Find a Fixed Point (Value equal to index) in a given array

Given an array of n distinct integers sorted in ascending order, write a function that returns a Fixed Point in the array, if there is any Fixed Point present in array, else returns -1. Fixed Point in an array is an index i such that arr[i] is equal to i. Note that integers in array can be negative. 
Examples: 

  Input: arr[] = {-10, -5, 0, 3, 7}
  Output: 3  // arr[3] == 3 

  Input: arr[] = {0, 2, 5, 8, 17}
  Output: 0  // arr[0] == 0 


  Input: arr[] = {-10, -5, 3, 4, 7, 9}
  Output: -1  // No Fixed Point

Method 1 (Linear Search) 
Linearly search for an index i such that arr[i] == i. Return the first such index found. Thanks to pm for suggesting this solution. 

C++

// C++ program to check fixed point
// in an array using linear search
#include <bits/stdc++.h>
using namespace std;

int linearSearch(int arr[], int n)
{
    int i;
    for (i = 0; i < n; i++) {
        if (arr[i] == i)
            return i;
    }

    /* If no fixed point present then return -1 */
    return -1;
}

/* Driver code */
int main()
{
    int arr[] = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Fixed Point is " << linearSearch(arr, n);
    return 0;
}

// This is code is contributed by rathbhupendra

C

// C program to check fixed point
// in an array using linear search
#include <stdio.h>

int linearSearch(int arr[], int n)
{
    int i;
    for (i = 0; i < n; i++) {
        if (arr[i] == i)
            return i;
    }

    /* If no fixed point present then return -1 */
    return -1;
}

/* Driver program to check above functions */
int main()
{
    int arr[] = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Fixed Point is %d", linearSearch(arr, n));
    getchar();
    return 0;
}

Java

// Java program to check fixed point
// in an array using linear search

class Main {
    static int linearSearch(int arr[], int n)
    {
        int i;
        for (i = 0; i < n; i++) {
            if (arr[i] == i)
                return i;
        }

        /* If no fixed point present
           then return -1 */
        return -1;
    }
    // main function
    public static void main(String args[])
    {
        int arr[] = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
        int n = arr.length;
        System.out.println("Fixed Point is "
                           + linearSearch(arr, n));
    }
}

Python3

# Python program to check fixed point
# in an array using linear search
def linearSearch(arr, n):
    for i in range(n):
        if arr[i] is i:
            return i
    # If no fixed point present then return -1
    return -1

# Driver program to check above functions
arr = [-10, -1, 0, 3, 10, 11, 30, 50, 100]
n = len(arr)
print("Fixed Point is " + str(linearSearch(arr, n)))

# This code is contributed by Pratik Chhajer

C#

// C# program to check fixed point
// in an array using linear search
using System;

class GFG {
    static int linearSearch(int[] arr, int n)
    {
        int i;
        for (i = 0; i < n; i++) {
            if (arr[i] == i)
                return i;
        }

        /* If no fixed point present
        then return -1 */
        return -1;
    }
    // Driver code
    public static void Main()
    {
        int[] arr = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
        int n = arr.Length;
        Console.Write("Fixed Point is " + linearSearch(arr, n));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to check fixed point
// in an array using linear search

function linearSearch($arr, $n)
{
    for($i = 0; $i < $n; $i++)
    {
        if($arr[$i] == $i)
            return $i;
    }

    // If no fixed point present then
    // return -1
    return -1;
}

    // Driver Code
    $arr = array(-10, -1, 0, 3, 10,
                  11, 30, 50, 100);
    $n = count($arr);
    echo "Fixed Point is ".
            linearSearch($arr, $n);

// This code is contributed by Sam007
?>

Javascript

<script>

// JavaScript program to check fixed point
// in an array using linear search

    function linearSearch(arr, n)
    {
        let i;
        for(i = 0; i < n; i++)
        {
            if(arr[i] == i)
                return i;
        }

        /* If no fixed point present
        then return -1 */
        return -1;
    }

// Driver Code
    let arr = [-10, -1, 0, 3, 10, 11, 30, 50, 100];
    let n = arr.length;
    document.write("Fixed Point is "
                + linearSearch(arr, n));

</script>

Output:

Fixed Point is 3

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

Method 2 (Binary Search)
First check whether middle element is Fixed Point or not. If it is, then return it; otherwise if the index of middle + 1 element is less than or equal to the value at the high index, then Fixed Point(s) might lie on the right side of the middle point (obviously only if there is a Fixed Point). Similarly, check if the index of middle – 1 element is greater than or equal to the value at the low index, then Fixed Point(s) might lie on the left side of the middle point. 

C++

// C++ program to check fixed point
// in an array using binary search
#include <bits/stdc++.h>
using namespace std;

int binarySearch(int arr[], int low, int high)
{
    if (high >= low) {
        int mid = low + (high - low) / 2;
        if (mid == arr[mid])
            return mid;
        int res = -1;
        if (mid + 1 <= arr[high])
            res = binarySearch(arr, (mid + 1), high);
        if (res != -1)
            return res;
        if (mid - 1 >= arr[low])
            return binarySearch(arr, low, (mid - 1));
    }

    /* Return -1 if there is no Fixed Point */
    return -1;
}

/* Driver code */
int main()
{
    int arr[10] = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Fixed Point is " << binarySearch(arr, 0, n - 1);
    return 0;
}
// This code is contributed by Ashutosh Singh

C

// C program to check fixed point
// in an array using binary search
#include <stdio.h>

int binarySearch(int arr[], int low, int high)
{
    if (high >= low) {
        int mid = low + (high - low) / 2;
        if (mid == arr[mid])
            return mid;
        int res = -1;
        if (mid + 1 <= arr[high])
            res = binarySearch(arr, (mid + 1), high);
        if (res != -1)
            return res;
        if (mid - 1 >= arr[low])
            return binarySearch(arr, low, (mid - 1));
    }

    /* Return -1 if there is no Fixed Point */
    return -1;
}

/* Driver program to check above functions */
int main()
{
    int arr[10] = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Fixed Point is %d", binarySearch(arr, 0, n - 1));
    getchar();
    return 0;
}

Java

// Java program to check fixed point
// in an array using binary search

class Main {

    static int binarySearch(int arr[], int low, int high)
    {
        if (high >= low) {
            int mid = low + (high - low) / 2;
            if (mid == arr[mid])
                return mid;
            int res = -1;
            if (mid + 1 <= arr[high])
                res = binarySearch(arr, (mid + 1), high);
            if (res != -1)
                return res;
            if (mid - 1 >= arr[low])
                return binarySearch(arr, low, (mid - 1));
        }

        /* Return -1 if there is no Fixed Point */
        return -1;
    }

    // main function
    public static void main(String args[])
    {
        int arr[] = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
        int n = arr.length;
        System.out.println("Fixed Point is "
                           + binarySearch(arr, 0, n - 1));
    }
}

Python3

# Python program to check fixed point
# in an array using binary search
def binarySearch(arr, low, high):
    if high >= low :

        mid = low + (high - low)//2
        if mid == arr[mid]:
            return mid
        res = -1
        if mid + 1 <= arr[high]:
            res = binarySearch(arr, (mid + 1), high)
        if res !=-1:
            return res
        if mid-1 >= arr[low]:
            return binarySearch(arr, low, (mid -1))


    # Return -1 if there is no Fixed Point
    return -1




# Driver program to check above functions */
arr = [-10, -1, 0, 3, 10, 11, 30, 50, 100]
n = len(arr)
print("Fixed Point is " + str(binarySearch(arr, 0, n-1)))

C#

// C# program to check fixed point
// in an array using binary search
using System;

class GFG {

    static int binarySearch(int[] arr, int low, int high)
    {
        if (high >= low) {
            int mid = low + (high - low) / 2;
            if (mid == arr[mid])
                return mid;
            int res = -1;
            if (mid + 1 <= arr[high])
                res = binarySearch(arr, (mid + 1), high);
            if (res != -1)
                return res;
            if (mid - 1 >= arr[low])
                return binarySearch(arr, low, (mid - 1));
        }

        /* Return -1 if there is no Fixed Point */
        return -1;
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { -10, -1, 0, 3, 10, 11, 30, 50, 100 };
        int n = arr.Length;
        Console.Write("Fixed Point is " + binarySearch(arr, 0, n - 1));
    }
}

PHP

<?php
// PHP program to check fixed point
// in an array using binary search


function binarySearch($arr, $low, $high)
{
    if($high >= $low)
     {
       $mid = (int)($low + ($high - $low)/2);
       if($mid == $arr[$mid])
      return $mid;
       $res = -1;
       if($mid+1 <= $arr[$high])
      $res = binarySearch($arr, ($mid + 1), $high);
       if($res!=-1)
          return $res;
       if($mid-1 >= $arr[$low])
      return binarySearch($arr, $low, ($mid -1));
     }

     /* Return -1 if there is no Fixed Point */
     return -1;
}



    // Driver Code
    $arr = array(-10, -1, 0, 3, 10,
                  11, 30, 50, 100);
    $n = count($arr);
    echo "Fixed Point is: "
        . binarySearch($arr, 0, $n - 1);

?>

Javascript

<script>

// javascript program to check fixed point
// in an array using binary search



function binarySearch(arr,low,high)
{
    if(high >= low)
     {
       let mid = math.floor(low + (high - low)/2);
       if(mid == arr[mid])
      return mid;
       let res = -1;
       if(mid+1 <= arr[high])
      res = binarySearch(arr, (mid + 1), high);
       if(res!=-1)
          return res;
       if(mid-1 >= arr[low])
      return binarySearch(arr, low, (mid -1));
     }

     /* Return -1 if there is no Fixed Point */
     return -1;
}



    // Driver program
    let arr = [-10, -1, 0, 3, 10, 11, 30, 50, 100];
    let n = arr.length;
    document.write("Fixed Point is "+ binarySearch(arr, 0, n-1));

</script>

Output: 

Fixed Point is 3

Algorithmic Paradigm: Divide & Conquer 

Time Complexity: O(log n)

Auxiliary Space: O(log n) (As implicit stack is used for recursive calls)

Find a Fixed Point (Value equal to index) in a given array | Duplicates Allowed
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