CareerCup

public static String remove2Zeroes (final String input) {
        //Example: a3409jd00dk000d
        //Output: a3409jddk000d

        int i =0;
        int len =input.length();
        char c = input.charAt(0);
        int cnt = 0;
        StringBuffer buf = new StringBuffer();
        while( i < len) {
            c = input.charAt(i);
            if( c != '0') {
                if( cnt != 2) {
                    for( int j = 0; j < cnt; ++j) {
                        buf.append('0');
                    }
                }
                buf.append(c);
                cnt = 0;
            }
            else {
                cnt++;
            }
            ++i;
        }

        //Ending zeroes
        if( cnt != 2) {
            for( int j = 0; j < cnt; ++j) {
                buf.append('0');
            }
        }

        return buf.toString();
    }

Here is my solution. Notice that there might be more than one sub arrays on the solution. This solution just take the last one found largest sub-array that satisfy the question (i.e. the one with start index largest). If you want to keep all of them, use HashMap<Integer,Integer> to keep the start/end index pairs of the matched.

public static void main( String ... args) {
        int arr[]= { -2, -3, 4, -1, -2, 1, 5, -3, 2, -2, -3, 6, -1, -2, 1, 5, -3, 2};
        largestSubArrayMatchSum(arr, 2);
    }

    public static void largestSubArrayMatchSum( int[] arr, int num) {

        int sum = 0;
        int i = 0;
        int max = 0;
        int start= 0;
        int j = 0;
        while( j < arr.length) {
            sum += arr[i];
            if( sum >= num) {
                if( sum == num) {
                    int prev = max;
                    max = Math.max(max, i );
                    if( max > prev) {
                        start = j;
                    }
                };
            }
            ++i;
            if( i == arr.length) {
                ++j;
                i = 0;
            }
        }

        if( max != 0) {
            //Found.
            System.out.println("sub-array based-0 index [from - to]:[" + start + " - " + max + "]");
        }
    }

Traverse the linked list and put the node in the stack until null.
Now pop out nodes from the stack and count until n is reached. That the node to look for. If stack is empty before n reached, return NULL