Find maximum difference between nearest left and right smaller elements - GeeksforGeeks

Find maximum difference between nearest left and right smaller elements - GeeksforGeeks

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

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

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 side
     1. 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 S   

Find all smaller element on right side
     3. 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])

// 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;

}

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