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1989. Maximum Number of People That Can Be Caught in Tag
Description
You are playing a game of tag with your friends. In tag, people are divided into two teams: people who are "it", and people who are not "it". The people who are "it" want to catch as many people as possible who are not "it".
You are given a 0indexed integer array team
containing only zeros (denoting people who are not "it") and ones (denoting people who are "it"), and an integer dist
. A person who is "it" at index i
can catch any one person whose index is in the range [i  dist, i + dist]
(inclusive) and is not "it".
Return the maximum number of people that the people who are "it" can catch.
Example 1:
Input: team = [0,1,0,1,0], dist = 3 Output: 2 Explanation: The person who is "it" at index 1 can catch people in the range [idist, i+dist] = [13, 1+3] = [2, 4]. They can catch the person who is not "it" at index 2. The person who is "it" at index 3 can catch people in the range [idist, i+dist] = [33, 3+3] = [0, 6]. They can catch the person who is not "it" at index 0. The person who is not "it" at index 4 will not be caught because the people at indices 1 and 3 are already catching one person.
Example 2:
Input: team = [1], dist = 1 Output: 0 Explanation: There are no people who are not "it" to catch.
Example 3:
Input: team = [0], dist = 1 Output: 0 Explanation: There are no people who are "it" to catch people.
Constraints:
1 <= team.length <= 10^{5}
0 <= team[i] <= 1
1 <= dist <= team.length
Solutions

class Solution { public int catchMaximumAmountofPeople(int[] team, int dist) { int ans = 0; int n = team.length; for (int i = 0, j = 0; i < n; ++i) { if (team[i] == 1) { while (j < n && (team[j] == 1  i  j > dist)) { ++j; } if (j < n && Math.abs(i  j) <= dist) { ++ans; ++j; } } } return ans; } }

class Solution { public: int catchMaximumAmountofPeople(vector<int>& team, int dist) { int ans = 0; int n = team.size(); for (int i = 0, j = 0; i < n; ++i) { if (team[i]) { while (j < n && (team[j]  i  j > dist)) { ++j; } if (j < n && abs(i  j) <= dist) { ++ans; ++j; } } } return ans; } };

class Solution: def catchMaximumAmountofPeople(self, team: List[int], dist: int) > int: ans = j = 0 n = len(team) for i, x in enumerate(team): if x: while j < n and (team[j] or i  j > dist): j += 1 if j < n and abs(i  j) <= dist: ans += 1 j += 1 return ans

func catchMaximumAmountofPeople(team []int, dist int) (ans int) { n := len(team) for i, j := 0, 0; i < n; i++ { if team[i] == 1 { for j < n && (team[j] == 1  ij > dist) { j++ } if j < n && abs(ij) <= dist { ans++ j++ } } } return } func abs(x int) int { if x < 0 { return x } return x }