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.
#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;
}
int main(void)
{
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;
}
Output :
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.