The problem I am talking about is this .

Problem statment:

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

The solution I got was a little messy, any cleaner solutions are welcome.

1. No other helper methods.
2. Do not use Java.util.Arrays.copyOf or any other utility under Arrays
3. Do not use collections.

Below is my solution:

The problem I am talking about is this:

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

1. No other helper methods.
2. Do not use Java.util.Arrays.copyOf or any other utility under Arrays
3. Do not use collections.

The solution I got was a little messy, and any cleaner solutions are welcome.

The problem I am talking about is this.

Problem statment:

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

The solution I got was a little messy, any cleaner solutions are welcome.

Conditions for solving:

1. No other helper methods.
2. Do not use Java.util.Arrays.copyOf or any other utility under Arrays
3. Do not use collections.

Below is my solution:

public int maxMirror(int[] nums) {
 final int len=nums.length;
 if(len==0)
 return 0;
 int maxCount=1;
 boolean flag=false;
 
 for(int i=0;i<len;i++)
 {
 int tempCount=1;
 int count=i;
 
 for(int j=len-1;j>=0&&(count<len);j--)
 {
 if((nums[count]==nums[j])&&!(flag))
 {
 flag=true;
 count++;
 continue;
 }
 if((nums[count]==nums[j])&&(flag))
 {
 tempCount++;
 count++;
 maxCount=(tempCount>maxCount)?tempCount:maxCount;
 continue;
 }
 if(!(nums[i]==nums[j])&&(flag))
 {
 flag=false;
 count=i;
 tempCount=1;
 continue;
 }
 if((j==count)||(j-count)==1)
 {
 flag=false;
 break;
 }
 
 }
 } 
 return maxCount; 
 
}

The problem I am talking about is this.

Problem statment:

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

The solution I got was a little messy, any cleaner solutions are welcome.

Conditions for solving:

1. No other helper methods.
2. Do not use Java.util.Arrays.copyOf or any other utility under Arrays
3. Do not use collections.

Below is my solution:

public int maxMirror(int[] nums) {
 final int len=nums.length;
 if(len==0)
 return 0;
 int maxCount=1;
 boolean flag=false;
 
 for(int i=0;i<len;i++)
 {
 int tempCount=1;
 int count=i;
 flag=false;

 for(int j=len-1;j>=0&&(count<len);j--)
 {
 if((nums[count]==nums[j])&&!(flag))
 {
 flag=true;
 count++;
 continue;
 }
 if((nums[count]==nums[j])&&(flag))
 {
 tempCount++;
 count++;
 maxCount=(tempCount>maxCount)?tempCount:maxCount;
 continue;
 }
 if(!(nums[i]==nums[j])&&(flag))
 {
 flag=false;
 count=i;
 tempCount=1;
 continue;
 }
 if((j==count)||(j-count)==1)
 {
 flag=false;
 break;
 }
 
 }
 } 
 return maxCount; 
 
}