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

• Difficulty Level : Medium
• Last Updated : 24 Nov, 2021

Given an array, find an element before which all elements are smaller than it, and after which all are greater than it. Return the 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:
Explanation: All elements on left of arr are smaller than it
and all elements on right are greater.

Input: arr[] = {5, 1, 4, 4};
Output: -1
Explanation : No such index exits.

Expected time complexity: 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 the left and all elements on right. The 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 the 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 a maximum element from 0 to i-1 in the input array.
3. Traverse input array from right to left and fill rightMin[] such that rightMin[i] contains a minimum element from to n-1 to i+1 in the 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 the above approach is to use only one extra array and traverse input array only twice. The first traversal is the same as above and fills leftMax[]. Next traversal traverses from the right and keeps track of the minimum. The second traversal also finds the required element.

Below image is a dry run of the above approach: Below is the implementation of the above approach.

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;

// Function to return the index of the element which is greater than
// all left elements and smaller than all right elements.
int findElement(int arr[], int n)
{
// leftMax[i] stores maximum of arr[0..i-1]
int leftMax[n];
leftMax = 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;
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 = 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
}

Python3

# Python3 program to find the element which is greater than
# all left elements and smaller than all right elements.

def findElement(arr, n):

# leftMax[i] stores maximum of arr[0..i-1]
leftMax = [None] * n
leftMax = arr

# Fill leftMax[]1..n-1]
for i in range(1, n):
leftMax[i] = max(leftMax[i-1], arr[i-1])

# Initialize minimum from right
rightMin = [None]*n
rightMin[n-1] = arr[n-1]

# Fill rightMin
for i in range(n-2, -1, -1):
rightMin[i] = min(rightMin[i+1], arr[i])
# Traverse array from right
for i in range(1, n-1):

# Check if we found a required element
# for ith element, it should be more than maximum of of array
# elements [0....i-1] and should be less than the minimum of
# [i+1.....n-1] array elements
if leftMax[i-1] <= arr[i] and arr[i] <= rightMin[i+1]:
return i

# If there was no element matching criteria
return -1

# Driver program
if __name__ == "__main__":

arr = [5, 1, 4, 3, 6, 8, 10, 7, 9]
n = len(arr)
print("Index of the element is", findElement(arr, n))

# This code is contributed by Rituraj Jain

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 = 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 = 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
?>

Javascript

<script>

// Javascript program to find the element
// which is greater than all left elements
// and smaller than all right elements.

// Function to return the index of 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]
var leftMax = Array(n).fill(0);
leftMax = Number.MIN_VALUE;

// Fill leftMax1..n-1]
for(i = 1; i < n; i++)
leftMax[i] = Math.max(leftMax[i - 1],
arr[i - 1]);

// Initialize minimum from right
var rightMin = Number.MAX_VALUE;

// 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 = Math.min(rightMin, arr[i]);
}

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

// Driver code
var arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ];
var n = arr.length;

document.write("Index of the element is " +
findElement(arr, n));

// This code is contributed by aashish1995

</script>

Output:

Index of the element is 4

Time Complexity: O(n)
Auxiliary Space: O(n)
Thanks to Gaurav Ahirwar for suggesting the above solution.

Space Optimized Approach:

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 a[], int n)
{
// Base case
if (n == 1 || n == 2) {
return -1;
}

// 1.element is the possible candidate for the solution
// of the problem.
// 2.idx is the index of the possible
// candidate.
// 3.maxx is the value which is maximum on the
// left side of the array.
// 4.bit tell whether the loop is
// terminated from the if condition or from the else
// condition when loop do not satisfied the condition.
// 5.check is the variable which tell whether the
// element is updated or not

int element = a, maxx = a, bit = -1, check = 0;
int idx = -1;

// The extreme two of the array can not be the solution
// Therefore iterate the loop from i = 1 to < n-1
for (int i = 1; i < (n - 1);) {

// here we find the possible candidate where Element
// with left side smaller and right side greater.
// when the if condition fail we check and update in
// else condition.

if (a[i] < maxx && i < (n - 1)) {
i++;
bit = 0;
}

// here we update the possible element if the
// element is greater than the maxx (maximum element
// so far). In while loop we sur-pass the value which
// is greater than the element
else {
if (a[i] >= maxx) {
element = a[i];
idx = i;
check = 1;
maxx = a[i];
}
if (check == 1) {
i++;
}
bit = 1;
while (a[i] >= element && i < (n - 1)) {
if (a[i] > maxx) {
maxx = a[i];
}
i++;
}
check = 0;
}
}

// checking for the last value and whether the loop is
// terminated from else or if block.

if (element <= a[n - 1] && bit == 1) {
return idx;
}
else {
return -1;
}
}

// Driver Code
int main()
{
int arr[] = { 5, 1, 4, 3, 6, 8, 10, 7, 9 };
int n = sizeof arr / sizeof arr;

// Function Call
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.
class GFG{

static int findElement(int []a, int n)
{

// Base case
if (n == 1 || n == 2)
{
return -1;
}

// 1.element is the possible candidate for
// the solution of the problem.
// 2.idx is the index of the possible
// candidate.
// 3.maxx is the value which is maximum on the
// left side of the array.
// 4.bit tell whether the loop is
// terminated from the if condition or from
// the else condition when loop do not
// satisfied the condition.
// 5.check is the variable which tell whether the
// element is updated or not
int element = a, maxx = a,
bit = -1, check = 0;
int idx = -1;

// The extreme two of the array can
// not be the solution. Therefore
// iterate the loop from i = 1 to < n-1
for(int i = 1; i < (n - 1);)
{

// Here we find the possible candidate
// where Element with left side smaller
// and right side greater. When the if
// condition fail we check and update in
// else condition.
if (a[i] < maxx && i < (n - 1))
{
i++;
bit = 0;
}

// Here we update the possible element
// if the element is greater than the
// maxx (maximum element so far). In
// while loop we sur-pass the value which
// is greater than the element
else
{
if (a[i] >= maxx)
{
element = a[i];
idx = i;
check = 1;
maxx = a[i];
}
if (check == 1)
{
i++;
}
bit = 1;

while (a[i] >= element && i < (n - 1))
{
if (a[i] > maxx)
{
maxx = a[i];
}
i++;
}
check = 0;
}
}

// Checking for the last value and whether
// the loop is terminated from else or
// if block.
if (element <= a[n - 1] && bit == 1)
{
return idx;
}
else
{
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 avanitrachhadiya2155

Python3

# Python3 program to find the element which
# is greater than all left elements and
# smaller than all right elements.
def findElement (a, n):

# Base case
if (n == 1 or n == 2):
return -1

# 1. element is the possible candidate
# for the solution of the problem
# 2. idx is the index of the
# possible candidate
# 3. maxx is the value which is maximum
# on the left side of the array
# 4. bit tell whether the loop is
# terminated from the if condition or
# from the else condition when loop do
# not satisfied the condition.
# 5. check is the variable which tell
# whether the element is updated or not
element, maxx, bit = a, a, -1
check = 0
idx = -1

# The extreme of the array can't be
# the solution Therefore iterate
# the loop from i = 1 to < n-1
i = 1
while (i < (n - 1)):

# Here we find the possible candidate
# where element with left side smaller
# and right side greater. when the if
# condition fail we check and update
# in else condition
if (a[i] < maxx and i < (n - 1)):
i += 1
bit = 0

# Here we update the possible element
# if the element is greater than the
# maxx (maximum element so far). In
# while loop we sur-pass the value
# which is greater than the element
else:
if (a[i] >= maxx):
element = a[i]
idx = i
check = 1
maxx = a[i]

if (check == 1):
i += 1

bit = 1
while (a[i] >= element and i < (n - 1)):
if (a[i] > maxx):
maxx = a[i]

i += 1

check = 0

# Checking for the last value and whether
# the loop is terminated from else or
# if block
if (element <= a[n - 1] and bit == 1):
return idx
else:
return -1

# Driver Code
if __name__ == '__main__':

arr = [ 5, 1, 4, 3,
6, 8, 10, 7, 9 ]
n = len(arr)

# Function call
print("Index of the element is",
findElement(arr, n))

# This code is contributed by himanshu77

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 []a, int n)
{

// Base case
if (n == 1 || n == 2)
{
return -1;
}

// 1.element is the possible candidate for
// the solution of the problem.
// 2.idx is the index of the possible
// candidate.
// 3.maxx is the value which is maximum on the
// left side of the array.
// 4.bit tell whether the loop is
// terminated from the if condition or from
// the else condition when loop do not
// satisfied the condition.
// 5.check is the variable which tell whether the
// element is updated or not
int element = a, maxx = a,
bit = -1, check = 0;
int idx = -1;

// The extreme two of the array can
// not be the solution. Therefore
// iterate the loop from i = 1 to < n-1
for(int i = 1; i < (n - 1);)
{

// Here we find the possible candidate
// where Element with left side smaller
// and right side greater. When the if
// condition fail we check and update in
// else condition.
if (a[i] < maxx && i < (n - 1))
{
i++;
bit = 0;
}

// Here we update the possible element
// if the element is greater than the
// maxx (maximum element so far). In
// while loop we sur-pass the value which
// is greater than the element
else
{
if (a[i] >= maxx)
{
element = a[i];
idx = i;
check = 1;
maxx = a[i];
}
if (check == 1)
{
i++;
}
bit = 1;

while (a[i] >= element && i < (n - 1))
{
if (a[i] > maxx)
{
maxx = a[i];
}
i++;
}
check = 0;
}
}

// Checking for the last value and whether
// the loop is terminated from else or
// if block.
if (element <= a[n - 1] && bit == 1)
{
return idx;
}
else
{
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;

// Function Call
Console.Write("Index of the element is " +
findElement(arr, n));
}
}

// This code is contributed by rutvik_56

Javascript

<script>
// javascript program to find the element
// which is greater than all left
// elements and smaller than all
// right elements.
function findElement(a , n)
{

// Base case
if (n == 1 || n == 2)
{
return -1;
}

// 1.element is the possible candidate for
// the solution of the problem.
// 2.idx is the index of the possible
// candidate.
// 3.maxx is the value which is maximum on the
// left side of the array.
// 4.bit tell whether the loop is
// terminated from the if condition or from
// the else condition when loop do not
// satisfied the condition.
// 5.check is the variable which tell whether the
// element is updated or not
var element = a, maxx = a, bit = -1, check = 0;
var idx = -1;

// The extreme two of the array can
// not be the solution. Therefore
// iterate the loop from i = 1 to < n-1
for (i = 1; i < (n - 1);) {

// Here we find the possible candidate
// where Element with left side smaller
// and right side greater. When the if
// condition fail we check and update in
// else condition.
if (a[i] < maxx && i < (n - 1)) {
i++;
bit = 0;
}

// Here we update the possible element
// if the element is greater than the
// maxx (maximum element so far). In
// while loop we sur-pass the value which
// is greater than the element
else {
if (a[i] >= maxx) {
element = a[i];
idx = i;
check = 1;
maxx = a[i];
}
if (check == 1) {
i++;
}
bit = 1;

while (a[i] >= element && i < (n - 1)) {
if (a[i] > maxx) {
maxx = a[i];
}
i++;
}
check = 0;
}
}

// Checking for the last value and whether
// the loop is terminated from else or
// if block.
if (element <= a[n - 1] && bit == 1)
{
return idx;
}
else
{
return -1;
}

}

// Driver code
var arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ];
var n = arr.length;
document.write("Index of the element is " + findElement(arr, n));

// This code is contributed by gauravrajput1
</script>
Output
Index of the element is 4

Time Complexity: O(n)
Auxiliary Space: O(1)