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Added script in JAVA for how many times an array has been rotated in O(log N) #2448

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Merged
siriak merged 3 commits into from
Oct 4, 2021
Merged

Added script in JAVA for how many times an array has been rotated in O(log N) #2448

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Added script in JAVA for how many times an array has been rotated in …
nirmalya8 Oct 3, 2021
8408205
Merge branch 'TheAlgorithms:master' into master
nirmalya8 Oct 4, 2021
dc25cd5
Added the no duplicates constraint
nirmalya8 Oct 4, 2021
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66 changes: 66 additions & 0 deletions Searches/howManyTimesRotated.java
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/*
Problem Statement:
Given an array, find out how many times it has to been rotated
from its initial sorted position.
Input-Output:
Eg. [11,12,15,18,2,5,6,8]
It has been rotated: 4 times
(One rotation means putting the first element to the end)
Note: The array cannot contain duplicates

Logic:
The position of the minimum element will give the number of times the array has been rotated
from its initial sorted position.
Eg. For [2,5,6,8,11,12,15,18], 1 rotation gives [5,6,8,11,12,15,18,2], 2 rotations [6,8,11,12,15,18,2,5] and so on.
Finding the minimum element will take O(N) time but, we can use Binary Search to find the mimimum element, we can reduce the complexity to O(log N).
If we look at the rotated array, to identify the minimum element (say a[i]), we observe that a[i-1]>a[i]<a[i+1].

Some other test cases:
1. [1,2,3,4] Number of rotations: 0 or 4(Both valid)
2. [15,17,2,3,5] Number of rotations: 3
*/

import java.util.*;

class howManyTimesRotated
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] a = new int[n];
for(int i = 0; i < n; i++)
a[i] = sc.nextInt();

System.out.println("The array has been rotated "+rotated(a)+" times");
sc.close();

}

public static int rotated(int[] a)
{
int low = 0;
int high = a.length-1;
int mid=0; // low + (high-low)/2 = (low + high)/2
int i=0;
while(low<=high)
{
mid = low + (high-low)/2;
i++;
if(a[mid]<a[mid-1] && a[mid]<a[mid+1])
{
break;
}
else if(a[mid]>a[mid-1] && a[mid]<a[mid+1])
{
high = mid+1;
}
else if(a[mid]>a[mid-1] && a[mid]>a[mid+1])
{
low = mid-1;
}
}

return mid;
}
}