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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).


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 the implementation of above approach.

C/C++

// 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;
}

Java

// Java program to find the element which is greater than
// all left elements and smaller than all right elements.
import java.io.*;
import java.util.*;

public class GFG {
static int findElement(int[] arr, int n)
{
// leftMax[i] stores maximum of arr[0..i-1]
int[] leftMax = new int[n];
leftMax[0] = Integer.MIN_VALUE;

// Fill leftMax[]1..n-1]
for (int i = 1; i < n; i++) leftMax[i] = Math.max(leftMax[i - 1], arr[i - 1]); // Initialize minimum from right int rightMin = Integer.MAX_VALUE; // 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 = Math.min(rightMin, arr[i]);
}

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

}

// Driver code
public static void main(String args[])
{
int[] arr = {5, 1, 4, 3, 6, 8, 10, 7, 9};
int n = arr.length;
System.out.println(“Index of the element is ” +
findElement(arr, n));
}



// This code is contributed
// by rachana soma
}

C#

// C# program to find the element which is greater than
// all left elements and smaller than all right elements.
using System;
  
class GFG
{
static int findElement(int[] arr, int n)
{
    // leftMax[i] stores maximum of arr[0..i-1]
    int[] leftMax = new int[n];
    leftMax[0] = int.MinValue;
  
    // Fill leftMax[]1..n-1]
    for (int i = 1; i < n; i++)
        leftMax[i] = Math.Max(leftMax[i - 1], arr[i - 1]);
  
    // Initialize minimum from right
    int rightMin = int.MaxValue;
  
    // 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 = Math.Min(rightMin, arr[i]);
    }
  
    // If there was no element matching criteria
    return -1;
}
  
// Driver program
public static void Main()
{
    int[] arr = {5, 1, 4, 3, 6, 8, 10, 7, 9};
    int n = arr.Length;
    Console.Write( "Index of the element is " + findElement(arr, n));
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP program to find the element 
// which is greater than all left 
// elements and smaller than all 
// right elements. 
  
function findElement($arr, $n
    // leftMax[i] stores maximum 
    // of arr[0..i-1] 
    $leftMax = array(0); 
    $leftMax[0] = PHP_INT_MIN; 
  
    // Fill leftMax[]1..n-1] 
    for ($i = 1; $i < $n; $i++) 
        $leftMax[$i] = max($leftMax[$i - 1],
                               $arr[$i - 1]); 
  
    // Initialize minimum from right 
    $rightMin = PHP_INT_MAX; 
  
    // Traverse array from right 
    for ($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 Code 
$arr = array(5, 1, 4, 3, 6, 8, 10, 7, 9); 
$n = count($arr);
echo "Index of the element is "
           findElement($arr, $n); 
  
// This code is contributed
// by Sach_Code
?>

Output:

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



This article is attributed to GeeksforGeeks.org

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