Daily Coding Problem: Sublist sum

My solution to a “Daily Coding Problem” that I received in my mail today.

Given a list of numbers L, implement a method sum(i, j) which returns the sum from the sublist L[i:j] (including i, excluding j).

For example, given L = [1, 2, 3, 4, 5], sum(1, 3) should return sum([2, 3]), which is 5.

You can assume that you can do some pre-processing. sum() should be optimized over the pre-processing step.

Here’s my solution in Typescript,

oneFortyNine(l: number[], i: number,j: number): number {
    if(l == null) {
        return 0;
    }
    if(l.length == 0 || i >= l.length || j >= l.length || j < i) {
        return 0;
    }
    //ok, we aren't avoid checking for i and j being zero
    //this code also assumes i < j
    if(i == (j - 1)) {  
        return l[i];
    }
    return l[i] + this.oneFortyNine(l, i+1, j);
}

I actually, don’t understand the pre-processing part of this code? As in pre-processing how? Also, this problem was labelled as “hard”, so while the tests give the right solution, I wonder do I have the right solution? This was quite easy, and if I had done it iteratively, it would be even easier.

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