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MaxSpan challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p189576
Array-3 > maxSpan
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Consider the leftmost and righmost appearances of some value in an array. We'll say that the "span" is the number of elements between the two inclusive. A single value has a span of 1. Returns the largest span found in the given array. (Efficiency is not a priority.)

maxSpan([1, 2, 1, 1, 3]) → 4//why 4??
maxSpan([1, 4, 2, 1, 4, 1, 4]) → 6//why 6?
maxSpan([1, 4, 2, 1, 4, 4, 4]) → 6

i have not understood above problem description?
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gudii9
Asked:
gudii9
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1 Solution
 
Gerwin Jansen, EE MVETopic Advisor Commented:
First example: the number 1 has the largest span (1,2,1,1). 4 wide.

Second example: numbers 1 and 4 have the largest span.

Third example: the number 4 has the largest span.
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gudii9Author Commented:
i see the point now
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Gerwin Jansen, EE MVETopic Advisor Commented:
OK ;)
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gudii9Author Commented:
maxSpan([1, 2, 1, 1, 3]) → 4

how above is 4 it looks 3 to me as there are 3 ones there?
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Gerwin Jansen, EE MVETopic Advisor Commented:
The '4' span is:
1, 2, 1, 1
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gudii9Author Commented:
The '4' span is:
1, 2, 1, 1
'

i got it.

Not getting idea on psedo code for this challenge?
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gudii9Author Commented:
can i write solution with simgle loop?
psedo code:

0. loop through array
1. find left most appearance of a number and note the index
2. find if that number exist within the loop again from backwards
3. if yes find the index of that.
4. find the span length which is difference of end index- start index
5. return span length
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gudii9Author Commented:
public int maxSpan(int[] nums) {
		int span=1;
		int counter=0;
	    if (nums.length>0) {
	    	for (int i = 0; i < nums.length; i++) {
				for (int k = nums.length-1; k >0; k--) {
					if(nums[k]==nums[i]){
						counter=k-i+1;
						if(counter>span){
							span=counter;
						}
						
					}
				}
			}
	    }
			
		 else {
			span=0;

		}
		return span;
		
}
		

Open in new window


above pass all tests
Expected      Run            
maxSpan([1, 2, 1, 1, 3]) → 4      4      OK      
maxSpan([1, 4, 2, 1, 4, 1, 4]) → 6      6      OK      
maxSpan([1, 4, 2, 1, 4, 4, 4]) → 6      6      OK      
maxSpan([3, 3, 3]) → 3      3      OK      
maxSpan([3, 9, 3]) → 3      3      OK      
maxSpan([3, 9, 9]) → 2      2      OK      
maxSpan([3, 9]) → 1      1      OK      
maxSpan([3, 3]) → 2      2      OK      
maxSpan([]) → 0      0      OK      
maxSpan([1]) → 1      1      OK      
other tests
OK      

public class MaxSpan {

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

	
	public static int maxSpan(int[] nums) {
		int span=1;
		int counter=0;
	    if (nums.length>0) {
	    	for (int i = 0; i < nums.length; i++) {
				for (int k = nums.length-1; k >0; k--) {
					if(nums[k]==nums[i]){
						counter=k-i+1;
						if(counter>span){
							span=counter;
						}
						
					}
				}
			}
	    }
			
		 else {
			span=0;

		}
		return span;
		
}
}

Open in new window

value is4



any improvement, optimization, refinement to my code?
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Gerwin Jansen, EE MVETopic Advisor Commented:
Passes all test - great! Efficiency was not a priority for this one so I'd keep it like this  ;)
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