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 functionint 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 functionint 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 1import 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 functionarr = [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 |
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> |
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.?> |
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
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
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 Codeint 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;
} |
Java
public class Main {
// Function to find the first index of the element x in the array arr[]
static int searchElement(int[] arr, int x) {
for (int i = 0; i < arr.length; i++) {
if (arr[i] == x) {
return i;
// Return the index when the element is found
}
}
return -1;
// Return -1 if the element is not found in the array
}
public static void main(String[] args) {
int[] arr = { 8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 };
int x = 3;
int result = searchElement(arr, x);
System.out.println(result);
// Print the result
}
} |
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))
|
C#
using System;
class Program
{ // Function to find the first index of the element x in the array arr[]
static int SearchElement(int[] arr, int x)
{
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] == x)
{
return i;
// Return the index when the element is found
}
}
return -1;
// Return -1 if the element is not found in the array
}
static void Main(string[] args)
{
int[] arr = { 8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 };
int x = 3;
int result = SearchElement(arr, x);
Console.WriteLine(result);
// Print the result
}
} |
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)); |
Output
7
Time Complexity: O(n) where n is the length of the array.
Auxiliary Space: O(1)