maxMirror challenge

1 2 1 is a 'mirror' that is 3 characters long

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?

not able to get idea on approach to this challenge?

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.

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.

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?

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;
}

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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;
	}

}

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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

any improvements or alternate approaches?