Find a partition point in array

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++

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// 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;
}

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Java

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// 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 

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Python3

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# 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

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C#

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// 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.

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PHP

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<?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
?>

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Output:

5

Time complexity: O(n2)

Efficient solution take O(n) time.

  1. Create an auxiliary array ‘GE[]’. GE[] should store the element which is greater than A[i] and is on left side of A[i].
  2. Create an another Auxliary array ‘SE[]’. SE[i] should store the element which is smaller than A[i] and is on right side of A[i].
  3. Find element in array that hold condition GE[i-1] < A[i] < SE[i+1].

Below is the implementation of above idea :

C++

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// 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 then all
    // elements at it's left and smaller the 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;
}

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Java

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// 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 then all
        // elements at it's left and smaller the 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.

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C#

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// 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 then all
        // elements at it's left and smaller the 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 .

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PHP

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<?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
?>

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Output:

5

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

This article is contributed by Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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