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maxMirror challenge

Posted on 2016-08-14
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Last Modified: 2016-08-23
Hi,

I am working on below challenge

http://codingbat.com/prob/p196409

Array-3 > maxMirror
prev  |  next  |  chance
We'll say that a "mirror" section in an array is a group of contiguous elements such that somewhere in the array, the same group appears in reverse order. For example, the largest mirror section in {1, 2, 3, 8, 9, 3, 2, 1} is length 3 (the {1, 2, 3} part). Return the size of the largest mirror section found in the given array.

maxMirror([1, 2, 3, 8, 9, 3, 2, 1]) → 3
maxMirror([1, 2, 1, 4]) → 3
maxMirror([7, 1, 2, 9, 7, 2, 1]) → 2

i have not completely understood the description.

How below is 3

maxMirror([1, 2, 1, 4]) → 3
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Question by:gudii9
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10 Comments
 
LVL 37

Expert Comment

by:Gerwin Jansen
Comment Utility
1 2 1 is a 'mirror' that is 3 characters long
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LVL 7

Author Comment

by:gudii9
Comment Utility
maxMirror([1, 2, 3, 8, 9, 3, 2, 1]) → 3//1,2,3  has mirror image of 3,2,1 with respect to mirror 8,9

maxMirror([7, 1, 2, 9, 7, 2, 1]) → 2// 1,2 has mirror image of 2,1 with respect to mirror 9.7

maxMirror([1, 2, 1, 4]) → 1,2,1 is base image right where is the mirror image and with respect to what mirror?
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Assisted Solution

by:Gerwin Jansen
Gerwin Jansen earned 250 total points
Comment Utility
Last one has no real mirror. Real description carefully. 121 and 121 are the same from left to right and from right to left.... So length is 3
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Accepted Solution

by:
d-glitch earned 250 total points
Comment Utility
The center can be outside of the sequence, or inside.
Write the array forward and reverse, and look for the longest common sequence.

FWD = [7, 1, 2, 9, 7, 2, 1]         REV = [1, 2, 7, 9, 2, 1, 7]
Note that there are actually two sequences of length 2.

FWD = [1, 2, 1, 4]                      REV = [4, 1, 2, 1]
Here there is one sequence of length 3.
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Author Comment

by:gudii9
Comment Utility
not able to get idea on approach to this challenge?
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LVL 37

Expert Comment

by:Gerwin Jansen
Comment Utility
Pseudo code! Write down what you would do to recognise a 'mirror'.

Look at the first example, I would start at the beginning, I find a 1, then also start and the end and walk back to be begin, see if I can find another 1. If not or if it's the same, then it's no mirror. If there is another 1, then at the beginning, increase one step and repeat from the end where you've found the other 1.
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LVL 27

Expert Comment

by:d-glitch
Comment Utility
You could also turn this into pseudo code:
Write the array forward and reverse, and look for the longest common sequence.
It doesn't say so it in the challenge, but every element in an array will be a mirror sequence of length 1.
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Author Comment

by:gudii9
Comment Utility
It doesn't say so it in the challenge, but every element in an array will be a mirror sequence of length 1.
i was not clear on above. can you please elaborate on it? How every element in array is mirror sequence of length 1?
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LVL 7

Author Comment

by:gudii9
Comment Utility
public int maxMirror(int[] nums) {

		int len = nums.length;
		int total = 0;
		int highest = 0;
		for (int x = 0; x < len; x++) {
			total = 0;
			for (int j = len - 1; x + total < len && j > -1; j--) {
				if (nums[x + total] == nums[j]) {
					total++;
				} else {
					if (total > 0) {
						highest = Math.max(total, highest);
						total = 0;
					}
				}
			}
			highest = Math.max(total, highest);
		}
		return highest;
}

Open in new window


above passed all tests
Expected      Run            
maxMirror([1, 2, 3, 8, 9, 3, 2, 1]) → 3      3      OK      
maxMirror([1, 2, 1, 4]) → 3      3      OK      
maxMirror([7, 1, 2, 9, 7, 2, 1]) → 2      2      OK      
maxMirror([21, 22, 9, 8, 7, 6, 23, 24, 6, 7, 8, 9, 25, 7, 8, 9]) → 4      4      OK      
maxMirror([1, 2, 1, 20, 21, 1, 2, 1, 2, 23, 24, 2, 1, 2, 1, 25]) → 4      4      OK      
maxMirror([1, 2, 3, 2, 1]) → 5      5      OK      
maxMirror([1, 2, 3, 3, 8]) → 2      2      OK      
maxMirror([1, 2, 7, 8, 1, 7, 2]) → 2      2      OK      
maxMirror([1, 1, 1]) → 3      3      OK      
maxMirror([1]) → 1      1      OK      
maxMirror([]) → 0      0      OK      
maxMirror([9, 1, 1, 4, 2, 1, 1, 1]) → 3      3      OK      
maxMirror([5, 9, 9, 4, 5, 4, 9, 9, 2]) → 7      7      OK      
maxMirror([5, 9, 9, 6, 5, 4, 9, 9, 2]) → 2      2      OK      
maxMirror([5, 9, 1, 6, 5, 4, 1, 9, 5]) → 3      3      OK      
other tests
OK      

for every i in forward direction checking every j in reverse. if equal increasing the total variable the highests

package com.solution;

public class MaxMirror {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] ar = { 1, 2, 3, 8, 9, 3, 2, 1 };
		int[] ar1 = { 1, 2, 1, 4};
		//System.out.println("value===>" + maxMirror(ar));
		System.out.println("value===>" + maxMirror(ar1));
	}

	public static int maxMirror(int[] nums) {
		int len = nums.length;
		int total = 0;
		int highest = 0;
		for (int x = 0; x < len; x++) {
			total = 0;
			for (int j = len - 1; x + total < len && j > -1; j--) {
				if (nums[x + total] == nums[j]) {
					total++;
				} else {
					if (total > 0) {
						highest = Math.max(total, highest);
						total = 0;
					}
				}
			}
			highest = Math.max(total, highest);
		}
		return highest;
	}

}

Open in new window


value===>3




i       j     highest
0     3      

0     2      

0     1      
0     0   3

1    3    

1     2      

1     1      

1     0       3



2    3    

2     2      

2     1      

2     0       3





3   3    

3     2      

3     1      

3     0       3
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LVL 7

Author Comment

by:gudii9
Comment Utility
any improvements or alternate approaches?
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