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# Jump Search

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.

## C++

 // C++ program to implement Jump Search #include 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 functionint 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#includeint 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 Searchimport 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 functionarr = [ 0, 1, 1, 2, 3, 5, 8, 13, 21,    34, 55, 89, 144, 233, 377, 610 ]x = 55n = len(arr) # Find the index of 'x' using Jump Searchindex = jumpSearch(arr, x, n) # Print the index where 'x' is locatedprint("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);    }}

## PHP

 = \$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 locatedecho "Number ".\$x." is at index " .\$index;return 0;?>

## Javascript



Output:

Number 55 is at index 10

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

1. Better than a linear search for arrays where the elements are uniformly distributed.
2. Jump search has a lower time complexity compared to a linear search for large arrays.
3. 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.
4. It is easier to implement compared to other search algorithms like binary search or ternary search.
5. 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