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

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:

DPi, 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)
?>

Javascript

<script>

    // JavaScript program to find minimum removals
    // to make max-min <= K

    let MAX = 100;
    let dp = new Array(MAX);
    for(let i = 0; i < MAX; i++)
    {
        dp[i] = new Array(MAX);
    }
    // function to check all possible combinations
    // of removal and return the minimum one
    function countRemovals(a,i,j,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
    function removals(a,n,k)
    {
        // sort the array
        a.sort(function(a, b){return a - b});

        // fill all stated with -1
        // when only one element
        for(let i = 0; i < MAX; i++)
        {
            for(let j = 0; j < MAX; j++)
            {
                dp[i][j] = -1;
            }
        }
        if (n == 1)
            return 0;
        else
            return countRemovals(a, 0, n - 1, k);
    }

    // Driver Code

    let a = [ 1, 3, 4, 9, 10, 11, 12, 17, 20 ];
    let n = a.length;
    let k = 4;
    document.write(removals(a, n, k));

</script>

Output

5

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 find 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) overall 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 satisfies 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 satisfies
        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 satisfies 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 satisfies
        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 satisfies 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
          # satisfies
          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

C#

// C# program for the above approach
using System;

class GFG{

// Function to find rightmost index
// which satisfies 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 satisfies
        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
    Array.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
static void Main()
{
    int[] a = { 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 sanjoy_62

Javascript

<script>
    // javascript program for the above approach

    // Function to find rightmost index
    // which satisfies the condition
    // arr[j] - arr[i] <= k
    function findInd(key , i , n , k , arr) {
        var 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 satisfies
            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
    function removals(arr , n , k) {
        var i, j, ans = n - 1;

        // Sort the given array
        arr.sort((a,b)=>a-b);

        // 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

    var a = [ 1, 3, 4, 9, 10, 11, 12, 17, 20 ];
    var n = a.length;
    var k = 4;

    document.write(removals(a, n, k));

// This code contributed by Rajput-Ji
</script>

Output

5

Time Complexity :O(nlogn) 
Auxiliary Space: O(n)

Approach:

1. 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 j) and find the minimum element on its left (index i) such that arr[j] – arr[i] <= k and store it in dp[j].
2. 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) overall 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));
    }
}

Python3

# Python3 program for the above approach

# To sort the array and return the answer
def removals(arr, n, k):

  # sort the array
  arr.sort()
  dp = [0 for i in range(n)]

  # Fill all stated with -1
  # when only one element
  for i in range(n):
    dp[i] = -1

  # As dp[0] = 0 (base case) so min
  # no of elements to be removed are
  # n-1 elements
  ans = n - 1
  dp[0] = 0

  for i in range(1, n):
    dp[i] = i
    j = dp[i - 1]

    while (j != i and arr[i] - arr[j] > k):
      j += 1

    dp[i] = min(dp[i], j)
    ans = min(ans, (n - (i - j + 1)))

  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 rohan07

C#

// C# program for the above approach
using System;

class GFG{

// To sort the array and return the answer
static int removals(int[] arr, int n, int k)
{

    // Sort the array
    Array.Sort(arr);
    int[] dp = new int[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;

    // 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] = Math.Min(dp[i], j);
        ans = Math.Min(ans, (n - (i - j + 1)));
    }
    return ans;
}

// Driver code
static public void Main()
{
    int[] a = { 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 lokeshpotta20

Javascript

<script>

// JavaScript program for the above approach

// To sort the array and return the answer
function removals(arr, n, k)
{

  // sort the array
  arr.sort((a,b)=>a-b);
  var dp = Array(n);

  // fill all stated with -1
  // when only one element
  for(var 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
  var ans = n - 1;
  dp[0] = 0;
  for (var i = 1; i < n; i++)
  {
    dp[i] = i;
    var j = dp[i - 1];
    while (j != i && arr[i] - arr[j] > k)
    {
      j++;
    }
    dp[i] = Math.min(dp[i], j);
    ans = Math.min(ans, (n - (i - j + 1)));
  }
  return ans;
}

// Driver code
var a = [1, 3, 4, 9, 10, 11, 12, 17, 20];
var n = a.length;
var k = 4;
document.write( removals(a, n, k));

</script>

Output

5

Time Complexity: O(nlog n). As outer loop is going to make n iterations. 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)

Space Optimised Approach

The solution could be further optimized. It can be done in Auxiliary Space: O(1). The idea is to first sort the array in ascending order. The keep 2 pointers, say i and j, where j iterates from index 1 to index n-1 and keeps track of ending subarray with the difference in max and min less than k, and i keeps the track of starting index of the subarray. If we find that a[j] – a[i[ <=k (means the i to j subarray is valid) we update the length from i to j in a variable to track of max length so far. Else, we update the starting index i with i+1. 

At first it seems that subarrays from i to j are ignored, but obviously their lengths are less than i to j subarray, so we don’t care about them.

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

int removal(int a[], int n, int k)
{
    // Sort the Array; Time complexity:O(nlogn)
    sort(a, a + n);

    // to store the max length of
    // array with difference <= k
    int maxLen = INT_MIN;
    int i = 0;
    // J goes from 0 to n-1
    // Thus time complexity of below loop is O(n)
    for (int j = 0; j < n && i < n; j++) {
        // if the subarray from i to j index is valid
        // the store it's length
        if (a[j] - a[i] <= k) {
            maxLen = max(maxLen, j - i + 1);
        }
        // if subarray difference exceeds k
        // change starting position, i.e. i
        else {
            i++;
        }
    }
    // no. of elements need to be removed is
    // Length of array - max subarray with diff <= k
    int removed = n - maxLen;
    return removed;
}

//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 << removal(a, n, k);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG {

    public static int removal(int[] a, int n, int k){
        //Sort the Array; Time complexity:O(nlogn)
        Arrays.sort(a);

        // to store the max length of
        // array with difference <= k
        int max = Integer.MIN_VALUE;
        int i=0;
        // J goes from 0 to n-1.
        // Thus time complexity of below loop is O(n)
        for(int j=0;j<n && i<n;j++){
            // if the subarray from i to j index is valid
            // the store it's length
            if(a[j]-a[i] <= k){
                max = Math.max(max, j-i+1);
            }
            // if subarray difference exceeds k
            // change starting position, i.e. i
            else{
                i++;
            }
        }
        // no. of elements need to be removed is
        // Length of array - max subarray with diff <= k
        int removed = n-max;
        return removed;
    }

  //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(removal(a, n, k));
    }
}

Python

# Python program for the above approach
def removal(a, n, k):
    # sort the array
    a.sort()
    # to store the max length of
    # array with difference <= k
    maxLen = 0
    # pointer to keep track of starting of each subarray
    i = 0
    for j in range(0, n):
        # if the subarray from i to j index is valid
        # the store it's length
        if a[j]-a[i] <= k:
            maxLen = max(maxLen, j-i+1)
        # if subarray difference exceeds k
        # change starting position, i.e. i
        else:
            i = i+1

        if i >= n:
            break
    remove = n-maxLen
    return remove


# Driver Code
a = [1, 3, 4, 9, 10, 11, 12, 17, 20]
n = len(a)
k = 4

print(removal(a, n, k))

C#

// C# program for the above approach
using System;

class GFG {

    static int removal(int[] a, int n, int k)
    {

        // Sort the Array; Time complexity:O(nlogn)
        Array.Sort(a);

        // to store the max length of
        // array with difference <= k
        int max = Int32.MinValue;
        int i = 0;

        // J goes from 1 to n-1 in n-2 iterations
        // Thus time complexity of below loop is O(n)
        for (int j = 0; j < n && i < n; j++)
        {

            // if the subarray from i to j index is valid
            // the store it's length
            if (a[j] - a[i] <= k) {
                max = Math.Max(max, j - i + 1);
            }

            // if subarray difference exceeds k
            // change starting position, i.e. i
            else {
                i++;
            }
        }

        // no. of elements need to be removed is
        // Length of array - max subarray with diff <= k
        int removed = n - max;
        return removed;
    }

    // Driver Code
    public static void Main()
    {
        int[] a = { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
        int n = a.Length;
        int k = 4;
        Console.Write(removal(a, n, k));
    }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>

// JavaScript program for the above approach
function removal(a, n, k)
{

    // sort the array
    a.sort((a,b)=>a-b)

    // to store the max length of
    // array with difference <= k
    let maxLen = 0;

    // pointer to keep track of starting of each subarray
    let i = 0;
    for(let j = 0; j < n; j++)
    {

        // if the subarray from i to j index is valid
        // the store it's length
        if(a[j]-a[i] <= k){
            maxLen = Math.max(maxLen, j-i+1);
        }

        // if subarray difference exceeds k
        // change starting position, i.e. i
        else i++;
        if(i >= n)break;
    }

    let remove = n-maxLen;
    return remove;
}

// Driver Code
let a = [1, 3, 4, 9, 10, 11, 12, 17, 20];
let n = a.length;
let k = 4;

document.write(removal(a, n, k),"</br>")

// This code is contributed by shinjanpatra

</script>

Output

5

Time Complexity: O(nlogn) For sorting the array, we require O(nlogn) time, and no extra space. And for calculating we only traverse the loop once, thus it has O(n) time complexity. So, overall time complexity is O(nlogn).
Auxiliary Space: O(1)

Approach: 

Here we are applying sliding window, our will maintain a[max]-a[min]<=k. First, ascending order should be used to sort the array. The maintain two pointers, let’s call them I and j, j iterating from index 1 to index n-1 and keeping track of the ending subarray with the difference in max and min smaller than k, and I keeping track of the initial index of the subarray. We update the length from I to j in a variable to track the maximum length thus far if we discover that a[j] – a[i]=k (meaning the I to j subarray is legitimate). If not, we add i+1 to the beginning index i.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

int removal(int a[], int n, int k)
{
    // Code here
    // Sort the Array; Time complexity:O(nlogn)
    sort(a, a + n);
    int diff= 0; // to store difference of max(a) and min(a)
    int ans = 0; // to store maximum length of array with length <=k

    // use sliding window to iterate array, maintaining two
    // pointers for desired subarray
    // which starting from index j and ending with index i
    for (int i = 0, j = 0; i < n; i++) {
        int diff = a[i] - a[j];
        // as array is already sorted, if difference exceeds
        // k we will move starting pointer forward
        while (i >= j && diff > k) {
            diff = a[i] - a[++j];
        }
        ans = max(ans, i - j + 1);
    }
    return n - 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 << removal(a, n, k);
    return 0;
}

// This code is contributed by Gaurav Garg

Java

// JAVA program for the above approach
import java.util.*;
class GFG {
  public static int removal(int[] a, int n, int k)
  {

    // Code here
    // Sort the Array; Time complexity:O(nlogn)
    Arrays.sort(a);
    int diff
      = 0; // to store difference of max(a) and min(a)
    int ans = 0; // to store maximum length of array
    // with length <=k

    // use sliding window to iterate array, maintaining
    // two pointers for desired subarray which starting
    // from index j and ending with index i
    for (int i = 0, j = 0; i < n; i++)
    {
      diff = a[i] - a[j];

      // as array is already sorted, if difference
      // exceeds k we will move starting pointer
      // forward
      while (i >= j && diff > k) {
        diff = a[i] - a[++j];
      }
      ans = Math.max(ans, i - j + 1);
    }
    return n - ans;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int a[]
      = new int[] { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
    int n = a.length;
    int k = 4;
    System.out.print(removal(a, n, k));
  }
}

// This code is contributed by Taranpreet

Python3

# Python program for the above approach
def removal(a, n, k):

  # Sort the Array
  a.sort()

  diff = 0  # to store difference of max(a) and min(a)
  ans = 0  # to store maximum length of array with length <= k

  j = 0

  # use sliding window to iterate array, maintaining
  # two pointers for desired subarray which starting
  # from index j and ending with index i
  for i in range(n):
    diff = a[i] - a[j]

    # as array is already sorted, if difference
    # exceeds k we will move starting pointer
    # forward
    while(i >= j and diff > k):
      j += 1
      diff = a[i] - a[j]

      ans = max(ans, i-j+1)

      return n-ans

    a = [1, 3, 4, 9, 10, 11, 12, 17, 20]
    n = len(a)
    k = 4

    print(removal(a, n, k))

    # This code is contributed by lokeshmvs21.

C#

// C# program for the above approach

using System;
using System.Collections;

public class GFG {

    static int removal(int[] a, int n, int k)
    {
        // Sort the Array
        Array.Sort(a);
        int diff
            = 0; // to store difference of max(a) and min(a)
        int ans = 0; // to store maximum length of array
                     // with length <=k

        // use sliding window to iterate array, maintaining
        // two pointers for desired subarray which starting
        // from index j and ending with index i
        for (int i = 0, j = 0; i < n; i++) {
            diff = a[i] - a[j];

            // as array is already sorted, if difference
            // exceeds k we will move starting pointer
            // forward
            while (i >= j && diff > k) {
                diff = a[i] - a[++j];
            }
            ans = Math.Max(ans, i - j + 1);
        }
        return n - ans;
    }

    static public void Main()
    {

        // Code
        int[] a
            = new int[] { 1, 3, 4, 9, 10, 11, 12, 17, 20 };
        int n = a.Length;
        int k = 4;

        Console.Write(removal(a, n, k));
    }
}

// This code is contributed by lokesh (lokeshmvs21).

Javascript

function removal(a, n, k)
  {

    // Code here
    // Sort the Array; Time complexity:O(nlogn)
    a.sort();
    let diff = 0; // to store difference of max(a) and min(a)
    let ans = 0; // to store maximum length of array
    // with length <=k

    // use sliding window to iterate array, maintaining
    // two pointers for desired subarray which starting
    // from index j and ending with index i
    for (let i = 0, j = 0; i < n; i++)
    {
      diff = a[i] - a[j];

      // as array is already sorted, if difference
      // exceeds k we will move starting pointer
      // forward
      while (i >= j && diff > k) {
        diff = a[i] - a[++j];
      }
      ans = Math.max(ans, i - j);
    }
    return n - ans;
  }

  // Driver Code
    let a= [ 1, 3, 4, 9, 10, 11, 12, 17, 20 ];
    let n = a.length;
    let k = 4;
    document.write(removal(a, n, k));

  // This code is contributed by aadityaburujwale.
</script>

Output

5

Time complexity: O(nlogn)
Auxiliary Space: O(1)