Given array of integers, the task is to find the maximum absolute difference between nearest left and right smaller element of every element in array.

**Note : ** If there is no smaller element on right side or left side of any element then we take zero as smaller element. For example for leftmost element, nearest smaller element on left side is considered as 0. Similarly for rightmost elements, smaller element on right side is considered as 0.

Examples:

Input : arr[] = {2, 1, 8} Output : 1 Left smaller LS[] {0, 0, 1} Right smaller RS[] {1, 0, 0} Maximum Diff of abs(LS[i] - RS[i]) = 1 Input : arr[] = {2, 4, 8, 7, 7, 9, 3} Output : 4 Left smaller LS[] = {0, 2, 4, 4, 4, 7, 2} Right smaller RS[] = {0, 3, 7, 3, 3, 3, 0} Maximum Diff of abs(LS[i] - RS[i]) = 7 - 3 = 4 Input : arr[] = {5, 1, 9, 2, 5, 1, 7} Output : 1

A **simple solution** is to find nearest left and right smaller elements for every element and then update the maximum difference between left and right smaller element , this take O(n^2) time.

An **efficient solution** takes O(n) time. We use a stack. The idea is based on the approach discussed in next greater element article. The interesting part here is we compute both left smaller and right smaller using same function.

Let input array be 'arr[]' and size of array be 'n'Find all smaller element on left side1. Create a new empty stack S and an array LS[] 2. For every element 'arr[i]' in the input arr[], where 'i' goes from 0 to n-1. a) while S is nonempty and the top element of S is greater than or equal to 'arr[i]': pop S b) if S is empty: 'arr[i]' has no preceding smaller value LS[i] = 0 c) else: the nearest smaller value to 'arr[i]' is top of stack LS[i] = s.top() d) push 'arr[i]' onto SFind all smaller element on right side3. First reverse array arr[]. After reversing the array, right smaller become left smaller. 4. Create an array RRS[] and repeat steps 1 and 2 to fill RRS (in-place of LS). 5. Initialize result as -1 and do following for every element arr[i]. In the reversed array right smaller for arr[i] is stored at RRS[n-i-1] return result = max(result, LS[i]-RRS[n-i-1])

Below is implementation of above idea

## C/C++

// C++ program to find the difference b/w left and // right smaller element of every element in array #include<bits/stdc++.h> using namespace std; // Function to fill left smaller element for every // element of arr[0..n-1]. These values are filled // in SE[0..n-1] void leftSmaller(int arr[], int n, int SE[]) { // Create an empty stack stack<int>S; // Traverse all array elements // compute nearest smaller elements of every element for (int i=0; i<n; i++) { // Keep removing top element from S while the top // element is greater than or equal to arr[i] while (!S.empty() && S.top() >= arr[i]) S.pop(); // Store the smaller element of current element if (!S.empty()) SE[i] = S.top(); // If all elements in S were greater than arr[i] else SE[i] = 0; // Push this element S.push(arr[i]); } } // Function returns maximum difference b/w Left & // right smaller element int findMaxDiff(int arr[], int n) { int LS[n]; // To store left smaller elements // find left smaller element of every element leftSmaller(arr, n, LS); // find right smaller element of every element // first reverse the array and do the same process int RRS[n]; // To store right smaller elements in // reverse array reverse(arr, arr + n); leftSmaller(arr, n, RRS); // find maximum absolute difference b/w LS & RRS // In the reversed array right smaller for arr[i] is // stored at RRS[n-i-1] int result = -1; for (int i=0 ; i< n ; i++) result = max(result, abs(LS[i] - RRS[n-1-i])); // return maximum difference b/w LS & RRS return result; } // Driver program int main() { int arr[] = {2, 4, 8, 7, 7, 9, 3}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Maximum diff : " << findMaxDiff(arr, n) << endl; return 0; }

## Python

# Python program to find the difference b/w left and # right smaller element of every element in array # Function to fill left smaller element for every # element of arr[0..n-1]. These values are filled # in SE[0..n-1] def leftsmaller(arr, n, SE): # create an empty stack sta = [] # Traverse all array elements # compute nearest smaller elements of every element for i in range(n): # Keep removing top element from S while the top # element is greater than or equal to arr[i] while(sta != [] and sta[len(sta)-1] >= arr[i]): sta.pop() # Store the smaller element of current element if(sta != []): SE[i]=sta[len(sta)-1] # If all elements in S were greater than arr[i] else: SE[i]=0 # push this element sta.append(arr[i]) # Function returns maximum difference b/w Left & # right smaller elemen def findMaxDiff(arr, n): ls=[0]*n # to store left smaller elements rs=[0]*n # to store right smaller elements # find left smaller elements of every element leftsmaller(arr, n, ls) # find right smaller element of every element # by sending reverse of array leftsmaller(arr[::-1], n, rs) # find maximum absolute difference b/w LS & RRS # In the reversed array right smaller for arr[i] is # stored at RRS[n-i-1] res = -1 for i in range(n): res = max(res, abs(ls[i] - rs[n-1-i])) # return maximum difference b/w LS & RRS return res # Driver Program if __name__=='__main__': arr = [2, 4, 8, 7, 7, 9, 3] print "Maximum Diff :", findMaxDiff(arr, len(arr)) #Contributed By: Harshit Sidhwa

Output:

Maximum Diff : 4

Time complexity : O(n)

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