Size of smallest subarray to be removed to make count of array elements greater and smaller than K equal

Last Updated : 17 Jan, 2022

Given an integer K and an array arr[] consisting of N integers, the task is to find the length of the subarray of smallest possible length to be removed such that the count of array elements smaller than and greater than K in the remaining array are equal.

Examples:

Input: arr[] = {5, 7, 2, 8, 7, 4, 5, 9}, K = 5
Output: 2
Explanation:
Smallest subarray required to be removed is {8, 7}, to make the largest resultant array {5, 7, 2, 4, 5, 9} satisfy the given condition.

Input: arr[] = {12, 16, 12, 13, 10}, K = 13
Output: 3
Explanation:
smallest subarray required to be removed is {12, 13, 10} to make the largest resultant array {12, 16} satisfy the given condition.

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays, and traverse the remaining array, to keep count of array elements that are strictly greater than and smaller than integer K. Then, select the smallest subarray whose deletion gives an array having equal number of smaller and greater elements.

Time Complexity: O(N²)
Auxiliary Space: O(N²)

Efficient Approach: The idea is to use Hashing with some modification to the array to solve it in O(N) time. The given array can have 3 types of elements:

element = K: change element to 0 (because we need elements, that are strictly greater than K or smaller than K)
element > K: change element to 1
element < K: change element to -1

Now, calculate the sum of all array elements and store it in a variable, say total_sum. Now, the total_sum can have three possible ranges of values:

If total_sum = 0: all 1s are cancelled by -1s. So, equal number of greater and smaller elements than K are already present. No deletion operation is required. Hence, print 0 as the required answer.
If total_sum > 0: some 1s are left uncancelled by -1s. i.e. array has more number of greater elements than K and less number of smaller elements than K. So, find the smallest subarray of sum = total_sum as it is the smallest subarray to be deleted.
If total_sum < 0: some -1s are left uncancelled by 1s. i.e. array has more number of smaller elements than k and less number of greater elements than K. So, find the smallest subarray of sum = total_sum as it is the smallest subarray to be deleted.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function ot find the length
// of the smallest subarray
int smallSubarray(int arr[], int n,
                  int total_sum)
{
    // Stores (prefix Sum, index)
    // as (key, value) mappings
    unordered_map<int, int> m;
    int length = INT_MAX;
    int prefixSum = 0;
 
    // Iterate till N
    for (int i = 0; i < n; i++) {
 
        // Update the prefixSum
        prefixSum += arr[i];
 
        // Update the length
        if (prefixSum == total_sum) {
            length = min(length, i + 1);
        }
 
        // Put the latest index to
        // find the minimum length
        m[prefixSum] = i;
 
        if (m.count(prefixSum - total_sum)) {
 
            // Update the length
            length
                = min(length,
                      i - m[prefixSum - total_sum]);
        }
    }
 
    // Return the answer
    return length;
}
 
// Function to find the length of
// the largest subarray
int smallestSubarrayremoved(int arr[], int n,
                            int k)
{
 
    // Stores the sum of all array
    // elements after modification
    int total_sum = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Change greater than k to 1
        if (arr[i] > k) {
            arr[i] = 1;
        }
 
        // Change smaller than k to -1
        else if (arr[i] < k) {
            arr[i] = -1;
        }
 
        // Change equal to k to 0
        else {
            arr[i] = 0;
        }
 
        // Update total_sum
        total_sum += arr[i];
    }
 
    // No deletion required, return 0
    if (total_sum == 0) {
        return 0;
    }
 
    else {
 
        // Delete smallest subarray
        // that has sum = total_sum
        return smallSubarray(arr, n,
                             total_sum);
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 12, 16, 12, 13, 10 };
    int K = 13;
 
    int n = sizeof(arr) / sizeof(int);
 
    cout << smallestSubarrayremoved(
        arr, n, K);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function ot find the length
// of the smallest subarray
static int smallSubarray(int arr[], int n,
                         int total_sum)
{
     
    // Stores (prefix Sum, index)
    // as (key, value) mappings
    Map<Integer,
        Integer> m = new HashMap<Integer,
                                 Integer>();
                                  
    int length = Integer.MAX_VALUE;
    int prefixSum = 0;
     
    // Iterate till N
    for(int i = 0; i < n; i++)
    {
         
        // Update the prefixSum
        prefixSum += arr[i];
         
        // Update the length
        if (prefixSum == total_sum)
        {
            length = Math.min(length, i + 1);
        }
         
        // Put the latest index to
        // find the minimum length
        m.put(prefixSum, i);
         
        if (m.containsKey(prefixSum - total_sum))
        {
             
            // Update the length
            length = Math.min(length,
                              i - m.get(prefixSum - 
                                        total_sum));
        }
    }
     
    // Return the answer
    return length;
}
 
// Function to find the length of
// the largest subarray
static int smallestSubarrayremoved(int arr[], int n,
                                   int k)
{
     
    // Stores the sum of all array
    // elements after modification
    int total_sum = 0;
     
    for(int i = 0; i < n; i++)
    {
         
        // Change greater than k to 1
        if (arr[i] > k)
        {
            arr[i] = 1;
        }
         
        // Change smaller than k to -1
        else if (arr[i] < k)
        {
            arr[i] = -1;
        }
         
        // Change equal to k to 0
        else
        {
            arr[i] = 0;
        }
         
        // Update total_sum
        total_sum += arr[i];
    }
     
    // No deletion required, return 0
    if (total_sum == 0)
    {
        return 0;
    }
    else
    {
         
        // Delete smallest subarray
        // that has sum = total_sum
        return smallSubarray(arr, n, total_sum);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 12, 16, 12, 13, 10 };
    int K = 13;
    int n = arr.length;
     
    System.out.println(
        smallestSubarrayremoved(arr, n, K));
}
}
 
// This code is contributed by chitranayal

Python3

# Python3 program to implement
# the above approach
import sys
 
# Function ot find the length
# of the smallest subarray
def smallSubarray(arr, n, total_sum):
 
    # Stores (prefix Sum, index)
    # as (key, value) mappings
    m = {}
    length = sys.maxsize
    prefixSum = 0
 
    # Iterate till N
    for i in range(n):
 
        # Update the prefixSum
        prefixSum += arr[i]
 
        # Update the length
        if(prefixSum == total_sum):
            length = min(length, i + 1)
 
        # Put the latest index to
        # find the minimum length
        m[prefixSum] = i
 
        if((prefixSum - total_sum) in m.keys()):
 
            # Update the length
            length = min(length,
                         i - m[prefixSum - total_sum])
 
    # Return the answer
    return length
 
# Function to find the length of
# the largest subarray
def smallestSubarrayremoved(arr, n, k):
 
    # Stores the sum of all array
    # elements after modification
    total_sum = 0
 
    for i in range(n):
 
        # Change greater than k to 1
        if(arr[i] > k):
            arr[i] = 1
 
        # Change smaller than k to -1
        elif(arr[i] < k):
            arr[i] = -1
 
        # Change equal to k to 0
        else:
            arr[i] = 0
 
        # Update total_sum
        total_sum += arr[i]
 
    # No deletion required, return 0
    if(total_sum == 0):
        return 0
    else:
         
        # Delete smallest subarray
        # that has sum = total_sum
        return smallSubarray(arr, n,
                             total_sum)
                              
# Driver Code
arr = [ 12, 16, 12, 13, 10 ]
K = 13
 
n = len(arr)
 
# Function call
print(smallestSubarrayremoved(arr, n, K))
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function ot find the length
// of the smallest subarray
static int smallSubarray(int []arr, int n,
                         int total_sum)
{
     
    // Stores (prefix Sum, index)
    // as (key, value) mappings
    Dictionary<int,
               int> m = new Dictionary<int,
                                       int>();
                                  
    int length = int.MaxValue;
    int prefixSum = 0;
     
    // Iterate till N
    for(int i = 0; i < n; i++)
    {
         
        // Update the prefixSum
        prefixSum += arr[i];
         
        // Update the length
        if (prefixSum == total_sum)
        {
            length = Math.Min(length, i + 1);
        }
         
        // Put the latest index to
        // find the minimum length
        if (m.ContainsKey(prefixSum))
            m[prefixSum] = i;
        else
            m.Add(prefixSum, i);
         
        if (m.ContainsKey(prefixSum - total_sum))
        {
             
            // Update the length
            length = Math.Min(length,
                              i - m[prefixSum - 
                                    total_sum]);
        }
    }
     
    // Return the answer
    return length;
}
 
// Function to find the length of
// the largest subarray
static int smallestSubarrayremoved(int []arr, 
                                   int n, int k)
{
     
    // Stores the sum of all array
    // elements after modification
    int total_sum = 0;
     
    for(int i = 0; i < n; i++)
    {
         
        // Change greater than k to 1
        if (arr[i] > k)
        {
            arr[i] = 1;
        }
         
        // Change smaller than k to -1
        else if (arr[i] < k)
        {
            arr[i] = -1;
        }
         
        // Change equal to k to 0
        else
        {
            arr[i] = 0;
        }
         
        // Update total_sum
        total_sum += arr[i];
    }
     
    // No deletion required, return 0
    if (total_sum == 0)
    {
        return 0;
    }
    else
    {
         
        // Delete smallest subarray
        // that has sum = total_sum
        return smallSubarray(arr, n, total_sum);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 12, 16, 12, 13, 10 };
    int K = 13;
    int n = arr.Length;
     
    Console.WriteLine(
            smallestSubarrayremoved(arr, n, K));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript program to implement
// the above approach
 
// Function ot find the length
// of the smallest subarray
function smallSubarray(arr,n,total_sum)
{
    // Stores (prefix Sum, index)
    // as (key, value) mappings
    let m = new Map();
                                   
    let length = Number.MAX_VALUE;
    let prefixSum = 0;
      
    // Iterate till N
    for(let i = 0; i < n; i++)
    {
          
        // Update the prefixSum
        prefixSum += arr[i];
          
        // Update the length
        if (prefixSum == total_sum)
        {
            length = Math.min(length, i + 1);
        }
          
        // Put the latest index to
        // find the minimum length
        m.set(prefixSum, i);
          
        if (m.has(prefixSum - total_sum))
        {
              
            // Update the length
            length = Math.min(length,
                              i - m.get(prefixSum -
                                        total_sum));
        }
    }
      
    // Return the answer
    return length;
}
 
// Function to find the length of
// the largest subarray
function smallestSubarrayremoved(arr,n,k)
{
    // Stores the sum of all array
    // elements after modification
    let total_sum = 0;
      
    for(let i = 0; i < n; i++)
    {
          
        // Change greater than k to 1
        if (arr[i] > k)
        {
            arr[i] = 1;
        }
          
        // Change smaller than k to -1
        else if (arr[i] < k)
        {
            arr[i] = -1;
        }
          
        // Change equal to k to 0
        else
        {
            arr[i] = 0;
        }
          
        // Update total_sum
        total_sum += arr[i];
    }
      
    // No deletion required, return 0
    if (total_sum == 0)
    {
        return 0;
    }
    else
    {
          
        // Delete smallest subarray
        // that has sum = total_sum
        return smallSubarray(arr, n, total_sum);
    }
}
 
// Driver Code
let arr=[12, 16, 12, 13, 10];
let K = 13;
let n = arr.length;
document.write(smallestSubarrayremoved(arr, n, K));
 
 
// This code is contributed by avanitrachhadiya2155
</script>

Output:

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

Minimum length of the subarray required to be replaced to make frequency of array elements equal to N / M

jigyansu

Improve

Article Tags :

Practice Tags :

Size of smallest subarray to be removed to make count of array elements greater and smaller than K equal

C++

Java

Python3

C#

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?