int cache[1000] = {0};
cache[1]=1;

#include<stdio.h>

// the cache stores the stopping time for all integers for efficiency
int *cache;

/*
 * returns the stopping time, the amount of times for an integer to reach 1 via the Collatz Conjecture
 * */

int stoppingTime(int n) {
	int tmp = n&1?n*3+1:n>>1;
	if(n > 1000000) return 1 + stoppingTime(tmp);
	if(cache[n-1]) return cache[n-1];
	return (cache[n-1] = 1 + stoppingTime(tmp));
}

int main() {
	cache = malloc(sizeof(int) * 1000000);

	// initialise the result for n=1
	cache[0]=1;

var d = new Date();
var a = Array(1000001).fill();
a[0] = a[1] = 1;
for (var i = 2; i <= 1000000; i++) {
    var c = -1, j = i;
    while (c++, (j = j % 2 ? j * 3 + 1 : j / 2) > i);
    a[i] = a[j] + c;
}
console.log(a.indexOf(a.reduce((p, c) => Math.max(p, c))));
new Date() - d;

var d = new Date(),
    a = Array(1000001).fill(),
    max = 0,
    winner = 0;
a[0] = a[1] = 1;
for (var i = 2; i <= 1000000; i++) {
    var c = -1, j = i;
    while (c++, (j = j % 2 ? j * 3 + 1 : j / 2) > i);
    a[i] = a[j] + c;
    if (a[i] > max) max = a[i], winner = i;
}
console.log(winner);
new Date() - d;

#include<stdio.h>
int cache[1000000];
int stoppingTime(long n) {
	long tmp = n&1?n*3+1:n>>1;
	if(n > 1000000) return 1 + stoppingTime(tmp);
	if (cache[n-1]) return cache[n-1];
	return (cache[n-1] = 1 + stoppingTime(tmp));
}
int main() {
	cache[0]=1;
	int max=0;
	int j=1;
	for(int i=1; i<1000000; i++) {
		if(stoppingTime(i) > max) {
			max = cache[i-1];
			j = i;
		}
	}
	printf("%d\n", j);
	return 0;
}