Minimum removals from array to make max - min <= K

Minimum removals from array to make max – min <= K

Difficulty Level : Easy
Last Updated : 04 Mar, 2021

Given N integers and K, find the minimum number of elements that should be removed such that A_max-A_min<=K. After the removal of elements, A_max and A_min is considered among the remaining elements.

Examples:

Input : a[] = {1, 3, 4, 9, 10, 11, 12, 17, 20} 
          k = 4 
Output : 5 
Explanation: Remove 1, 3, 4 from beginning 
and 17, 20 from the end.

Input : a[] = {1, 5, 6, 2, 8}  K=2
Output : 3
Explanation: There are multiple ways to 
remove elements in this case.
One among them is to remove 5, 6, 8.
The other is to remove 1, 2, 5

Approach: Sort the given elements. Using the greedy approach, the best way is to remove the minimum element or the maximum element and then check if A_max-A_min <= K. There are various combinations of removals that have to be considered. We write a recurrence relation to try every possible combination. There will be two possible ways of removal, either we remove the minimum or we remove the maximum. Let(i…j) be the index of elements left after removal of elements. Initially, we start with i=0 and j=n-1 and the number of elements removed is 0 at the beginning. We only remove an element if a[j]-a[i]>k, the two possible ways of removal are (i+1…j) or (i…j-1). The minimum of the two is considered.
Let DP_{i, j} be the number of elements that need to be removed so that after removal a[j]-a[i]<=k.

Recurrence relation for sorted array:

DP_{i, j = 1+ (min(countRemovals(i+1, j), countRemovals(i, j-1))}

Below is the implementation of the above idea:

C++

// CPP program to find minimum removals
// to make max-min <= K
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 100
int dp[MAX][MAX];
 
// function to check all possible combinations
// of removal and return the minimum one
int countRemovals(int a[], int i, int j, int k)
{
    // base case when all elements are removed
    if (i >= j)
        return 0;
 
    // if condition is satisfied, no more
    // removals are required
    else if ((a[j] - a[i]) <= k)
        return 0;
 
    // if the state has already been visited
    else if (dp[i][j] != -1)
        return dp[i][j];
 
    // when Amax-Amin>d
    else if ((a[j] - a[i]) > k) {
 
        // minimum is taken of the removal
        // of minimum element or removal
        // of the maximum element
        dp[i][j] = 1 + min(countRemovals(a, i + 1, j, k),
                           countRemovals(a, i, j - 1, k));
    }
    return dp[i][j];
}
 
// To sort the array and return the answer
int removals(int a[], int n, int k)
{
    // sort the array
    sort(a, a + n);
 
    // fill all stated with -1
    // when only one element
    memset(dp, -1, sizeof(dp));
    if (n == 1)
        return 0;
    else
        return countRemovals(a, 0, n - 1, k);
}
 
// Driver Code
int main()
{
    int a[] = { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 4;
    cout << removals(a, n, k);
    return 0;
}

Java

// Java program to find minimum removals
// to make max-min <= K
import java.util.Arrays;
 
class GFG
{
    static int MAX=100;
    static int dp[][]=new int[MAX][MAX];
     
    // function to check all possible combinations
    // of removal and return the minimum one
    static int countRemovals(int a[], int i, int j, int k)
    {
        // base case when all elements are removed
        if (i >= j)
            return 0;
     
        // if condition is satisfied, no more
        // removals are required
        else if ((a[j] - a[i]) <= k)
            return 0;
     
        // if the state has already been visited
        else if (dp[i][j] != -1)
            return dp[i][j];
     
        // when Amax-Amin>d
        else if ((a[j] - a[i]) > k) {
     
            // minimum is taken of the removal
            // of minimum element or removal
            // of the maximum element
            dp[i][j] = 1 + Math.min(countRemovals(a, i + 1, j, k),
                                    countRemovals(a, i, j - 1, k));
        }
        return dp[i][j];
    }
     
    // To sort the array and return the answer
    static int removals(int a[], int n, int k)
    {
        // sort the array
        Arrays.sort(a);
     
        // fill all stated with -1
        // when only one element
        for(int[] rows:dp)
        Arrays.fill(rows,-1);
        if (n == 1)
            return 0;
        else
            return countRemovals(a, 0, n - 1, k);
    }
     
    // Driver code
    public static void main (String[] args) 
    {
        int a[] = { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
        int n = a.length;
        int k = 4;
        System.out.print(removals(a, n, k));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python program to find 
# minimum removals to
# make max-min <= K
MAX = 100
dp = [[0 for i in range(MAX)] 
         for i in range(MAX)]
for i in range(0, MAX) :
    for j in range(0, MAX) :
        dp[i][j] = -1
 
# function to check all 
# possible combinations
# of removal and return
# the minimum one
def countRemovals(a, i, j, k) :
 
    global dp
     
    # base case when all 
    # elements are removed
    if (i >= j) :
        return 0
 
    # if condition is satisfied, 
    # no more removals are required
    elif ((a[j] - a[i]) <= k) :
        return 0
 
    # if the state has 
    # already been visited
    elif (dp[i][j] != -1) :
        return dp[i][j]
 
    # when Amax-Amin>d
    elif ((a[j] - a[i]) > k) :
 
        # minimum is taken of 
        # the removal of minimum
        # element or removal of 
        # the maximum element
        dp[i][j] = 1 + min(countRemovals(a, i + 1, 
                                         j, k),
                           countRemovals(a, i, 
                                         j - 1, k))
    return dp[i][j]
 
# To sort the array 
# and return the answer
def removals(a, n, k) :
 
    # sort the array
    a.sort()
 
    # fill all stated 
    # with -1 when only
    # one element
    if (n == 1) :
        return 0
    else :
        return countRemovals(a, 0, 
                             n - 1, k)
 
# Driver Code
a = [1, 3, 4, 9, 10, 
     11, 12, 17, 20]
n = len(a)
k = 4
print (removals(a, n, k))
 
# This code is contributed by 
# Manish Shaw(manishshaw1)

C#

// C# program to find minimum 
// removals to make max-min <= K
using System;
 
class GFG
{
    static int MAX = 100;
    static int [,]dp = new int[MAX, MAX];
     
    // function to check all 
    // possible combinations
    // of removal and return
    // the minimum one
    static int countRemovals(int []a, int i,
                             int j, int k)
    {
        // base case when all
        // elements are removed
        if (i >= j)
            return 0;
     
        // if condition is satisfied, 
        // no more removals are required
        else if ((a[j] - a[i]) <= k)
            return 0;
     
        // if the state has
        // already been visited
        else if (dp[i, j] != -1)
            return dp[i, j];
     
        // when Amax-Amin>d
        else if ((a[j] - a[i]) > k)
        {
     
            // minimum is taken of the 
            // removal of minimum element 
            // or removal of the maximum 
            // element
            dp[i, j] = 1 + Math.Min(countRemovals(a, i + 1, 
                                                  j, k),
                                    countRemovals(a, i, 
                                                  j - 1, k));
        }
        return dp[i, j];
    }
     
    // To sort the array and
    // return the answer
    static int removals(int []a, 
                        int n, int k)
    {
        // sort the array
        Array.Sort(a);
     
        // fill all stated with -1
        // when only one element
        for(int i = 0; i < MAX; i++) 
        {
            for(int j = 0; j < MAX; j++)
                dp[i, j] = -1;
        }
        if (n == 1)
            return 0;
        else
            return countRemovals(a, 0, 
                                 n - 1, k);
    }
     
    // Driver code
    static void Main() 
    {
        int []a = new int[]{ 1, 3, 4, 9, 10, 
                             11, 12, 17, 20 };
        int n = a.Length;
        int k = 4;
        Console.Write(removals(a, n, k));
    }
}
 
// This code is contributed by 
// ManishShaw(manishshaw1)

PHP

<?php
// PHP program to find 
// minimum removals to
// make max-min <= K
$MAX = 100;
$dp = array(array());
for($i = 0; $i < $MAX; $i++)
{
    for($j = 0; $j < $MAX; $j++)
        $dp[$i][$j] = -1;
}
 
// function to check all 
// possible combinations
// of removal and return
// the minimum one
function countRemovals($a, $i, 
                       $j, $k)
{
    global $dp;
     
    // base case when all 
    // elements are removed
    if ($i >= $j)
        return 0;
 
    // if condition is satisfied, 
    // no more removals are required
    else if (($a[$j] - $a[$i]) <= $k)
        return 0;
 
    // if the state has 
    // already been visited
    else if ($dp[$i][$j] != -1)
        return $dp[$i][$j];
 
    // when Amax-Amin>d
    else if (($a[$j] - $a[$i]) > $k) 
    {
 
        // minimum is taken of 
        // the removal of minimum
        // element or removal of 
        // the maximum element
        $dp[$i][$j] = 1 + min(countRemovals($a, $i + 1, 
                                                $j, $k),
                              countRemovals($a, $i, 
                                            $j - 1, $k));
    }
    return $dp[$i][$j];
}
 
// To sort the array 
// and return the answer
function removals($a, $n, $k)
{
    // sort the array
    sort($a);
 
    // fill all stated with -1
    // when only one element
    if ($n == 1)
        return 0;
    else
        return countRemovals($a, 0, 
                             $n - 1, $k);
}
 
// Driver Code
$a = array(1, 3, 4, 9, 10, 
           11, 12, 17, 20);
$n = count($a);
$k = 4;
echo (removals($a, $n, $k));
 
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

Output

Time Complexity :O(n²)
Auxiliary Space: O(n²)

The solution could be further optimized. The idea is to sort the array in increasing order and traverse through all the elements (let’s say index i) and finding the maximum element on its right (index j) such that arr[j] – arr[i] <= k. Thus the number of elements to be removed becomes n-(j-i+1). The minimum number of elements can be found by taking the minimum of n-(j-i-1) over all i. The value of index j can be found through a binary search between start = i+1 and end = n-1;

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find 
// rightmost index
// which satsfies the condition
// arr[j] - arr[i] <= k
int findInd(int key, int i, 
            int n, int k, int arr[])
{
    int start, end, mid, ind = -1;
     
    // Initialising start to i + 1
    start = i + 1;
     
    // Initialising end to n - 1
    end = n - 1;
     
    // Binary search implementation
    // to find the rightmost element
    // that satisfy the condition
    while (start < end)
    {
        mid = start + (end - start) / 2;
         
        // Check if the condition satsfies 
        if (arr[mid] - key <= k)
        {   
             
            // Store the position
            ind = mid;
             
            // Make start = mid + 1
            start = mid + 1;
        }
        else
        {
            // Make end = mid
            end = mid;
        }
    }
     
    // Return the rightmost position
    return ind;
}
 
// Function to calculate 
// minimum number of elements
// to be removed
int removals(int arr[], int n, int k)
{
    int i, j, ans = n - 1;
     
    // Sort the given array
    sort(arr, arr + n);
     
    // Iterate from 0 to n-1
    for (i = 0; i < n; i++)
    {
         
        // Find j such that 
        // arr[j] - arr[i] <= k
        j = findInd(arr[i], i, n, k, arr);
         
        // If there exist such j 
        // that satisfies the condition
        if (j != -1)
        {
            // Number of elements
            // to be removed for this
            // particular case is 
            // (j - i + 1)
            ans = min(ans, n - (j - i + 1));
        }
    }
     
    // Return answer
    return ans;
}
 
// Driver Code
int main()
{
    int a[] = {1, 3, 4, 9, 10, 
               11, 12, 17, 20};
    int n = sizeof(a) / sizeof(a[0]);
    int k = 4;
    cout << removals(a, n, k);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find rightmost index
// which satsfies the condition
// arr[j] - arr[i] <= k
static int findInd(int key, int i, 
                   int n, int k, int arr[])
{
    int start, end, mid, ind = -1;
     
    // Initialising start to i + 1
    start = i + 1;
     
    // Initialising end to n - 1
    end = n - 1;
     
    // Binary search implementation
    // to find the rightmost element
    // that satisfy the condition
    while (start < end)
    {
        mid = start + (end - start) / 2;
         
        // Check if the condition satsfies 
        if (arr[mid] - key <= k)
        { 
             
            // Store the position
            ind = mid;
             
            // Make start = mid + 1
            start = mid + 1;
        }
        else
        {
             
            // Make end = mid
            end = mid;
        }
    }
     
    // Return the rightmost position
    return ind;
}
 
// Function to calculate 
// minimum number of elements
// to be removed
static int removals(int arr[], int n, int k)
{
    int i, j, ans = n - 1;
     
    // Sort the given array
    Arrays.sort(arr);
     
    // Iterate from 0 to n-1
    for(i = 0; i < n; i++)
    {
         
        // Find j such that 
        // arr[j] - arr[i] <= k
        j = findInd(arr[i], i, n, k, arr);
         
        // If there exist such j 
        // that satisfies the condition
        if (j != -1)
        {
             
            // Number of elements
            // to be removed for this
            // particular case is 
            // (j - i + 1)
            ans = Math.min(ans, 
                           n - (j - i + 1));
        }
    }
     
    // Return answer
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int a[] = { 1, 3, 4, 9, 10, 
                11, 12, 17, 20 };
    int n = a.length;
    int k = 4;
     
    System.out.println(removals(a, n, k));
}
}
 
// This code is contributed by adityapande88

Python3

# Python program for the 
# above approach
 
# Function to find 
# rightmost index
# which satsfies the condition
# arr[j] - arr[i] <= k
def findInd(key, i, n, 
            k, arr):
   
     ind = -1
     
     # Initialising start
     # to i + 1
     start = i + 1
       
     # Initialising end 
     # to n - 1
     end = n - 1;
     
     # Binary search implementation
     # to find the rightmost element
     # that satisfy the condition
     while (start < end):
          mid = int(start +
                   (end - start) / 2)
         
          # Check if the condition 
          # satsfies
          if (arr[mid] - key <= k):
             
               # Store the position
               ind = mid
                 
               # Make start = mid + 1
               start = mid + 1
          else:
               # Make end = mid
               end = mid
                 
     # Return the rightmost position
     return ind
     
# Function to calculate 
# minimum number of elements
# to be removed
def removals(arr, n, k):
   
     ans = n - 1
     
     # Sort the given array
     arr.sort()
     
     # Iterate from 0 to n-1
     for i in range(0, n):
       
          # Find j such that
          # arr[j] - arr[i] <= k
          j = findInd(arr[i], i, 
                      n, k, arr)
           
          # If there exist such j
          # that satisfies the condition
          if (j != -1):
             
               # Number of elements
               # to be removed for this
               # particular case is
               # (j - i + 1)
               ans = min(ans, n -
                        (j - i + 1))
               
     # Return answer
     return ans
     
# Driver Code
a = [1, 3, 4, 9, 
     10,11, 12, 17, 20]
n = len(a)
k = 4
print(removals(a, n, k))
 
# This code is contributed by Stream_Cipher

Output

Time Complexity :O(nlogn)

Auxiliary Space: O(n)

Approach:

The solution could be further optimised. The idea is to sort the array in increasing order and traverse through all the elements (let’s say index j) and finding the minimum element on its left (index i) such that arr[j] – arr[i] <= k and store it in dp[j].
Thus the number of elements to be removed becomes n-(j-i+1). The minimum number of elements can be found by taking the minimum of n-(j-i-1) over all j. The value of index i can be found through its previous dp array element value.

Below is the implementation of the approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// To sort the array and return the answer
int removals(int arr[], int n, int k)
{
   
  // sort the array
  sort(arr, arr + n);
  int dp[n];
 
  // fill all stated with -1
  // when only one element
  for(int i = 0; i < n; i++)
    dp[i] = -1;
 
  // as dp[0] = 0 (base case) so min
  // no of elements to be removed are
  // n-1 elements
  int ans = n - 1; 
  dp[0] = 0;
  for (int i = 1; i < n; i++)
  {
    dp[i] = i;
    int j = dp[i - 1];
    while (j != i && arr[i] - arr[j] > k)
    {
      j++;
    }
    dp[i] = min(dp[i], j);
    ans = min(ans, (n - (i - j + 1)));
  }
  return ans;
}
 
// Driver code    
int main()
{
  int a[] = { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
  int n = sizeof(a) / sizeof(a[0]);
  int k = 4;
  cout<<removals(a, n, k);
  return 0;
}
 
// This code is contributed by Balu Nagar

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
   
    // To sort the array and return the answer
    static int removals(int arr[], int n, int k)
    {
        // sort the array
        Arrays.sort(arr);
 
        // fill all stated with -1
        // when only one element
        int dp[] = new int[n];
        Arrays.fill(dp, -1);
       
        // as dp[0] = 0 (base case) so min
        // no of elements to be removed are
        // n-1 elements
        int ans = n - 1; 
        dp[0] = 0;
       
        // Iterate from 1 to n - 1
        for (int i = 1; i < n; i++) {
            dp[i] = i;
            int j = dp[i - 1];
            while (j != i && arr[i] - arr[j] > k) {
                j++;
            }
            dp[i] = Integer.min(dp[i], j);
            ans = Integer.min(ans, (n - (i - j + 1)));
        }
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
        int n = a.length;
        int k = 4;
        System.out.print(removals(a, n, k));
    }
}

Output

Time Complexity: O(n). As outer loop is going to make n interations. And the inner loop iterates at max n times for all outer iterations. Because we start value of j from dp[i-1] and loops it until it reaches i and then for the next element we again start from the previous dp[i] value. So the total time complexity is O(n) if we don’t consider the complexity of the sorting as it is not considered in the above solution as well.

Auxiliary Space: O(n)

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