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 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 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
Similar Reads
Length of smallest subarray required to be removed to make remaining elements consecutive
Given an array arr[] consisting of N integers, the task is to find the length of the smallest subarray required to be removed to make the remaining array elements consecutive. Examples: Input: arr[] = {1, 2, 3, 7, 5, 4, 5}Output: 2Explanation:Removing the subarray {7, 5} from the array arr[] modifie
15+ min read
Smallest subarray having an element with frequency greater than that of other elements
Given an array arr of positive integers, the task is to find the smallest length subarray of length more than 1 having an element occurring more times than any other element. Examples: Input: arr[] = {2, 3, 2, 4, 5} Output: 2 3 2 Explanation: The subarray {2, 3, 2} has an element 2 which occurs more
15+ min read
Find the index of the smallest element to be removed to make sum of array divisible by K
Given an array arr[] of size N and a positive integer K, the task is to find the index of the smallest array element required to be removed to make the sum of remaining array divisible by K. If multiple solutions exist, then print the smallest index. Otherwise, print -1. Examples: Input: arr[ ] = {6
8 min read
Length of smallest subarray to be removed to make sum of remaining elements divisible by K
Given an array arr[] of integers and an integer K, the task is to find the length of the smallest subarray that needs to be removed such that the sum of remaining array elements is divisible by K. Removal of the entire array is not allowed. If it is impossible, then print "-1". Examples: Input: arr[
11 min read
Smallest subarray from a given Array with sum greater than or equal to K | Set 2
Given an array A[] consisting of N positive integers and an integer K, the task is to find the length of the smallest subarray with a sum greater than or equal to K. If no such subarray exists, print -1. Examples: Input: arr[] = {3, 1, 7, 1, 2}, K = 11Output: 3Explanation:The smallest subarray with
15+ min read
Minimum length of the subarray required to be replaced to make frequency of array elements equal to N / M
Given an array arr[] of size N consisting of only the first M natural numbers, the task is to find the minimum length of the subarray that is required to be replaced such that the frequency of array elements is N / M. Note: N is a multiple of M. Examples: Input: M = 3, arr[] = {1, 1, 1, 1, 2, 3}Outp
10 min read
Smallest subarray such that all elements are greater than K
Given an array of N integers and a number K, the task is to find the length of the smallest subarray in which all the elements are greater than K. If there is no such subarray possible, then print -1. Examples: Input: a[] = {3, 4, 5, 6, 7, 2, 10, 11}, K = 5 Output: 1 The subarray is {10} Input: a[]
4 min read
Number of subarrays with at-least K size and elements not greater than M
Given an array of N elements, K is the minimum size of the sub-array, and M is the limit of every element in the sub-array, the task is to count the number of sub-arrays with a minimum size of K and all the elements are lesser than or equal to M. Examples: Input: arr[] = {-5, 0, -10}, K = 1, M = 15O
7 min read
Count non-overlapping Subarrays of size K with equal alternate elements
Given an array arr[] of length N, the task is to find the count of non-overlapping subarrays of size K such that the alternate elements are equal. Examples: Input: arr[] = {2, 4, 2, 7}, K = 3Output: 1Explanation: Given subarray {2, 4, 2} is a valid array because the elements in even position(index n
7 min read
Length of smallest subarray to be removed such that the remaining array is sorted
Given an array arr[] consisting of N integers, the task is to print the length of the smallest subarray to be removed from arr[] such that the remaining array is sorted. Examples: Input: arr[] = {1, 2, 3, 10, 4, 2, 3, 5}Output: 3Explanation:The smallest subarray to be remove is {10, 4, 2} of length
8 min read