maxMirror challenge

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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gudii9Asked:
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Gerwin Jansen, EE MVETopic Advisor Commented:
1 2 1 is a 'mirror' that is 3 characters long
gudii9Author Commented:
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?
Gerwin Jansen, EE MVETopic Advisor Commented:
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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d-glitchCommented:
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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gudii9Author Commented:
not able to get idea on approach to this challenge?
Gerwin Jansen, EE MVETopic Advisor Commented:
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.
d-glitchCommented:
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.
gudii9Author Commented:
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?
gudii9Author Commented:
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
gudii9Author Commented:
any improvements or alternate approaches?
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