Verbose Solution - Java - O(n) time over all calls - O(1) space
Hop over the count of the numbers on even positions until n > 0 \nwhen n <= 0 return the odd position number next to the current location. \n\n\nclass RLEIterat
Hop over the count of the numbers on even positions until `n > 0` \nwhen `n <= 0` return the odd position number next to the current location. \n\n```\nclass RLEIterator {\n    int index; \n    int[] a; \n    \n    public RLEIterator(int[] A) {\n        a = A; \n        index = 0; \n    }\n    \n    public int next(int n) {\n        for(int i = index; i < a.length; i+=2, index=i){\n            if(n <= a[i]){\n                a[i] -= n;\n                return a[i+1]; \n            }else{\n                n -= a[i]; \n            }\n        }\n        return -1; \n    }\n}\n```
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