Take every k-th element from the (1 .. n) natural numbers series

Question

For example, we have series 1, 2, 3, 4, 5. We take every 3 element => 3, 1, 5, 2, 4 (chosen element shouldn\'t remain, we can take while series is not empty). Naive implemen

Answer 1

Here is an implementation with an array representation of the binary tree, only storing the size of the left sub-tree as node value. The input array is not actually stored, but silently assumed to be the leaves at the bottom level, below the binary tree:

function josephusPermutation(size, step) {
    var len = 1 << 32 - Math.clz32(size-1), // Smallest power of 2 >= size
        tree = Array(len).fill(0), // Create tree in array representation 
        current = 0, 
        skip = step - 1,
        result = Array(size).fill(0),
        goRight, leftSize, order, i, j;

    // Initialise tree with sizes of left subtrees as node values
    (function init(i) {
        if (i >= len) return +(i - len < size); // Only count when within size
        var left = tree[i] = init(i*2); // recursive, only store left-size
        return left + (left ? init(i*2+1) : 0); // return sum of left and right 
    })(1);

    for (j = 0; j < result.length; j++, size--) {
        current = (current + skip) % size; // keep within range
        order = current;
        for (i = 1; i < len; i = i*2+goRight) {
            leftSize = tree[i];
            goRight = order >= leftSize;
            if (goRight) {
                order -= leftSize; // Moving rightward, counting what is at left side.
            } else {
                tree[i]--; // we will remove value at left side
            }
        }
        result[j] = 1 + i - len;
    }
    return result;
}

var sequence = josephusPermutation(100000, 123456);
console.log(sequence.join(','));