//then I turned to Discuss and found this post: https://discuss.leetcode.com/topic/55870/share-my-solutions-for-contest-2

//instead of literally removing half of the elements in each scan, this solution is just moving the pointer to point to next start position

//So brilliant!

public int lastRemaining(int n) {

int remaining = n;

int start = 1;

int step = 2;

boolean forward = true;

while (remaining > 1) {

remaining /= 2;

if (forward) {

start = start + step * remaining - step / 2;

} else {

start = start - step * remaining + step / 2;

}

step *= 2;

forward = !forward;

}

return start;

}

//I tried brute force, all producing the correct output, but got TLE by OJ.

public static int lastRemaining_brute_force_TLE(int n) {

List<Integer> list = new ArrayList();

for (int i = 0; i < n; i++) {

list.add(i + 1);

}

boolean forward = true;

while (list.size() > 1) {

int size = list.size() / 2;

if (list.size() == 1) {

return list.get(0);

}

if (forward) {

if (list.size() == 1) {

return list.get(0);

for (int i = 0; i <= size && i < list.size(); i++) {

list.remove(i);

}

forward = false;

} else {

if (list.size() == 1) {

return list.get(0);

}

for (int i = list.size() - 1, count = 0; i >= 0 && count <= size; count++) {

list.remove(i);

i -= 2;

}

forward = true;

return list.get(0);

public static void main(String... strings) {

System.out.println(lastRemaining_brute_force_TLE(5204));

System.out.println(lastRemaining_brute_force_TLE(5058));

}