# Find the element before which all the elements are smaller than it, and after which all are greater

Given an array, find an element before which all elements are smaller than it, and after which all are greater than it. Return index of the element if there is such an element, otherwise return -1.

Examples:

```Input:   arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9};
Output:  Index of element is 4
All elements on left of arr[4] are smaller than it
and all elements on right are greater.

Input:   arr[] = {5, 1, 4, 4};
Output:  Index of element is -1
```

Expected time complexity is O(n).

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Simple Solution is to consider every element one by one. For every element, compare it with all elements on left and all elements on right. Time complexity of this solution is O(n2).

An Efficient Solution can solve this problem in O(n) time using O(n) extra space. Below is detailed solution.

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

Further Optimization to above approach is to use only one extra array and traverse input array only twice. First traversal is same as above and fills leftMax[]. Next traversal traverses from right and keeps track of minimum. The second traversal also finds the required element.

Below is C++ implementation of above approach.

```// C++ program to find the element which is greater than
// all left elements and smaller than all right elements.
#include <bits/stdc++.h>
using namespace std;

int findElement(int arr[], int n)
{
// leftMax[i] stores maximum of arr[0..i-1]
int leftMax[n];
leftMax[0] = INT_MIN;

// Fill leftMax[]1..n-1]
for (int i = 1; i < n; i++)
leftMax[i] = max(leftMax[i-1], arr[i-1]);

// Initialize minimum from right
int rightMin = INT_MAX;

// Traverse array from right
for (int i=n-1; i>=0; i--)
{
// Check if we found a required element
if (leftMax[i] < arr[i] && rightMin > arr[i])
return i;

// Update right minimum
rightMin = min(rightMin, arr[i]);
}

// If there was no element matching criteria
return -1;
}

// Driver program
int main()
{
int arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9};
int n = sizeof arr / sizeof arr[0];
cout << "Index of the element is " << findElement(arr, n);
return 0;
}
```

Index of the element is 4.

Time Complexity: O(n)
Auxiliary Space: O(n)

Thanks to Gaurav Ahirwar for suggesting above solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3 Average Difficulty : 3/5.0
Based on 61 vote(s)

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.