse()
  
    # HashMap to keep track of current values  
  
    mp = {}  
  
    while (len(pq) > 0) :  
    
        temp = pq[0]  
        pq.pop(0)
        mp[temp[0]] = temp[1]  
  
        # if for every index there is at-least  
        # 1 value we calculate the answer  
        if (len(mp) == n) :
        
            min_value = sys.maxsize 
  
            for x in mp :
                min_value = min(min_value, mp[x])
  
            ans = min(ans, temp[0] - min_value - 1)
  
    # Returning the calculated answer  
      
    return ans
    
# Input Array  
arr = [ 1, 2, 3, 4, 5, 10, 7 ]
K = 5 

# Size of the array  
N = len(arr)

# Printing final output  
print(calculateMinDiff(arr, K, N)) 

# This code is contributed by divyesh072019.

// C# implementation of the above approach  
using System;
using System.Collections.Generic;
class GFG 
{

  // Function to calculate Minimum  
  // difference between maximum and  
  // minimum value of the array  
  static int calculateMinDiff(int[] arr, int k , int n)  
  {  

    // Variable to store minimum difference  
    int ans = Int32.MaxValue;  

    // PriorityQueue which is used as a multiset  
    // to store all possible values  
    List<Tuple<int,int>> pq = new List<Tuple<int,int>>(); 

    // 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 Tuple<int,int>( arr[i] / k, i ));  

      // Adding number as it is  
      pq.Add(new Tuple<int,int>( arr[i], i ));  

      // Adding number after multiplying it by k  
      pq.Add(new Tuple<int,int>(arr[i] * k, i ));  
    } 

    pq.Sort();
    pq.Reverse();

    // HashMap to keep track of current values  
    Dictionary<int, int> mp = new Dictionary<int, int>();

    while (pq.Count > 0)  
    {  
      Tuple<int,int> temp = pq[0];  
      pq.RemoveAt(0); 
      mp[temp.Item1] = temp.Item2;  

      // if for every index there is at-least  
      // 1 value we calculate the answer  
      if (mp.Count == n) 
      {  
        int min_value = Int32.MaxValue; 

        foreach(KeyValuePair<int, int> x in mp) 
        {
          min_value=Math.Min(min_value, x.Value); 
        }

        ans = Math.Min(ans, temp.Item1 - min_value - 1);  
      }  
    }  

    // Returning the calculated answer  

    return ans;  
  }

  // Driver code
  static void Main() 
  {

    // Input Array  
    int[] arr = { 1, 2, 3, 4, 5, 10, 7 };  
    int K = 5; 

    // Size of the array  
    int N = arr.Length; 

    // Printing final output
    Console.WriteLine(calculateMinDiff(arr, K, N));
  }
}

// This code is contributed by divyeshrabadiya07.

<script>

// JavaScript implementation of the above approach


// Function to calculate Minimum
// difference between maximum and
// minimum value of the array
function calculateMinDiff(arr, k, n) {

    // Variable to store minimum difference
    let ans = Number.MAX_SAFE_INTEGER;

    // PriorityQueue which is used as a multiset
    // to store all possible values
    let pq = new Array();

    // Iterate through all the values and
    // push it to the priority queue
    for (let i = 0; i < n; i++) {

        // If the number is divisible by k
        // divide it by K and push to priority queue
        if (arr[i] % k == 0)
            pq.push(new Array(Math.floor(arr[i] / k), i));

        // pushing number as it is
        pq.push(new Array(arr[i], i));

        // pushing number after multiplying it by k
        pq.push(new Array(arr[i] * k, i));
    }

    pq.sort((a, b) => a[0] - b[0]);
    pq.reverse();

    // HashMap to keep track of current values
    let mp = new Map();

    while (pq.length > 0) {
        let temp = pq[0];
        pq.shift();
        mp.set(temp[0], temp[1]);

        // if for every index there is at-least
        // 1 value we calculate the answer
        if (mp.size == n) {
            let min_value = Number.MAX_SAFE_INTEGER;

            for (let x of mp) {
                min_value = Math.min(min_value, x[1]);
            }

            ans = Math.min(ans, temp[1] - min_value);
        }
    }

    // Returning the calculated answer

    return ans;
}

// Driver code

// Input Array
let arr = [1, 2, 3, 4, 5, 10, 7];
let K = 5;

// Size of the array
let N = arr.length;

// Printing final output
document.write(calculateMinDiff(arr, K, N));

// This code is contributed by gfgking

</script>

5

Time Complexity: O(NlogN)
Auxiliary Space: O(N), where N is the size of the input array. This is because the priority queue and the map are the two data structures that are being used to store the elements of the input array, and the size of these data structures can be at most equal to the size of the input array.

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• Sorting
• Competitive Programming
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