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2086. Minimum Number of Food Buckets to Feed the Hamsters

Description

You are given a 0-indexed string hamsters where hamsters[i] is either:

  • 'H' indicating that there is a hamster at index i, or
  • '.' indicating that index i is empty.

You will add some number of food buckets at the empty indices in order to feed the hamsters. A hamster can be fed if there is at least one food bucket to its left or to its right. More formally, a hamster at index i can be fed if you place a food bucket at index i - 1 and/or at index i + 1.

Return the minimum number of food buckets you should place at empty indices to feed all the hamsters or -1 if it is impossible to feed all of them.

 

Example 1:

Input: hamsters = "H..H"
Output: 2
Explanation: We place two food buckets at indices 1 and 2.
It can be shown that if we place only one food bucket, one of the hamsters will not be fed.

Example 2:

Input: hamsters = ".H.H."
Output: 1
Explanation: We place one food bucket at index 2.

Example 3:

Input: hamsters = ".HHH."
Output: -1
Explanation: If we place a food bucket at every empty index as shown, the hamster at index 2 will not be able to eat.

 

Constraints:

  • 1 <= hamsters.length <= 105
  • hamsters[i] is either'H' or '.'.

Solutions

  • class Solution {
    public int minimumBuckets(String street) {
        int n = street.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (street.charAt(i) == 'H') {
                if (i + 1 < n && street.charAt(i + 1) == '.') {
                    ++ans;
                    i += 2;
                } else if (i > 0 && street.charAt(i - 1) == '.') {
                    ++ans;
                } else {
                    return -1;
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int minimumBuckets(string street) {
        int n = street.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (street[i] == 'H') {
                if (i + 1 < n && street[i + 1] == '.') {
                    ++ans;
                    i += 2;
                } else if (i && street[i - 1] == '.') {
                    ++ans;
                } else {
                    return -1;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def minimumBuckets(self, street: str) -> int:
        ans = 0
        i, n = 0, len(street)
        while i < n:
            if street[i] == 'H':
                if i + 1 < n and street[i + 1] == '.':
                    i += 2
                    ans += 1
                elif i and street[i - 1] == '.':
                    ans += 1
                else:
                    return -1
            i += 1
        return ans


  • func minimumBuckets(street string) int {
	ans, n := 0, len(street)
	for i := 0; i < n; i++ {
		if street[i] == 'H' {
			if i+1 < n && street[i+1] == '.' {
				ans++
				i += 2
			} else if i > 0 && street[i-1] == '.' {
				ans++
			} else {
				return -1
			}
		}
	}
	return ans
}

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