Open in App
Not now

# Efficient search in an array where difference between adjacent is 1

• Difficulty Level : Easy
• Last Updated : 14 Dec, 2022

Given an array of n integers. Each array element is obtained by adding either +1 or -1 to previous element i.e absolute difference between any two consecutive elements is 1. The task is to search an element index with the minimum number of comparison (less than simple element by element search). If the element is present multiple time, then print the smallest index. If the element is not present print -1.
Examples:

Input : arr[] = {5, 4, 5, 6, 5, 4, 3, 2}
x = 4.
Output : 1
The first occurrence of element x is at
index 1.
Input : arr[] = { 5, 4, 5, 6, 4, 3, 2, 3 }
x = 9.
Output : -1
Element x is not present in arr[]

Let element to be search is x. At any index i, if arr[i] != x, the possibility of x to be present is at location i + abs(arr[i] – a), since each element is obtained by adding either +1 or -1 to the previous element. There is no possibility of having el between i and i + abs(arr[i] – a). So directly jump to i + abs(arr[i] – a), if arr[i] != x.

Algorithm to solve the problem:
1. Start from index = 0.
2. Compare arr[index] and x.
a) If both are equal, return index.
b) If not, set index = index + abs(arr[index] - x).
3. Repeat step 2.

Below is the implementation of the above idea :

## C++

 // C++ program to search an element in an array // where each element is obtained by adding // either +1 or -1 to previous element. #include using namespace std;   // Return the index of the element to be searched. int search(int arr[], int n, int x) {     // Searching x in arr[0..n-1]     int i = 0;     while (i <= n-1)     {         // Checking if element is found.         if (arr[i] == x)             return i;           // Else jumping to abs(arr[i]-x).         i += abs(arr[i]-x);     }       return -1; }   // Driven Program int main() {     int arr[] =  {5, 4, 5, 6, 5, 4, 3, 2};     int n = sizeof(arr)/sizeof(arr[0]);     int x = 4;       cout << search(arr, n, x) << endl;       return 0; }

## Java

 // Java program to search an element // in an array where each element is // obtained by adding either +1 or // -1 to previous element. import java.io.*; class GFG {       // Return the index of the // element to be searched. static int search(int arr[], int n, int x) {     // Searching x in arr[0..n-1]     int i = 0;     while (i <= n-1)     {         // Checking if element is found.         if (arr[i] == x)             return i;           // Else jumping to abs(arr[i]-x).         i += Math.abs(arr[i]-x);     }       return -1; }   // Driver code public static void main (String[] args) {     int arr[] = {5, 4, 5, 6, 5, 4, 3, 2};     int n = arr.length;     int x = 6;       System.out.println(search(arr, n, x)); } }   // This code is contributed by Anant Agarwal.

## Python3

 # Python program to search an element in # an array where each element is obtained # by adding either +1 or -1 to previous element   # Return the index of the element to be searched def search(arr, n, x):       # Searching x in arr[0..n-1]     i = 0     while (i <= n-1):               # Checking if element is found.         if (arr[i] == x):             return i           # Else jumping to abs(arr[i]-x).         i += abs(arr[i] - x)           return -1   # Driver code arr = [5, 4, 5, 6, 5, 4, 3, 2] n = len(arr) x = 4   print(search(arr, n, x))   # This code is contributed by Anant Agarwal.

## C#

 // C# program to search an element in // an array where each element is // obtained by adding either + 1 or // -1 to previous element. using System;   class GFG {       // Return the index of the // element to be searched. static int search(int []arr, int n,                   int x) {           // Searching x in arr[0.. n - 1]     int i = 0;     while (i <= n - 1)     {         // Checking if element is found         if (arr[i] == x)             return i;           // Else jumping to abs(arr[i] - x)         i += Math.Abs(arr[i] - x);     }       return -1; }   // Driver code public static void Main () {     int []arr = {5, 4, 5, 6, 5, 4, 3, 2};     int n = arr.Length;     int x = 4;       Console.WriteLine(search(arr, n, x)); } }   // This code is contributed by vt_m.



## Javascript



Output

1

Time Complexity: O(n), where n is the size of the given array
Auxiliary Space: O(1), as no extra space is required

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