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Thread: Logic to find adjacency of numbers 0 & 1 in an array

  1. #1
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    Default Logic to find adjacency of numbers 0 & 1 in an array

    [redacted]

    I should mention here that number of elements can be from 1....10'000. And it has to be done as O(N)

    I attempted at the problem like so,

    public static int adjacencies(int[] A) {        
        int count = 0;
        boolean found = false;
     
        for(int i = 0; i < A.length-1; i++) {
            if (A[i] == A[i+1]) {
                System.out.println(A[i] + " " + A[i+1]);
                count++;
            }
            else if (((i + 2) <  A.length-1) && !found) {
                if ((A[i] == 0 && A[i+1] == 1 && A[i+2] == 0) || (A[i] == 1 && A[i+1] == 0 && A[i+2] == 1)) {
                    found = true;
                    count = count + 2;
                }
                else if ((A[i] == 1 && A[i+1] == 0 && A[i+2] == 0) || (A[i] == 0 && A[i+1] == 1 && A[i+2] == 1)) {
                    found = true;
                    count = count + 1;
                }
            }
        }
     
        return count;
    }
     
    public static void main (String[] args) {
        int[] A = new int[] {1,1,0,1,0,0};
        System.out.println(A);
        int count = adjacencies(A);
        System.out.println("Count: " + count);
    }

    To me it looks right and it seems to work. Can someone find a better way of doing this instead of this clunky way ?


  2. #2
    Grand Poobah
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    Default Re: Logic to find adjacency of numbers 0 & 1 in an array

    If all you are doing for now is count adjacent coins then use a for loop and a single if statement.
    for each coin (except the last one) {
        if current coin equals next coin {
            increase count
        }
    }
    Determining which coin to flip is a bit more complicated. You would start by find the longest sequence of coins that are the same.
    Improving the world one idiot at a time!

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