Given an array arr[] of N integers and an integer K, our task is to find the length of the longest subarray such that for all possible pairs in the subarray absolute difference between elements is less than or equal to K.
Examples:
Input: arr[] = {2, 4, 5, 5, 5, 3, 1}, K = 0
Output: 3
Explanation:
The possible subarray with difference in elements as 0 is {5, 5, 5} whose length is 3. Hence the output is 3.Input: arr[] = {1, 2, 3, 6, 7}, K = 2
Output: 3
Explanation:
The possible subarray with difference in elements at most 2 is {1, 2, 3} whose length is 3. Hence the output is 3.
Naive Approach:
To solve the problem mentioned above the naive method is to use The Brute Force approach that is to generate all the possible subarray of the given array and check if the difference between the maximum and minimum element of the subarray is at most K or not. If it is, then update the length of the current subarray with the maximum length. Print the maximum length of the subarray after all the operations.
Below is the implementation of the above approach:
// C++ implementation to find the Longest subarray // of the given array with absolute difference between // elements less than or equal to integer K #include <bits/stdc++.h> using namespace std;
int computeLongestSubarray( int arr[], int k, int n)
{ // maxLength is 1 because k >= 0,
// a single element, subarray will always
// have absolute difference zero
int maxLength = 1;
// Check for all possible subarrays
for ( int i = 0; i < n; i++)
{
// Initialization of minimum &
// maximum of current subarray
int minOfSub = arr[i];
int maxOfSub = arr[i];
for ( int j = i + 1; j < n; j++)
{
// Update the values for minimum & maximum
if (arr[j] > maxOfSub)
maxOfSub = arr[j];
if (arr[j] < minOfSub)
minOfSub = arr[j];
// Check if current subarray satisfies
// the given condition
if ((maxOfSub - minOfSub) <= k)
{
int currLength = j - i + 1;
// Update the value for maxLength
if (maxLength < currLength)
maxLength = currLength;
}
}
}
// Return the final result
return maxLength;
} // Driver Code int main()
{ int arr[] = { 1, 2, 3, 6, 7 };
int k = 2;
int n = sizeof (arr) / sizeof (arr[0]);
int maxLength = computeLongestSubarray(arr, k, n);
cout << (maxLength);
} // This code is contributed by chitranayal |
// Java implementation to find the Longest subarray // of the given array with absolute difference between // elements less than or equal to integer K class GFG {
public static int computeLongestSubarray( int arr[],
int k)
{
// maxLength is 1 because k >= 0,
// a single element, subarray will always
// have absolute difference zero
int maxLength = 1 ;
// Check for all possible subarrays
for ( int i = 0 ; i < arr.length; i++) {
// Initialization of minimum &
// maximum of current subarray
int minOfSub = arr[i];
int maxOfSub = arr[i];
for ( int j = i + 1 ; j < arr.length; j++) {
// Update the values for minimum & maximum
if (arr[j] > maxOfSub)
maxOfSub = arr[j];
if (arr[j] < minOfSub)
minOfSub = arr[j];
// Check if current subarray satisfies
// the given condition
if ((maxOfSub - minOfSub) <= k) {
int currLength = j - i + 1 ;
// Update the value for maxLength
if (maxLength < currLength)
maxLength = currLength;
}
}
}
// Return the final result
return maxLength;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 6 , 7 };
int k = 2 ;
int maxLength = computeLongestSubarray(arr, k);
System.out.println(maxLength);
}
} |
# Python3 implementation to find the # Longest subarray of the given array # with absolute difference between # elements less than or equal to integer K def computeLongestSubarray (arr, k, n):
# maxLength is 1 because k >= 0,
# a single element, subarray will always
# have absolute difference zero
maxLength = 1
# Check for all possible subarrays
for i in range (n):
# Initialization of minimum &
# maximum of current subarray
minOfSub = arr[i]
maxOfSub = arr[i]
for j in range (i + 1 , n):
# Update the values for
# minimum & maximum
if (arr[j] > maxOfSub):
maxOfSub = arr[j]
if (arr[j] < minOfSub):
minOfSub = arr[j]
# Check if current subarray
# satisfies the given condition
if ((maxOfSub - minOfSub) < = k):
currLength = j - i + 1
# Update the value for maxLength
if (maxLength < currLength):
maxLength = currLength
# Return the final result
return maxLength
# Driver Code if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 6 , 7 ]
k = 2
n = len (arr)
maxLength = computeLongestSubarray(arr, k, n)
print (maxLength)
# This code is contributed by himanshu77 |
// C# implementation to find the longest subarray // of the given array with absolute difference between // elements less than or equal to integer K using System;
class GFG
{ public static int computelongestSubarray( int []arr,
int k)
{
// maxLength is 1 because k >= 0,
// a single element, subarray will always
// have absolute difference zero
int maxLength = 1;
// Check for all possible subarrays
for ( int i = 0; i < arr.Length; i++)
{
// Initialization of minimum &
// maximum of current subarray
int minOfSub = arr[i];
int maxOfSub = arr[i];
for ( int j = i + 1; j < arr.Length; j++)
{
// Update the values for minimum & maximum
if (arr[j] > maxOfSub)
maxOfSub = arr[j];
if (arr[j] < minOfSub)
minOfSub = arr[j];
// Check if current subarray satisfies
// the given condition
if ((maxOfSub - minOfSub) <= k)
{
int currLength = j - i + 1;
// Update the value for maxLength
if (maxLength < currLength)
maxLength = currLength;
}
}
}
// Return the readonly result
return maxLength;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 6, 7 };
int k = 2;
int maxLength = computelongestSubarray(arr, k);
Console.WriteLine(maxLength);
}
} // This code is contributed by shikhasingrajput |
<script> // JavaScript implementation to find the Longest subarray // of the given array with absolute difference between // elements less than or equal to integer K function computeLongestSubarray(arr,k)
{ // maxLength is 1 because k >= 0,
// a single element, subarray will always
// have absolute difference zero
let maxLength = 1;
// Check for all possible subarrays
for (let i = 0; i < arr.length; i++) {
// Initialization of minimum &
// maximum of current subarray
let minOfSub = arr[i];
let maxOfSub = arr[i];
for (let j = i + 1; j < arr.length; j++) {
// Update the values for minimum & maximum
if (arr[j] > maxOfSub)
maxOfSub = arr[j];
if (arr[j] < minOfSub)
minOfSub = arr[j];
// Check if current subarray satisfies
// the given condition
if ((maxOfSub - minOfSub) <= k) {
let currLength = j - i + 1;
// Update the value for maxLength
if (maxLength < currLength)
maxLength = currLength;
}
}
}
// Return the final result
return maxLength;
} // Driver Code
let arr=[1, 2, 3, 6, 7]; let k = 2; let maxLength = computeLongestSubarray(arr, k); document.write(maxLength); // This code is contributed by avanitrachhadiya2155 </script> |
3
Time Complexity: O(n2)
Auxiliary Space: O(n)
Efficient Approach:
To optimize the above approach the idea is to use Heap Data Structure. Initialize a minHeap that will store the indices of the current subarray such that the elements are in ascending order, where the smallest appears at the top and a maxHeap that will store the indices of the current subarray such that the elements are in descending order, where the largest element appears at the top. Then iterate over the entire array and for each iteration check if:
- All the subarray elements satisfy the condition of maxOfSub-minOfSub <= k, then we compare maxLength so far to the length of current subarray and update maxLength to maximum of either maxLength or current subarray length.
- If the condition is not satisfied then increase the beginning pointer for subarray by 1 and remove all indices from minHeap and maxHeap which are not included in the new subarray.
- After each iteration, we increase our subarray length by incrementing the end pointer.
Below is the implementation of the above approach:
// C++ implementation to find the Longest // subarray of the given array with absolute // difference between elements less than or equal // to integer K using Heaps #include <bits/stdc++.h> using namespace std;
// Function to compute the Longest subarray int computeLongestSubarray(vector< int > arr, int k)
{ // Stores the maximum length subarray so far
int maxLength = 0;
deque< int > maxHeap;
deque< int > minHeap;
// Marks to the beginning and end
// pointer for current subarray
int beg = 0;
int end = 0;
while (end < arr.size()) {
// Stores the current element being
// added to the subarray
int currEl = arr[end];
// Remove indices of all elements smaller
// than or equal to current from maxHeap
while (!maxHeap.empty()
&& arr[maxHeap.back()] <= currEl)
maxHeap.pop_back();
// Add current element's index to maxHeap
maxHeap.push_back(end);
// Remove indices of all elements larger
// than or equal to current from minHeap
while (!minHeap.empty()
&& arr[minHeap.back()] >= currEl)
minHeap.pop_back();
// Add current element's index to minHeap
minHeap.push_back(end);
// Index of maximum of current subarray
int maxOfSub = arr[maxHeap.front()];
// Index of minimum of current subarray
int minOfSub = arr[minHeap.front()];
// check if the largest possible difference
// between a pair of elements <= k
if (maxOfSub - minOfSub <= k) {
// Length of current subarray
int currLength = end - beg + 1;
// Update maxLength
if (maxLength < currLength)
maxLength = currLength;
}
else {
// If current subarray doesn't satisfy
// the condition then remove the starting
// element from subarray that satisfy
// increment the beginning pointer
beg += 1;
// Remove elements from heaps that
// are not in the subarray anymore
while (!minHeap.empty()
&& minHeap.front() < beg)
minHeap.pop_front();
while (!maxHeap.empty()
&& maxHeap.front() < beg)
maxHeap.pop_front();
}
end += 1;
}
// Return the final answer
return maxLength;
} // Driver code int main()
{ vector< int > arr = { 1, 2, 3, 6, 7 };
int k = 2;
int maxLength = computeLongestSubarray(arr, k);
cout << maxLength;
return 0;
} // This code is contributed by adityamaharshi21 |
// Java implementation to find the Longest // subarray of the given array with absolute // difference between elements less than or equal // to integer K using Heaps import java.util.*;
class GFG {
public static int computeLongestSubarray( int arr[],
int k)
{
// Stores the maximum length subarray so far
int maxLength = 0 ;
Deque<Integer> maxHeap = new LinkedList<>();
Deque<Integer> minHeap = new LinkedList<>();
// Marks to the beginning and end
// pointer for current subarray
int beg = 0 , end = 0 ;
while (end < arr.length) {
// Stores the current element being
// added to the subarray
int currEl = arr[end];
// Remove indices of all elements smaller
// than or equal to current from maxHeap
while (maxHeap.size() > 0 &&
arr[maxHeap.peekLast()] <= currEl)
maxHeap.removeLast();
// Add current element's index to maxHeap
maxHeap.addLast(end);
// Remove indices of all elements larger
// than or equal to current from minHeap
while (minHeap.size() > 0 &&
arr[minHeap.peekLast()] >= currEl)
minHeap.removeLast();
// Add current element's index to minHeap
minHeap.addLast(end);
// Index of maximum of current subarray
int maxOfSub = arr[maxHeap.peekFirst()];
// Index of minimum of current subarray
int minOfSub = arr[minHeap.peekFirst()];
// check if the largest possible difference
// between a pair of elements <= k
if (maxOfSub - minOfSub <= k) {
// Length of current subarray
int currLength = end - beg + 1 ;
// Update maxLength
if (maxLength < currLength)
maxLength = currLength;
}
else {
// If current subarray doesn't satisfy
// the condition then remove the starting
// element from subarray that satisfy
// increment the beginning pointer
beg++;
// Remove elements from heaps that
// are not in the subarray anymore
while (minHeap.size() > 0 &&
minHeap.peekFirst() < beg)
minHeap.removeFirst();
while (maxHeap.size() > 0 &&
maxHeap.peekFirst() < beg)
maxHeap.removeFirst();
}
end++;
}
// Return the final answer
return maxLength;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 6 , 7 };
int k = 2 ;
int maxLength = computeLongestSubarray(arr, k);
System.out.println(maxLength);
}
} |
# Python3 implementation to find the Longest # subarray of the given array with absolute # difference between elements less than or equal # to integer K using Heaps from collections import deque
def computeLongestSubarray(arr, k):
# Stores the maximum length subarray so far
maxLength = 0
maxHeap = []
minHeap = []
# Marks to the beginning and end
# pointer for current subarray
beg = 0
end = 0
while (end < len (arr)):
# print(end)
# Stores the current element being
# added to the subarray
currEl = arr[end]
# Remove indices of all elements smaller
# than or equal to current from maxHeap
while ( len (maxHeap) > 0 and arr[maxHeap[ - 1 ]] < = currEl):
del maxHeap[ - 1 ]
# Add current element's index to maxHeap
maxHeap.append(end)
# Remove indices of all elements larger
# than or equal to current from minHeap
while ( len (minHeap) > 0 and arr[minHeap[ - 1 ]] > = currEl):
# print(minHeap[-1])
del minHeap[ - 1 ]
# Add current element's index to minHeap
minHeap.append(end)
# Index of maximum of current subarray
maxOfSub = arr[maxHeap[ 0 ]]
# Index of minimum of current subarray
minOfSub = arr[minHeap[ 0 ]]
# check if the largest possible difference
# between a pair of elements <= k
if (maxOfSub - minOfSub < = k):
# Length of current subarray
currLength = end - beg + 1
# Update maxLength
if (maxLength < currLength):
maxLength = currLength
else :
# If current subarray doesn't satisfy
# the condition then remove the starting
# element from subarray that satisfy
# increment the beginning pointer
beg + = 1
# Remove elements from heaps that
# are not in the subarray anymore
while ( len (minHeap) > 0 and minHeap[ 0 ] < beg):
del minHeap[ 0 ]
while ( len (maxHeap) > 0 and maxHeap[ 0 ] < beg):
del maxHeap[ 0 ]
end + = 1
# Return the final answer
return maxLength
# Driver code
if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 6 , 7 ]
k = 2
maxLength = computeLongestSubarray(arr, k)
print (maxLength)
# This code is contributed by mohit kumar 29 |
// C# code to find the Longest subarray of the given array // with absolute difference between elements less than or // equal to integer K using Heaps using System;
class GFG {
public static int computeLongestSubarray( int [] arr,
int k)
{
// Stores the maximum length subarray so far
int maxLength = 0;
// Stores the maximum of current subarray
int maxOfSub = 0;
// Stores the minimum of current subarray
int minOfSub = 0;
// Initializes the value of maxOfSub to
// first element of array
maxOfSub = arr[0];
// Initializes the value of minOfSub to
// first element of array
minOfSub = arr[0];
// Stores the end pointer for current
// subarray
int end = 0;
// Stores the beginning pointer for
// current subarray
int beg = 0;
// Loop over the given array
while (end < arr.Length) {
// Stores the current element being
// added to the subarray
int currEl = arr[end];
// If maximum value in the array is less
// than current element,
if (maxOfSub < currEl) {
// Remove indices of all elements smaller
// than or equal to current from maxHeap
// and set maximum value as current element
maxOfSub = currEl;
}
// If minimum value in the array is greater
// than current element,
if (minOfSub > currEl) {
// Remove indices of all elements larger
// than or equal to current from minHeap
// and set minimum value as current element
minOfSub = currEl;
}
// check if the largest possible difference
// between a pair of elements <= k
if (maxOfSub - minOfSub <= k) {
// Length of current subarray
int currLength = end - beg + 1;
// Update maxLength
if (maxLength < currLength)
maxLength = currLength;
}
else {
// If current subarray doesn't satisfy the
// condition then remove the starting
// element from subarray that satisfy
// increment the beginning pointer
beg += 1;
// If index of max element currently in the
// subarray is less than beg
if (maxOfSub < beg)
maxOfSub = beg;
// If index of min element currently in the
// subarray is less than beg
if (minOfSub < beg)
minOfSub = beg;
}
end += 1;
}
// Return the final answer
return maxLength;
}
public static void Main()
{
int [] arr = { 1, 2, 3, 6, 7 };
int k = 2;
int maxLength = computeLongestSubarray(arr, k);
Console.WriteLine(maxLength);
}
} // This code is contributed by aadityamaharshi21. |
<script> // JavaScript implementation to find the Longest // subarray of the given array with absolute // difference between elements less than or equal // to integer K using Heaps function computeLongestSubarray(arr,k)
{ // Stores the maximum length subarray so far
let maxLength = 0;
let maxHeap = [];
let minHeap = [];
// Marks to the beginning and end
// pointer for current subarray
let beg = 0, end = 0;
while (end < arr.length) {
// Stores the current element being
// added to the subarray
let currEl = arr[end];
// Remove indices of all elements smaller
// than or equal to current from maxHeap
while (maxHeap.length > 0 &&
arr[maxHeap[maxHeap.length-1]] <= currEl)
maxHeap.pop();
// Add current element's index to maxHeap
maxHeap.push(end);
// Remove indices of all elements larger
// than or equal to current from minHeap
while (minHeap.length > 0 &&
arr[minHeap[minHeap.length-1]] >= currEl)
minHeap.pop();
// Add current element's index to minHeap
minHeap.push(end);
// Index of maximum of current subarray
let maxOfSub = arr[maxHeap[0]];
// Index of minimum of current subarray
let minOfSub = arr[minHeap[0]];
// check if the largest possible difference
// between a pair of elements <= k
if (maxOfSub - minOfSub <= k) {
// Length of current subarray
let currLength = end - beg + 1;
// Update maxLength
if (maxLength < currLength)
maxLength = currLength;
}
else {
// If current subarray doesn't satisfy
// the condition then remove the starting
// element from subarray that satisfy
// increment the beginning pointer
beg++;
// Remove elements from heaps that
// are not in the subarray anymore
while (minHeap.length > 0 &&
minHeap[0] < beg)
minHeap.shift();
while (maxHeap.length > 0 &&
maxHeap[0] < beg)
maxHeap.shift();
}
end++;
}
// Return the final answer
return maxLength;
} // Driver code let arr=[ 1, 2, 3, 6, 7 ]; let k = 2; let maxLength = computeLongestSubarray(arr, k); document.write(maxLength); // This code is contributed by rag2127 </script> |
3
Time Complexity: O(n) because every element of the array is added and removed from the heaps only once.
Auxiliary Space: O(n)