Last Updated : 18 Jan, 2024
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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 a block (to be jumped) of size m. Then we search in 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). The length of the array is 16. The 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 a linear search from index 8 to get the element 55.
Performance in comparison to linear and binary search:
If we compare it with linear and binary search then it comes out then it is better than linear search but not better than binary search.
The increasing order of performance is:
linear search < jump search < binary search
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.
Algorithm steps
- Jump Search is an algorithm for finding a specific value in a sorted array by jumping through certain steps in the array.
- The steps are determined by the sqrt of the length of the array.
- Here is a step-by-step algorithm for the jump search:
- Determine the step size m by taking the sqrt of the length of the array n.
- Start at the first element of the array and jump m steps until the value at that position is greater than the target value.
Once a value greater than the target is found, perform a linear search starting from the previous step until the target is found or it is clear that the target is not in the array.
If the target is found, return its index. If not, return -1 to indicate that the target was not found in the array.
Example 1 :
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
C
#include<stdio.h>
#include<math.h>
int
min(
int
a,
int
b){
if
(b>a)
return
a;
else
return
b;
}
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;
}
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]);
int
index = jumpsearch(arr, x, n);
if
(index >= 0)
printf
(
"Number is at %d index"
,index);
else
printf
(
"Number is not exist in the array"
);
return
0;
}
// This code is contributed by Susobhan Akhuli
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
;
for
(
int
minStep = Math.min(step, n)-
1
; arr[minStep] < x; minStep = Math.min(step, n)-
1
)
{
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.Sqrt(n);
// Finding the block where the element is
// present (if it is present)
int
prev = 0;
for
(
int
minStep = Math.Min(step, n)-1; arr[minStep] < x; minStep = Math.Min(step, n)-1)
{
prev = step;
step += (
int
)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);
}
}
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;
for
(int minStep = Math.Min(step, n)-1; arr[minStep] < x; minStep = Math.Min(step, n)-1)
{
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>
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;
?>
Output:
Number 55 is at index 10
Time Complexity : O(?n)
Auxiliary Space : O(1)
Advantages of Jump Search:
- Better than a linear search for arrays where the elements are uniformly distributed.
- Jump search has a lower time complexity compared to a linear search for large arrays.
- The number of steps taken in jump search is proportional to the square root of the size of the array, making it more efficient for large arrays.
- It is easier to implement compared to other search algorithms like binary search or ternary search.
- Jump search works well for arrays where the elements are in order and uniformly distributed, as it can jump to a closer position in the array with each iteration.
Important points:
- Works only with 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 the 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 just bigger than the smallest). So, in a system where binary search is costly, we use Jump Search.
References:
https://en.wikipedia.org/wiki/Jump_search
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