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# 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
JavaJava EEProgrammingProgramming Languages-OtherProgramming Theory Last Comment
gudii9

8/22/2022 - Mon
Gerwin Jansen

1 2 1 is a 'mirror' that is 3 characters long
gudii9

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?
SOLUTION
Gerwin Jansen

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d-glitch

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gudii9

not able to get idea on approach to this challenge?
Gerwin Jansen

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-glitch

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.
gudii9

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?
gudii9

``````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;
}
``````

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      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;
}

}
``````

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
gudii9