Fencepost (AKA An off-by-one error (OBOE), also commonly known as an OBOB (off-by-one bug).
Given an array, I would generally iterate through index 0 to array.length() (Half-open interval).
I've noticed that certain versions of merge-sort require the mid value to be (start+end)/2. And when you calculate the number of elements in the merge process then we sometimes refer to using (end - start) as number of elements or (end - mid + 1). I am not able to intuitively get this? I somehow have difficulty understanding this and feel like am mugging up each time I look at any new implementation.
Can someone provide an intuitive way to understanding how I can apply/recognize a fencepost problem? Is this the same for multi-dimensional arrays?
public static BigInteger mergeSort(int[] integerArray, int start, int end) {
    if (start >= end) { // less than equal to is important for cases when start = end = 0
        return BigInteger.valueOf(0);
    }
    int middle = (start + end) / 2;
    BigInteger numInv1 = mergeSort(integerArray, start, middle);
    BigInteger numInv2 = mergeSort(integerArray, middle + 1, end);
    BigInteger numInv3 = runMerge(integerArray, start, middle, end);
    return numInv1.add(numInv2).add(numInv3);
}
private static BigInteger runMerge(int[] integerArray,
                                   int start, int middle, int end) {
    BigInteger numInv = BigInteger.valueOf(0);
    int n1 = middle - start + 1;
    /*
    number of elements in 1st array is middle - start + 1. why ?
    */
    int n2 = end - middle;       // number of elements in 2nd array
    /*
    number of elements in 2nd array is end - middle. why ?
    */
    int []p = new int[n1];
    int []q = new int[n2];
    int i, j, k;
    for (i = 0; i < n1 ; i++) {
        p[i] = integerArray[start + i];
    }
    for (j = 0; j < n2; j++) {
        q[j] = integerArray[middle + j + 1];
        //Why do we do +1 here? because we use 2nd array for mid+1 till end elements
    }
    i = 0;
    j = 0;
    k = start;
    while ( i < n1 && j < n2) {
        if (p[i] <= q[j]) {
            integerArray[k++] = p[i++];
        } else {
            integerArray[k++] = q[j++];
            numInv = numInv.add(BigInteger.valueOf(n1-i));
        }
    }
    while ( i < n1 ) {
        integerArray[k++] = p[i++];
    }
    while ( j < n2 ) {
        integerArray[k++] = q[j++];
    }
    return numInv;
}
 
    