Leetcode Ugly Number II problem solution

class Solution:
    def nthUglyNumber(self, n: int) -> int:
        # DP cache
        dp = [0]*n
        # DP base case
        dp[0] = 1
        # DP step case
        i2 = i3 = i5 = 0
        for i in range(1, n):
            k = min(dp[i2]*2, dp[i3]*3, dp[i5]*5) # Next ugly number
            dp[i] = k
            if dp[i] == dp[i2]*2:
                i2 = i2+1
            if dp[i] == dp[i3]*3:
                i3 = i3+1
            if dp[i] == dp[i5]*5:
                i5 = i5+1
        return dp[n-1]

class Solution{
public int nthUglyNumber(int n) {
    ArrayList<Integer> l= new ArrayList<>();
    l.add(1);
    int i2=0,i3=0,i5=0;
    while(l.size()<n){
        int m2 = l.get(i2)*2;
        int m3 = l.get(i3)*3;
        int m5 = l.get(i5)*5;
        int min = Math.min(m2, Math.min(m3, m5));
        l.add(min);
        if(min==m2)
            i2++;
        if(min==m3)
            i3++;
        if(min==m5)
            i5++;
    }
    return l.get(l.size()-1);
}
}

class Solution {
public:
    int nthUglyNumber(int n) {
        priority_queue<long int, vector<long int>, greater<long int> > q;
        q.push(1);
        unordered_set<long int> walked;
        walked.insert(1);
        for(int i=0; i<n-1; i++){
            long int tmp=q.top();
            q.pop();
            if(tmp>=INT_MAX/2)continue;
            if(walked.find(tmp*2)==walked.end()){
                q.push(tmp*2);
                walked.insert(tmp*2);
            }
            if(walked.find(tmp*3)==walked.end()){
                q.push(tmp*3);
                walked.insert(tmp*3);
            }
            if(walked.find(tmp*5)==walked.end()){
                q.push(tmp*5);
                walked.insert(tmp*5);
            }
        }
        return q.top();
        
    }
};

#define min(x,y) (x < y ? x : y)

int nthUglyNumber(int n) {
  static int arr[1691] = {[0]= 1},
  i  = 1, i2 = 0, i3 = 0, i5 = 0,
  n2 = 2, n3 = 3, n5 = 5;
  
  for (; i < n; i++) {
    arr[i] = min(min(n2, n3), n5);
    if (arr[i] == n2) n2 = 2 * arr[++i2];
    if (arr[i] == n3) n3 = 3 * arr[++i3];
    if (arr[i] == n5) n5 = 5 * arr[++i5];
  }

  return arr[n-1];
}