Last Updated: 13 Dec, 2020
Longest Contiguous Subarray With Absolute Diff Bounded by a Limit
Easy
Asked in company
Given an array/list 'ARR' of integers and an integer 'LIMIT'. You are supposed to return the length of the longest subarray for which the absolute difference between the maximum and minimum element is less than or equal to the 'LIMIT'.
Note :
An array ‘B’ is a subarray of an array ‘A’ if ‘B’ that can be obtained by deletion of, several elements(possibly none) from the start of ‘A’ and several elements(possibly none) from the end of ‘A’.
Input Format :
The first line contains a single integer ‘T’ denoting the number of test cases. Then the 'T' test cases follow.
The first line of each test case contains two integers separated by single space ‘N’ and 'LIMIT' denoting the number of elements in the array/list.
The second line of each test case contains ‘N’ single space-separated integers denoting the elements of the array/list.
Output Format :
For each test case, print a single line that contains an integer that denotes the length of the longest contiguous subarray with absolute difference bounded by the 'LIMIT'.
Note :
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints :
1 <= 'T' <= 50
1 <= 'N' <= 10^4
0 <= 'ARR[i]' <= 10^5
0 <= 'LIMIT' <= 10^5
Where 'ARR[i]' denotes the ith elements of the given array/list.
Time Limit: 1 sec
Approaches
The basic idea of this approach is to iterate through all the possible subarrays of the given array and choose the longest one having the absolute difference of maximum and minimum element as less than or equal to the limit.
Consider the steps as follows :
- Initialize a variable maxLength to zero which denotes the length of the longest subarray having the absolute difference of maximum and minimum element less than or equal to the given limit.
- Start traversing in the array from start to end using a variable low which denotes the lowest index of the subarray such that 0 <= low <= N - 1.
- For each value of the low, start iterating in a nested loop using a variable high which denotes the highest index of the subarray such that low <= high <= N - 1. And initialise two variables to store the maximum and minimum element. maxElement = INT_MIN and minElement = INT_MAX.
- Now, for each subarray [low, higher], update the minimum and maximum element.
- If the absolute difference of minimum and the maximum element is less than or equal to the given limit the current subarray can be chosen.
- If the length of the current subarray is greater than the maximum subarray length so for then do the update as
- If the absolute difference of minimum and the maximum element is less than or equal to the given limit the current subarray can be chosen.
maxLength = max(maxLength, high - low + 1).The basic idea of this approach is to use a sliding window technique to solve this task. Since we need the longest subarray for which the absolute difference of minimum and the maximum element is less than or equal to the limit. We can use an ordered_map to store the elements of a subarray so that we can easily query for minimum and maximum elements. Consider the steps as follows:
- Create an empty ordered_map named window that stores the elements of the window.
- Initialize two variables low and high to zero which denotes the smallest index and largest index of elements in the window. Insert the first element in the window. Also, initialize a variable maxLength to one which stores the answer to the problem.
- Start iterating while high is less than the size of the given array.
- Get the maximum and minimum element of the window and store it in two variables maxLength and minElement respectively.
- If the absolute difference between maxLength and minLength is less than or equal to the limit, then update maxLength accordingly i.e. maxLength = max(maxLength, high - low + 1). And then increment the value of high. Insert the current element denoted by the “high” index into the window. The logic behind this is to add the next element into the window to increase its size.
- Otherwise, remove the element at index low and increment the value of low. The logic behind this is to remove the last element from the window since the absolute difference between the minimum and the maximum element is exceeding the limit.
- Return the “maxLength” which denotes the length of the longest subarray which is bounded by limit.
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