Find a partition point in array
Last Updated :
14 Mar, 2023
Given an unsorted array of integers. Find an element such that all the elements to its left are smaller and to its right are greater. Print -1 if no such element exists.
Note that there can be more than one such elements. For example an array which is sorted in increasing order all elements follow the property. We need to find only one such element.
Examples :
Input : A[] = {4, 3, 2, 5, 8, 6, 7}
Output : 5
Input : A[] = {5, 6, 2, 8, 10, 9, 8}
Output : -1
Simple solution takes O(n2). Idea is to pick each array element one by one and for each element we have to check it is greater than all the elements to its left side and smaller than all the elements to its right side.
Below is the implementation of above idea :
C++
// Simple C++ program to find a partition point in
// an array
#include <bits/stdc++.h>
using namespace std;
// Prints an element such than all elements on left
// are smaller and all elements on right are greater.
int FindElement(int A[], int n)
{
// traverse array elements
for (int i = 0; i < n; i++) {
// If we found that such number
int flag = 0;
// check All the elements on its left are smaller
for (int j = 0; j < i; j++)
if (A[j] >= A[i]) {
flag = 1;
break;
}
// check All the elements on its right are Greater
for (int j = i + 1; j < n; j++)
if (A[j] <= A[i]) {
flag = 1;
break;
}
// If flag == 0 indicates we found that number
if (flag == 0)
return A[i];
}
return -1;
}
// driver program to test above function
int main()
{
int A[] = { 4, 3, 2, 5, 8, 6, 7 };
int n = sizeof(A) / sizeof(A[0]);
cout << FindElement(A, n) << endl;
return 0;
}
// Simple Java program to find
// a partition point in an array
import java.io.*;
class GFG {
// Prints an element such than all elements
// on left are smaller and all elements on
// right are greater.
static int FindElement(int[] A, int n)
{
// traverse array elements
for (int i = 0; i < n; i++) {
// If we found that such number
int flag = 0;
// check All the elements on
// its left are smaller
for (int j = 0; j < i; j++)
if (A[j] >= A[i]) {
flag = 1;
break;
}
// check All the elements on
// its right are Greater
for (int j = i + 1; j < n; j++)
if (A[j] <= A[i]) {
flag = 1;
break;
}
// If flag == 0 indicates we
// found that number
if (flag == 0)
return A[i];
}
return -1;
}
// Driver code
static public void main(String[] args)
{
int[] A = {4, 3, 2, 5, 8, 6, 7};
int n = A.length;
System.out.println(FindElement(A, n));
}
}
// This code is contributed by vt_m
# Simple python 3 program to find a
# partition point in an array
# Prints an element such than all
# elements on left are smaller and
# all elements on right are greater.
def FindElement(A, n):
# traverse array elements
for i in range(0, n, 1):
# If we found that such number
flag = 0
# check All the elements on its
# left are smaller
for j in range(0, i, 1):
if (A[j] >= A[i]):
flag = 1
break
# check All the elements on its
# right are Greater
for j in range(i + 1, n, 1):
if (A[j] <= A[i]):
flag = 1
break
# If flag == 0 indicates we found
# that number
if (flag == 0):
return A[i]
return -1
# Driver Code
if __name__ == '__main__':
A = [4, 3, 2, 5, 8, 6, 7]
n = len(A)
print(FindElement(A, n))
# This code is contributed by
# Sanjit_Prasad
// Simple C# program to find a
// partition point in an array
using System;
class GFG {
// Prints an element such than all
// elements on left are smaller and all
// elements on right are greater.
static int FindElement(int[] A, int n)
{
// traverse array elements
for (int i = 0; i < n; i++) {
// If we found that such number
int flag = 0;
// check All the elements on
// its left are smaller
for (int j = 0; j < i; j++)
if (A[j] >= A[i]) {
flag = 1;
break;
}
// check All the elements on
// its right are Greater
for (int j = i + 1; j < n; j++)
if (A[j] <= A[i]) {
flag = 1;
break;
}
// If flag == 0 indicates we
// found that number
if (flag == 0)
return A[i];
}
return -1;
}
// Driver code
static public void Main()
{
int[] A = { 4, 3, 2, 5, 8, 6, 7 };
int n = A.Length;
Console.WriteLine(FindElement(A, n));
}
}
// This code is contributed by vt_m.
<?php
// Simple PHP program to find a partition
// point in an array
// Prints an element such than all elements
// on left are smaller and all elements on
// right are greater.
function FindElement($A, $n)
{
// traverse array elements
for ($i = 0; $i < $n; $i++)
{
// If we found that such number
$flag = 0;
// check All the elements on its
// left are smaller
for ( $j = 0; $j < $i; $j++)
if ($A[$j] >= $A[$i])
{
$flag = 1;
break;
}
// check All the elements on its
// right are Greater
for ( $j = $i + 1; $j < $n; $j++)
if ($A[$j] <= $A[$i])
{
$flag = 1;
break;
}
// If flag == 0 indicates we found
// that number
if ($flag == 0)
return $A[$i];
}
return -1;
}
// Driver Code
$A = array( 4, 3, 2, 5, 8, 6, 7 );
$n = count($A);
echo FindElement($A, $n);
// This code is contributed by
// Rajput-Ji
?>
<script>
// Simple JavaScript program to find
// a partition point in an array
// Prints an element such than all elements
// on left are smaller and all elements on
// right are greater.
function FindElement(A , n)
{
// traverse array elements
for (i = 0; i < n; i++) {
// If we found that such number
var flag = 0;
// check All the elements on
// its left are smaller
for (j = 0; j < i; j++)
if (A[j] >= A[i]) {
flag = 1;
break;
}
// check All the elements on
// its right are Greater
for (j = i + 1; j < n; j++)
if (A[j] <= A[i]) {
flag = 1;
break;
}
// If flag == 0 indicates we
// found that number
if (flag == 0)
return A[i];
}
return -1;
}
// Driver code
var A = [4, 3, 2, 5, 8, 6, 7];
var n = A.length;
document.write(FindElement(A, n));
// This code contributed by shikhasingrajput
</script>
Time complexity: O(n2)
Space Complexity: O(1)
Efficient solution take O(n) time.
- Create an auxiliary array 'GE[]'. GE[] should store the element which is greater than A[i] and is on left side of A[i].
- Create an another Auxiliary array 'SE[]'. SE[i] should store the element which is smaller than A[i] and is on right side of A[i].
- Find element in array that hold condition GE[i-1] < A[i] < SE[i+1].
Below is the implementation of above idea :
C++
// Simple C++ program to find
// a partition point in an array
#include <bits/stdc++.h>
using namespace std;
// Returns an element that has all
// the element to its left smaller and
// to its right greater
int FindElement(int A[], int n)
{
// Create an array 'SE[]' that will
// store smaller element on right side.
int SE[n];
// Create an another array 'GE[]' that will
// store greatest element on left side.
int GE[n];
// initialize first and last index of SE[], GE[]
GE[0] = A[0];
SE[n - 1] = A[n - 1];
// store greatest element from left to right
for (int i = 1; i < n; i++) {
if (GE[i - 1] < A[i])
GE[i] = A[i];
else
GE[i] = GE[i - 1];
}
// store smallest element from right to left
for (int i = n - 2; i >= 0; i--) {
if (A[i] < SE[i + 1])
SE[i] = A[i];
else
SE[i] = SE[i + 1];
}
// Now find a number which is greater than all
// elements at it's left and smaller than all
// then elements to it's right
for (int j = 0; j < n; j++)
{
if ((j == 0 && A[j] < SE[j + 1]) ||
(j == n - 1 && A[j] > GE[j - 1]) ||
(A[j] < SE[j + 1] && A[j] > GE[j - 1]))
return A[j];
}
return -1;
}
// Driver code
int main()
{
int A[] = {4, 3, 2, 5, 8, 6, 7};
int n = sizeof(A) / sizeof(A[0]);
cout << FindElement(A, n) << endl;
return 0;
}
// Simple java program to find a
// partition point in an array
import java.io.*;
class GFG {
// Returns an element that has all
// the element to its left smaller
// and to its right greater
static int FindElement(int[] A, int n)
{
// Create an array 'SE[]' that will
// store smaller element on right side.
int[] SE = new int[n];
// Create an another array 'GE[]' that
// will store greatest element on left side.
int[] GE = new int[n];
// initialize first and last index of SE[], GE[]
GE[0] = A[0];
SE[n - 1] = A[n - 1];
// store greatest element from left to right
for (int i = 1; i < n; i++)
{
if (GE[i - 1] < A[i])
GE[i] = A[i];
else
GE[i] = GE[i - 1];
}
// store smallest element from right to left
for (int i = n - 2; i >= 0; i--)
{
if (A[i] < SE[i + 1])
SE[i] = A[i];
else
SE[i] = SE[i + 1];
}
// Now find a number which is greater than all
// elements at it's left and smaller than all
// then elements to it's right
for (int j = 0; j < n; j++)
{
if ((j == 0 && A[j] < SE[j + 1]) ||
(j == n - 1 && A[j] > GE[j - 1]) ||
(A[j] < SE[j + 1] && A[j] > GE[j - 1]))
return A[j];
}
return -1;
}
// Driver code
static public void main(String[] args)
{
int[] A = {4, 3, 2, 5, 8, 6, 7};
int n = A.length;
System.out.println(FindElement(A, n));
}
}
// This code is contributed by vt_m.
# Python Program to implement
# the above approach
# Returns an element that has all
# the element to its left smaller and
# to its right greater
def FindElement(A, n) :
# Create an array 'SE[]' that will
# store smaller element on right side.
SE = [0] * n
# Create an another array 'GE[]' that will
# store greatest element on left side.
GE = [0] * n
# initialize first and last index of SE[], GE[]
GE[0] = A[0]
SE[n - 1] = A[n - 1]
# store greatest element from left to right
for i in range(1, n):
if (GE[i - 1] < A[i]) :
GE[i] = A[i]
else :
GE[i] = GE[i - 1]
# store smallest element from right to left
for i in range(n-2, -1, -1):
if (A[i] < SE[i + 1]) :
SE[i] = A[i]
else :
SE[i] = SE[i + 1]
# Now find a number which is greater than all
# elements at it's left and smaller than all
# then elements to it's right
for j in range(n):
if ((j == 0 and A[j] < SE[j + 1]) or
(j == n - 1 and A[j] > GE[j - 1]) or
(A[j] < SE[j + 1] and A[j] > GE[j - 1])):
return A[j]
return -1
# Driver code
A = [ 4, 3, 2, 5, 8, 6, 7 ]
n = len(A)
print(FindElement(A, n))
# This code is contributed by code_hunt.
// Simple C# program to find
// a partition point in an array
using System;
class GFG {
// Returns an element that has all
// the element to its left smaller
// and to its right greater
static int FindElement(int[] A, int n)
{
// Create an array 'SE[]' that will
// store smaller element on right side.
int[] SE = new int[n];
// Create an another array 'GE[]' that will
// store greatest element on left side.
int[] GE = new int[n];
// initialize first and last index of SE[], GE[]
GE[0] = A[0];
SE[n - 1] = A[n - 1];
// store greatest element from left to right
for (int i = 1; i < n; i++)
{
if (GE[i - 1] < A[i])
GE[i] = A[i];
else
GE[i] = GE[i - 1];
}
// store smallest element from right to left
for (int i = n - 2; i >= 0; i--)
{
if (A[i] < SE[i + 1])
SE[i] = A[i];
else
SE[i] = SE[i + 1];
}
// Now find a number which is greater than all
// elements at it's left and smaller than all
// then elements to it's right
for (int j = 0; j < n; j++)
{
if ((j == 0 && A[j] < SE[j + 1]) ||
(j == n - 1 && A[j] > GE[j - 1]) ||
(A[j] < SE[j + 1] && A[j] > GE[j - 1]))
return A[j];
}
return -1;
}
// Driver code
static public void Main()
{
int[] A = {4, 3, 2, 5, 8, 6, 7};
int n = A.Length;
Console.WriteLine(FindElement(A, n));
}
}
// This code is contributed by vt_m .
<?php
// PHP program to find a partition point
// in an array
// Prints an element such than all elements
// on left are smaller and all elements on
// right are greater.
function FindElement($A, $n)
{
// traverse array elements
for ($i = 0; $i < $n; $i++)
{
// If we found that such number
$flag = 0;
// check All the elements on
// its left are smaller
for ($j = 0; $j < $i; $j++)
if ($A[$j] >= $A[$i])
{
$flag = 1;
break;
}
// check All the elements on
// its right are Greater
for ($j = $i + 1; $j < $n; $j++)
if ($A[$j] <= $A[$i])
{
$flag = 1;
break;
}
// If flag == 0 indicates we
// found that number
if ($flag == 0)
return $A[$i];
}
return -1;
}
// Driver code
$A = array(4, 3, 2, 5, 8, 6, 7);
$n = sizeof($A);
echo(FindElement($A, $n));
// This code is contributed
// by Mukul Singh
?>
<script>
// Simple Javascript program to find
// a partition point in an array
// Returns an element that has all
// the element to its left smaller
// and to its right greater
function FindElement(A, n)
{
// Create an array 'SE[]' that will
// store smaller element on right side.
let SE = new Array(n);
// Create an another array 'GE[]' that will
// store greatest element on left side.
let GE = new Array(n);
// initialize first and last index of SE[], GE[]
GE[0] = A[0];
SE[n - 1] = A[n - 1];
// store greatest element from left to right
for (let i = 1; i < n; i++)
{
if (GE[i - 1] < A[i])
GE[i] = A[i];
else
GE[i] = GE[i - 1];
}
// store smallest element from right to left
for (let i = n - 2; i >= 0; i--)
{
if (A[i] < SE[i + 1])
SE[i] = A[i];
else
SE[i] = SE[i + 1];
}
// Now find a number which is greater than all
// elements at it's left and smaller than all
// then elements to it's right
for (let j = 0; j < n; j++)
{
if ((j == 0 && A[j] < SE[j + 1]) ||
(j == n - 1 && A[j] > GE[j - 1]) ||
(A[j] < SE[j + 1] && A[j] > GE[j - 1]))
return A[j];
}
return -1;
}
let A = [4, 3, 2, 5, 8, 6, 7];
let n = A.length;
document.write(FindElement(A, n));
</script>
Time complexity: O(n)
Auxiliary Space: O(n)
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