README_EN.md

1419. Minimum Number of Frogs Croaking

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Description

You are given the string croakOfFrogs, which represents a combination of the string "croak" from different frogs, that is, multiple frogs can croak at the same time, so multiple "croak" are mixed.

Return the minimum number of different frogs to finish all the croaks in the given string.

A valid "croak" means a frog is printing five letters 'c', 'r', 'o', 'a', and 'k' sequentially. The frogs have to print all five letters to finish a croak. If the given string is not a combination of a valid "croak" return -1.

 

Example 1:

Input: croakOfFrogs = "croakcroak"
Output: 1 
Explanation: One frog yelling "croak" twice.

Example 2:

Input: croakOfFrogs = "crcoakroak"
Output: 2 
Explanation: The minimum number of frogs is two. 
The first frog could yell "crcoakroak".
The second frog could yell later "crcoakroak".

Example 3:

Input: croakOfFrogs = "croakcrook"
Output: -1
Explanation: The given string is an invalid combination of "croak" from different frogs.

 

Constraints:

• 1 <= croakOfFrogs.length <= 105
• croakOfFrogs is either 'c', 'r', 'o', 'a', or 'k'.

Solutions

Python3

class Solution:
    def minNumberOfFrogs(self, croakOfFrogs: str) -> int:
        if len(croakOfFrogs) % 5 != 0:
            return -1
        idx = {c: i for i, c in enumerate('croak')}
        cnt = [0] * 5
        ans = x = 0
        for i in map(idx.get, croakOfFrogs):
            cnt[i] += 1
            if i == 0:
                x += 1
                ans = max(ans, x)
            else:
                if cnt[i - 1] == 0:
                    return -1
                cnt[i - 1] -= 1
                if i == 4:
                    x -= 1
        return -1 if x else ans

Java

class Solution {
    public int minNumberOfFrogs(String croakOfFrogs) {
        int n = croakOfFrogs.length();
        if (n % 5 != 0) {
            return -1;
        }
        int[] idx = new int[26];
        String s = "croak";
        for (int i = 0; i < 5; ++i) {
            idx[s.charAt(i) - 'a'] = i;
        }
        int[] cnt = new int[5];
        int ans = 0, x = 0;
        for (int k = 0; k < n; ++k) {
            int i = idx[croakOfFrogs.charAt(k) - 'a'];
            ++cnt[i];
            if (i == 0) {
                ans = Math.max(ans, ++x);
            } else {
                if (--cnt[i - 1] < 0) {
                    return -1;
                }
                if (i == 4) {
                    --x;
                }
            }
        }
        return x > 0 ? -1 : ans;
    }
}

C++

class Solution {
public:
    int minNumberOfFrogs(string croakOfFrogs) {
        int n = croakOfFrogs.size();
        if (n % 5 != 0) {
            return -1;
        }
        int idx[26]{};
        string s = "croak";
        for (int i = 0; i < 5; ++i) {
            idx[s[i] - 'a'] = i;
        }
        int cnt[5]{};
        int ans = 0, x = 0;
        for (char& c : croakOfFrogs) {
            int i = idx[c - 'a'];
            ++cnt[i];
            if (i == 0) {
                ans = max(ans, ++x);
            } else {
                if (--cnt[i - 1] < 0) {
                    return -1;
                }
                if (i == 4) {
                    --x;
                }
            }
        }
        return x > 0 ? -1 : ans;
    }
};

Go

func minNumberOfFrogs(croakOfFrogs string) int {
	n := len(croakOfFrogs)
	if n%5 != 0 {
		return -1
	}
	idx := [26]int{}
	for i, c := range "croak" {
		idx[c-'a'] = i
	}
	cnt := [5]int{}
	ans, x := 0, 0
	for _, c := range croakOfFrogs {
		i := idx[c-'a']
		cnt[i]++
		if i == 0 {
			x++
			ans = max(ans, x)
		} else {
			cnt[i-1]--
			if cnt[i-1] < 0 {
				return -1
			}
			if i == 4 {
				x--
			}
		}
	}
	if x > 0 {
		return -1
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

function minNumberOfFrogs(croakOfFrogs: string): number {
    const n = croakOfFrogs.length;
    if (n % 5 !== 0) {
        return -1;
    }
    const idx = (c: string): number => 'croak'.indexOf(c);
    const cnt: number[] = [0, 0, 0, 0, 0];
    let ans = 0;
    let x = 0;
    for (const c of croakOfFrogs) {
        const i = idx(c);
        ++cnt[i];
        if (i === 0) {
            ans = Math.max(ans, ++x);
        } else {
            if (--cnt[i - 1] < 0) {
                return -1;
            }
            if (i === 4) {
                --x;
            }
        }
    }
    return x > 0 ? -1 : ans;
}

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