Element with Smaller Left and Greater Right

Given an array arr[], find an element before which all elements are equal or smaller than it, and after which all the elements are equal or greater.

Note: Print -1, if no such element exists.

Examples:

Input: arr[] = [5, 1, 4, 3, 6, 8, 10, 7, 9]
Output: 6 
Explanation: 6 is present at index 4. All elements on the left of arr[4] are smaller than it and all elements on right are greater.

Input: arr[] = [5, 1, 4, 4]
Output: -1 
Explanation: No such element exists.

[Naive Approach] Using Nested Loops - O(n²) Time and O(1) Space

The idea is to consider every non-boundary element of the given array arr[] one by one. For each element, find the largest element in left of it, and the smallest element in the right of it using the nested loops. If the largest element in its left is smaller or equal and the smallest element in its right is greater or equal, print the current element, else move to next. At last if no such element is found, print -1.

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[Expected Approach 1] Using Prefix and Suffix Arrays - O(n) Time and O(n) Space

The idea is to precompute the largest values from index 0 to n - 1 and smallest values from n - 1 to 0. To do so,

• Create two arrays leftMax[] and rightMin[].
• Traverse input array from left to right and fill leftMax[] such that leftMax[i] contains the maximum element from 0 to i-1 in the input array.
• Traverse input array from right to left and fill rightMin[] such that rightMin[i] contains the minimum element from n-1 to i+1 in the input array.
• Traverse input array. For every element arr[i], check if arr[i] is greater than or equals to leftMax[i] and smaller than or equals to rightMin[i]. If true, return arr[i]

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[Expected Approach 2] Using Prefix Array - O(n) Time and O(n) Space

The idea is to create a single array leftMax[] to store the largest elements at left of each index. And for right minimum values, we can use an integer rightMin that will store the current minimum value. At each iteration update the value rightMin and compare it with current value arr[i].

Below image is a dry run of the above approach:

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