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Jump Search

  • Difficulty Level : Easy
  • Last Updated : 22 Mar, 2021

Like Binary Search, Jump Search is a searching algorithm for sorted arrays. The basic idea is to check fewer elements (than linear search) by jumping ahead by fixed steps or skipping some elements in place of searching all elements.
For example, suppose we have an array arr[] of size n and block (to be jumped) size m. Then we search at the indexes arr[0], arr[m], arr[2m]…..arr[km] and so on. Once we find the interval (arr[km] < x < arr[(k+1)m]), we perform a linear search operation from the index km to find the element x.
Let’s consider the following array: (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610). Length of the array is 16. Jump search will find the value of 55 with the following steps assuming that the block size to be jumped is 4. 
STEP 1: Jump from index 0 to index 4; 
STEP 2: Jump from index 4 to index 8; 
STEP 3: Jump from index 8 to index 12; 
STEP 4: Since the element at index 12 is greater than 55 we will jump back a step to come to index 8. 
STEP 5: Perform linear search from index 8 to get the element 55.
What is the optimal block size to be skipped? 
In the worst case, we have to do n/m jumps and if the last checked value is greater than the element to be searched for, we perform m-1 comparisons more for linear search. Therefore the total number of comparisons in the worst case will be ((n/m) + m-1). The value of the function ((n/m) + m-1) will be minimum when m = √n. Therefore, the best step size is m = √n.
 

C++




// C++ program to implement Jump Search
 
#include <bits/stdc++.h>
using namespace std;
 
int jumpSearch(int arr[], int x, int n)
{
    // Finding block size to be jumped
    int step = sqrt(n);
 
    // Finding the block where element is
    // present (if it is present)
    int prev = 0;
    while (arr[min(step, n)-1] < x)
    {
        prev = step;
        step += sqrt(n);
        if (prev >= n)
            return -1;
    }
 
    // Doing a linear search for x in block
    // beginning with prev.
    while (arr[prev] < x)
    {
        prev++;
 
        // If we reached next block or end of
        // array, element is not present.
        if (prev == min(step, n))
            return -1;
    }
    // If element is found
    if (arr[prev] == x)
        return prev;
 
    return -1;
}
 
// Driver program to test function
int main()
{
    int arr[] = { 0, 1, 1, 2, 3, 5, 8, 13, 21,
                34, 55, 89, 144, 233, 377, 610 };
    int x = 55;
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Find the index of 'x' using Jump Search
    int index = jumpSearch(arr, x, n);
 
    // Print the index where 'x' is located
    cout << "\nNumber " << x << " is at index " << index;
    return 0;
}
 
// Contributed by nuclode

Java




// Java program to implement Jump Search.
public class JumpSearch
{
    public static int jumpSearch(int[] arr, int x)
    {
        int n = arr.length;
 
        // Finding block size to be jumped
        int step = (int)Math.floor(Math.sqrt(n));
 
        // Finding the block where element is
        // present (if it is present)
        int prev = 0;
        while (arr[Math.min(step, n)-1] < x)
        {
            prev = step;
            step += (int)Math.floor(Math.sqrt(n));
            if (prev >= n)
                return -1;
        }
 
        // Doing a linear search for x in block
        // beginning with prev.
        while (arr[prev] < x)
        {
            prev++;
 
            // If we reached next block or end of
            // array, element is not present.
            if (prev == Math.min(step, n))
                return -1;
        }
 
        // If element is found
        if (arr[prev] == x)
            return prev;
 
        return -1;
    }
 
    // Driver program to test function
    public static void main(String [ ] args)
    {
        int arr[] = { 0, 1, 1, 2, 3, 5, 8, 13, 21,
                    34, 55, 89, 144, 233, 377, 610};
        int x = 55;
 
        // Find the index of 'x' using Jump Search
        int index = jumpSearch(arr, x);
 
        // Print the index where 'x' is located
        System.out.println("\nNumber " + x +
                            " is at index " + index);
    }
}

Python3




# Python3 code to implement Jump Search
import math
 
def jumpSearch( arr , x , n ):
     
    # Finding block size to be jumped
    step = math.sqrt(n)
     
    # Finding the block where element is
    # present (if it is present)
    prev = 0
    while arr[int(min(step, n)-1)] < x:
        prev = step
        step += math.sqrt(n)
        if prev >= n:
            return -1
     
    # Doing a linear search for x in
    # block beginning with prev.
    while arr[int(prev)] < x:
        prev += 1
         
        # If we reached next block or end
        # of array, element is not present.
        if prev == min(step, n):
            return -1
     
    # If element is found
    if arr[int(prev)] == x:
        return prev
     
    return -1
 
# Driver code to test function
arr = [ 0, 1, 1, 2, 3, 5, 8, 13, 21,
    34, 55, 89, 144, 233, 377, 610 ]
x = 55
n = len(arr)
 
# Find the index of 'x' using Jump Search
index = jumpSearch(arr, x, n)
 
# Print the index where 'x' is located
print("Number" , x, "is at index" ,"%.0f"%index)
 
# This code is contributed by "Sharad_Bhardwaj".

C#




// C# program to implement Jump Search.
using System;
public class JumpSearch
{
    public static int jumpSearch(int[] arr, int x)
    {
        int n = arr.Length;
 
        // Finding block size to be jumped
        int step = (int)Math.Floor(Math.Sqrt(n));
 
        // Finding the block where element is
        // present (if it is present)
        int prev = 0;
        while (arr[Math.Min(step, n)-1] < x)
        {
            prev = step;
            step += (int)Math.Floor(Math.Sqrt(n));
            if (prev >= n)
                return -1;
        }
 
        // Doing a linear search for x in block
        // beginning with prev.
        while (arr[prev] < x)
        {
            prev++;
 
            // If we reached next block or end of
            // array, element is not present.
            if (prev == Math.Min(step, n))
                return -1;
        }
 
        // If element is found
        if (arr[prev] == x)
            return prev;
 
        return -1;
    }
 
    // Driver program to test function
    public static void Main()
    {
        int[] arr = { 0, 1, 1, 2, 3, 5, 8, 13, 21,
                    34, 55, 89, 144, 233, 377, 610};
        int x = 55;
 
        // Find the index of 'x' using Jump Search
        int index = jumpSearch(arr, x);
 
        // Print the index where 'x' is located
        Console.Write("Number " + x +
                            " is at index " + index);
    }
}

PHP




<?php
// PHP program to implement Jump Search
 
function jumpSearch($arr, $x, $n)
{
    // Finding block size to be jumped
    $step = sqrt($n);
 
    // Finding the block where element is
    // present (if it is present)
    $prev = 0;
    while ($arr[min($step, $n)-1] < $x)
    {
        $prev = $step;
        $step += sqrt($n);
        if ($prev >= $n)
            return -1;
    }
 
    // Doing a linear search for x in block
    // beginning with prev.
    while ($arr[$prev] < $x)
    {
        $prev++;
 
        // If we reached next block or end of
        // array, element is not present.
        if ($prev == min($step, $n))
            return -1;
    }
    // If element is found
    if ($arr[$prev] == $x)
        return $prev;
 
    return -1;
}
 
// Driver program to test function
$arr = array( 0, 1, 1, 2, 3, 5, 8, 13, 21,
                34, 55, 89, 144, 233, 377, 610 );
$x = 55;
$n = sizeof($arr) / sizeof($arr[0]);
     
// Find the index of '$x' using Jump Search
$index = jumpSearch($arr, $x, $n);
 
// Print the index where '$x' is located
echo "Number ".$x." is at index " .$index;
return 0;
?>

Javascript




<script>
// Javascript program to implement Jump Search
 
function jumpSearch(arr, x, n)
{
    // Finding block size to be jumped
    let step = Math.sqrt(n);
   
    // Finding the block where element is
    // present (if it is present)
    let prev = 0;
    while (arr[Math.min(step, n)-1] < x)
    {
        prev = step;
        step += Math.sqrt(n);
        if (prev >= n)
            return -1;
    }
   
    // Doing a linear search for x in block
    // beginning with prev.
    while (arr[prev] < x)
    {
        prev++;
   
        // If we reached next block or end of
        // array, element is not present.
        if (prev == Math.min(step, n))
            return -1;
    }
    // If element is found
    if (arr[prev] == x)
        return prev;
   
    return -1;
}
 
// Driver program to test function
let arr = [0, 1, 1, 2, 3, 5, 8, 13, 21,
                34, 55, 89, 144, 233, 377, 610];
let x = 55;
let n = arr.length;
       
// Find the index of 'x' using Jump Search
let index = jumpSearch(arr, x, n);
   
// Print the index where 'x' is located
document.write(`Number ${x} is at index ${index}`);
   
// This code is contributed by _saurabh_jaiswal
</script>

Output: 
 

Number 55 is at index 10

Time Complexity : O(√n) 
Auxiliary Space : O(1)
Important points: 
 

  • Works only sorted arrays.
  • The optimal size of a block to be jumped is (√ n). This makes the time complexity of Jump Search O(√ n).
  • The time complexity of Jump Search is between Linear Search ( ( O(n) ) and Binary Search ( O (Log n) ).
  • Binary Search is better than Jump Search, but Jump search has an advantage that we traverse back only once (Binary Search may require up to O(Log n) jumps, consider a situation where the element to be searched is the smallest element or smaller than the smallest). So in a system where binary search is costly, we use Jump Search.

 



References: 
https://en.wikipedia.org/wiki/Jump_search
This article is contributed by Harsh Agarwal. 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.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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