Skip to content

1989. Maximum Number of People That Can Be Caught in Tag 👍

  • Time: $O(n)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution {
 public:
  int catchMaximumAmountofPeople(vector<int>& team, int dist) {
    int ans = 0;
    int i = 0;  // 0s index
    int j = 0;  // 1s index

    while (i < team.size() && j < team.size())
      if (i + dist < j || team[i] != 0) {
        // Find the next 0 that can be caught by 1.
        ++i;
      } else if (j + dist < i || team[j] != 1) {
        // Find the next 1 that can catch 0.
        ++j;
      } else {
        // team[j] catches team[i], so move both.
        ++ans;
        ++i;
        ++j;
      }

    return ans;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
  public int catchMaximumAmountofPeople(int[] team, int dist) {
    int ans = 0;
    int i = 0; // 0s index
    int j = 0; // 1s index

    while (i < team.length && j < team.length)
      if (i + dist < j || team[i] != 0) {
        // Find the next 0 that can be caught by 1.
        ++i;
      } else if (j + dist < i || team[j] != 1) {
        // Find the next 1 that can catch 0.
        ++j;
      } else {
        // team[j] catches team[i], so move both.
        ++ans;
        ++i;
        ++j;
      }

    return ans;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution:
  def catchMaximumAmountofPeople(self, team: list[int], dist: int) -> int:
    ans = 0
    i = 0  # 0s index
    j = 0  # 1s index

    while i < len(team) and j < len(team):
      if i + dist < j or team[i] != 0:
        # Find the next 0 that can be caught by 1.
        i += 1
      elif j + dist < i or team[j] != 1:
        # Find the next 1 that can catch 0.
        j += 1
      else:
        # team[j] catches team[i], so move both.
        ans += 1
        i += 1
        j += 1

    return ans