Search an element in an array where difference between adjacent elements is 1

Given an array where difference between adjacent elements is 1, write an algorithm to search for an element in the array and return the position of the element (return the first occurrence).
Examples :

Let element to be searched be x

Input: arr[] = {8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3}     
       x = 3
Output: Element 3 found at index 7

Input: arr[] =  {1, 2, 3, 4, 5, 4}
       x = 5
Output: Element 5 found at index 4 

A Simple Approach is to traverse the given array one by one and compare every element with given element ‘x’. If matches, then return index.

The above solution can be Optimized using the fact that difference between all adjacent elements is 1. The idea is to start comparing from the leftmost element and find the difference between current array element and x. Let this difference be ‘diff’. From the given property of array, we always know that x must be at-least ‘diff’ away, so instead of searching one by one, we jump ‘diff’. 

Below is the implementation of above idea.

C++

// C++ program to search an element in an array where
// difference between all elements is 1
#include <bits/stdc++.h>
using namespace std;

// x is the element to be searched in arr[0..n-1]
int search(int arr[], int n, int x)
{
    // Traverse the given array starting from
    // leftmost element
    int i = 0;
    while (i < n) {
        // If x is found at index i
        if (arr[i] == x)
            return i;

        // Jump the difference between current
        // array element and x
        i = i + abs(arr[i] - x);
    }

    cout << "number is not present!";
    return -1;
}

// Driver program to test above function
int main()
{
    int arr[] = { 8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 3;
    cout << "Element " << x << " is present at index "
         << search(arr, n, 3);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to search an element in an array where
// difference between all elements is 1
#include <stdio.h>
#include <stdlib.h>

// x is the element to be searched in arr[0..n-1]
int search(int arr[], int n, int x)
{
    // Traverse the given array starting from
    // leftmost element
    int i = 0;
    while (i < n) {
        // If x is found at index i
        if (arr[i] == x)
            return i;

        // Jump the difference between current
        // array element and x
        i = i + abs(arr[i] - x);
    }

    printf("number is not present!");
    return -1;
}

// Driver program to test above function
int main()
{
    int arr[] = { 8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 3;
    printf("Element %d is present at index ",
           search(arr, n, 3));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to search an element in an
// array where difference between all
// elements is 1

import java.io.*;

class GFG {

    // x is the element to be searched
    // in arr[0..n-1]
    static int search(int arr[], int n, int x)
    {

        // Traverse the given array starting
        // from leftmost element
        int i = 0;
        while (i < n)
        {

            // If x is found at index i
            if (arr[i] == x)
                return i;

            // Jump the difference between current
            // array element and x
            i = i + Math.abs(arr[i]-x);
        }

        System.out.println ("number is not" +
                                     " present!");

        return -1;
    }

    // Driver program to test above function
    public static void main (String[] args) {

        int arr[] = {8 ,7, 6, 7, 6, 5, 4, 3,
                                   2, 3, 4, 3 };
        int n = arr.length;
        int x = 3;
        System.out.println("Element " + x +
                        " is present at index "
                            + search(arr,n,3));
    }
}

//This code is contributed by vt_m.

Python 3

# Python 3 program to search an element
# in an array where difference between
# all elements is 1

# x is the element to be searched in
# arr[0..n-1]
def search(arr, n, x):

    # Traverse the given array starting
    # from leftmost element
    i = 0
    while (i < n):

        # If x is found at index i
        if (arr[i] == x):
            return i

        # Jump the difference between
        # current array element and x
        i = i + abs(arr[i] - x)

    print("number is not present!")
    return -1

# Driver program to test above function
arr = [8 ,7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 ]
n = len(arr)
x = 3
print("Element" , x , " is present at index ",
                             search(arr,n,3))

# This code is contributed by Smitha

C#

// C# program to search an element
// in an array where difference
// between all elements is 1
using System;

public class GFG
{

    // in arr[0..n-1]
    static int search(int []arr, int n,
                      int x)
    {

        // Traverse the given array starting
        // from leftmost element
        int i = 0;
        while (i < n)
        {

            // If x is found at index i
            if (arr[i] == x)
                return i;

            // Jump the difference between
            // current array element and x
            i = i + Math.Abs(arr[i] - x);
        }

        Console.WriteLine ("number is not" +
                           " present!");

        return -1;
    }

    // Driver code
    public static void Main()
    {

        int []arr = {8 ,7, 6, 7, 6, 5,
                     4,3, 2, 3, 4, 3 };
        int n = arr.Length;
        int x = 3;
        Console.WriteLine("Element " + x +
                        " is present at index "
                        + search(arr, n, 3));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to search an
// element in an array where
// difference between all
// elements is 1

// x is the element to be
// searched in arr[0..n-1]
function search($arr, $n, $x)
{
    // Traverse the given array
    // starting from leftmost
    // element
    $i = 0;
    while ($i < $n)
    {
        // If x is found at index i
        if ($arr[$i] == $x)
            return $i;

        // Jump the difference
        // between current
        // array element and x
        $i = $i + abs($arr[$i] - $x);
    }

    echo "number is not present!";
    return -1;
}

// Driver Code
$arr = array(8 ,7, 6, 7, 6, 5,
             4, 3, 2, 3, 4, 3);
$n = sizeof($arr);
$x = 3;
echo "Element " , $x , " is present " ,
      "at index ", search($arr, $n, 3);

// This code is contributed
// by nitin mittal.
?>

Javascript

<script>


// Javascript program to search an element in an array where
// difference between all elements is 1

// x is the element to be searched in arr[0..n-1]
function search(arr, n, x)
{

    // Traverse the given array starting from
    // leftmost element
    let i = 0;
    while (i<n)
    {
        // If x is found at index i
        if (arr[i] == x)
            return i;

        // Jump the difference between current
        // array element and x
        i = i + Math.abs(arr[i]-x);
    }

    document.write("number is not present!");
    return -1;
}

    // Driver program
    let arr = [8 ,7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 ];
    let n = arr.length;
    let x = 3;
    document.write( "Element " + x  + " is present at index "
         + search(arr,n,3));

    // This code is contributed by jana_sayantan.
</script>

Output

Element 3 is present at index 7

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

Searching in an array where adjacent differ by at most k
This article is contributed by Rishabh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 

Approach#2: Using linear search

The approach of this code is linear search. It iterates through the array and checks if the current element is equal to the target element. If the target element is found, it returns the index of that element. If the target element is not found, it returns -1.

Algorithm

1. Traverse through the array starting from the first element.
2. Compare each element with the given element x.
3. If the element is found, return its index.
4. If the end of array is reached and the element is not found, return -1

Python3

def search_element(arr, x):
    n = len(arr)
    for i in range(n):
        if arr[i] == x:
            return i
    return -1


arr = [8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3]
x = 3
print(search_element(arr, x))

Javascript

function search_element(arr, x) {
    // Get the length of the array
    let n = arr.length;

    // Iterate over the array and check if the element is equal to x
    for (let i = 0; i < n; i++) {
        if (arr[i] == x) {
            return i;
        }
    }

    // If the element is not found, return -1
    return -1;
}

let arr = [8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3];
let x = 3;
console.log(search_element(arr, x));

C++

#include <iostream>
using namespace std;

// Function to find the first index of the
// element x in the array arr[]
int search_element(int arr[], int n, int x)
{
    for (int i = 0; i < n; i++) {
        if (arr[i] == x) {
            return i;
        }
    }
    return -1;
}

// Driver Code
int main()
{
    int arr[] = { 8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 };
    int x = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << search_element(arr, n, x) << endl;
    return 0;
}

Output

7

Time Complexity: O(n) where n is the length of the array.
Auxiliary Space: O(1)