Formatted question description: https://leetcode.ca/all/1989.html
1989. Maximum Number of People That Can Be Caught in Tag
Level
Medium
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 0-indexed 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 [i-dist, i+dist] = [1-3, 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 [i-dist, i+dist] = [3-3, 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 <= 10^50 <= team[i] <= 11 <= dist <= team.length
Solution
Use a greedy approach. For each element 1, find the leftmost element 0 that is within distance dist of the element 1, and the element 0 is caught by the 1. Once an element 0 is caught by an element 1, it can no longer be caught again, so move on to the next element 0. If an element 1 cannot catch any element 0, then move on to the next element 1. Finally, return the number of elements 1 that can catch at least one element 0 each, which is the maximum number of people that can be caught in tag.
class Solution {
public int catchMaximumAmountofPeople(int[] team, int dist) {
int length = team.length;
int zero = 0, one = 0;
while (zero < length && team[zero] != 0)
zero++;
while (one < length && team[one] != 1)
one++;
int count = 0;
while (zero < length && one < length) {
while (one - zero > dist)
zero++;
int maxIndex = Math.min(one + dist, length - 1);
while (zero <= maxIndex && team[zero] != 0)
zero++;
if (zero <= maxIndex) {
count++;
zero++;
}
one++;
while (one < length && team[one] != 1)
one++;
}
return count;
}
}