67 lines
2.0 KiB
Java
67 lines
2.0 KiB
Java
/*
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Problem Statement:
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Given an array, find out how many times it has to been rotated
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from its initial sorted position.
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Input-Output:
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Eg. [11,12,15,18,2,5,6,8]
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It has been rotated: 4 times
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(One rotation means putting the first element to the end)
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Note: The array cannot contain duplicates
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Logic:
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The position of the minimum element will give the number of times the array has been rotated
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from its initial sorted position.
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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.
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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).
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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].
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Some other test cases:
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1. [1,2,3,4] Number of rotations: 0 or 4(Both valid)
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2. [15,17,2,3,5] Number of rotations: 3
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*/
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import java.util.*;
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class howManyTimesRotated
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{
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public static void main(String[] args)
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{
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Scanner sc = new Scanner(System.in);
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int n = sc.nextInt();
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int[] a = new int[n];
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for(int i = 0; i < n; i++)
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a[i] = sc.nextInt();
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System.out.println("The array has been rotated "+rotated(a)+" times");
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sc.close();
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}
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public static int rotated(int[] a)
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{
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int low = 0;
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int high = a.length-1;
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int mid=0; // low + (high-low)/2 = (low + high)/2
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int i=0;
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while(low<=high)
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{
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mid = low + (high-low)/2;
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i++;
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if(a[mid]<a[mid-1] && a[mid]<a[mid+1])
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{
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break;
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}
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else if(a[mid]>a[mid-1] && a[mid]<a[mid+1])
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{
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high = mid+1;
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}
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else if(a[mid]>a[mid-1] && a[mid]>a[mid+1])
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{
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low = mid-1;
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}
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}
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return mid;
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}
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}
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