# Solution For Minimum Number Of Frogs Croaking

The Minimum Number of Frogs Croaking problem on LeetCode is a medium-level problem that involves string manipulation. The problem statement is as follows:

You have a string s that represents a sequence of frogs croaking. Each ‘croak’ of a frog is represented by a character in the string. The four characters are ‘c’, ‘r’, ‘o’, and ‘a’. You need to determine the minimum number of frogs that are needed to complete the sequence. A valid sequence is one where the frogs croak in the order ‘croak’. In other words, for a valid sequence, the first ‘c’ must be followed by an ‘r’, then an ‘o’, and finally an ‘a’. Once a frog has croaked ‘a’, it is considered done and can no longer contribute to the sequence.

To solve this problem, we can use a hashmap to keep track of the number of frogs that are currently croaking and a variable to keep track of the maximum number of frogs that have been croaking at the same time. We iterate through the string s and for each character we check if it is a valid frog character. If it is, we update the hashmap accordingly. If at any point the sequence is not valid, we return -1. Finally, we return the maximum number of frogs that have been croaking at the same time.

Here is an implementation of the solution in Python:

`def minNumberOfFrogs(croakOfFrogs: str) -> int:`

count = {"c": 0, "r": 0, "o": 0, "a": 0}

max_count = 0

for c in croakOfFrogs:

if c not in count:

return -1

if c == "c":

count["c"] += 1

elif c == "r":

if count["c"] <= count["r"]:

return -1

count["r"] += 1

elif c == "o":

if count["r"] <= count["o"]:

return -1

count["o"] += 1

elif c == "a":

if count["o"] <= count["a"]:

return -1

count["a"] += 1

count["c"] -= 1

count["r"] -= 1

count["o"] -= 1

max_count = max(max_count, count["c"])

if sum(count.values()) > 0:

return -1

return max_count

The time complexity of this solution is O(n), where n is the length of the string s. This is because we only iterate through the string once and perform constant time operations for each character. The space complexity is also O(1), as the size of the hashmap is constant and independent of the length of the string.

## Step by Step Implementation For Minimum Number Of Frogs Croaking

class Solution { public int minNumberOfFrogs(String croakOfFrogs) { // keep track of the number of frogs croaking at each stage int c = 0, r = 0, o = 0, a = 0, k = 0; int frogs = 0; int maxFrogs = 0; for (char ch : croakOfFrogs.toCharArray()) { switch (ch) { case 'c': c++; break; case 'r': if (c == 0) return -1; c--; r++; break; case 'o': if (r == 0) return -1; r--; o++; break; case 'a': if (o == 0) return -1; o--; a++; break; case 'k': if (a == 0) return -1; a--; k++; break; default: return -1; } // if this is the start of a new frog croaking, increment frogs if (c == 1 && r == 0 && o == 0 && a == 0 && k == 0) { frogs++; } // if this is the end of a frog croaking, decrement frogs if (c == 0 && r == 0 && o == 0 && a == 0 && k == 1) { frogs--; } // keep track of the maximum number of frogs croaking at the same time maxFrogs = Math.max(maxFrogs, frogs); } // if any frogs are still croaking, return -1 if (c != 0 || r != 0 || o != 0 || a != 0 || k != 0) return -1; return maxFrogs; } }

class Solution: def minNumberOfFrogs(self, croakOfFrogs: str) -> int: # keep track of the number of frogs croaking at each stage # also keep track of the number of frogs needed overall frogs_croaking = [0] * 5 num_frogs = 0 # iterate through the croaking sequence for frog in croakOfFrogs: # if the frog is croaking, increase the number of frogs at that stage if frog == 'c': frogs_croaking[0] += 1 elif frog == 'r': frogs_croaking[1] += 1 elif frog == 'o': frogs_croaking[2] += 1 elif frog == 'a': frogs_croaking[3] += 1 elif frog == 'k': frogs_croaking[4] += 1 # if there are more frogs croaking at a stage than there are at the previous stage, # we need to add more frogs if frogs_croaking[0] < frogs_croaking[1]: num_frogs += frogs_croaking[1] - frogs_croaking[0] # if there are less frogs croaking at a stage than there are at the previous stage, # then there are not enough frogs and we return -1 elif frogs_croaking[0] > frogs_croaking[1]: return -1 # if the frog is at the last stage, we decrease the number of frogs croaking at that stage if frog == 'k': frogs_croaking[0] -= 1 frogs_croaking[1] -= 1 frogs_croaking[2] -= 1 frogs_croaking[3] -= 1 frogs_croaking[4] -= 1 # if the number of frogs croaking at the first stage is not 0, then there are not enough frogs if frogs_croaking[0] != 0: return -1 # return the number of frogs needed overall return num_frogs

var minNumberOfFrogs = function(croakOfFrogs) { // keep track of the number of frogs croaking at each stage // c: croaks, r: rests, o: starts croaking, a: croaks, k: ends croaking // we need at least one frog at each stage let frogs = { c: 0, r: 0, o: 0, a: 0, k: 0 }; // keep track of the total number of frogs let totalFrogs = 0; // iterate through the croakOfFrogs string for (let i = 0; i < croakOfFrogs.length; i++) { // current frog croaking let frog = croakOfFrogs[i]; // if the current frog is croaking 'c', then there is no need to do anything // else, we need to check if there are any frogs at the previous stage if (frog !== 'c') { // if there are no frogs at the previous stage, then this is not a valid croak sequence if (frogs[frog - 1] === 0) { return -1; } // otherwise, there is a frog at the previous stage, so we decrement the number of frogs at that stage frogs[frog - 1]--; } // increment the number of frogs at the current stage frogs[frog]++; // if the current frog is croaking 'k', then we increment the total number of frogs if (frog === 'k') { totalFrogs++; } } // if there are any frogs remaining at any stage, then this is not a valid croak sequence for (let key in frogs) { if (frogs[key] > 0) { return -1; } } // otherwise, return the total number of frogs return totalFrogs; };

int minNumberOfFrogs(string croakOfFrogs) { int c=0,r=0,o=0,a=0,k=0; //counters for each letter int frogs=0; //current number of frogs int maxFrogs=0; //maximum number of frogs for(int i=0;i0) //if there are any k's { k--; //decrement k counter frogs--; //one frog has croaked so decrement } else //otherwise { frogs++; //increment frog counter if(frogs>maxFrogs) //if current frog counter is greater than max maxFrogs=frogs; //set max to current frog counter } } else if(croakOfFrogs[i]=='r') //if letter is r r++; //increment r counter else if(croakOfFrogs[i]=='o') //if letter is o o++; //increment o counter else if(croakOfFrogs[i]=='a') //if letter is a a++; //increment a counter else if(croakOfFrogs[i]=='k') //if letter is k k++; //increment k counter if(c public int MinNumberOfFrogs(string croakOfFrogs) { int frogs = 0; int maxFrogs = 0; int c = 0; int r = 0; int o = 0; int a = 0; int k = 0; foreach(char ch in croakOfFrogs) { switch(ch) { case 'c': c++; frogs++; if(frogs > maxFrogs) maxFrogs = frogs; break; case 'r': if(c == 0) return -1; c--; r++; break; case 'o': if(r == 0) return -1; r--; o++; break; case 'a': if(o == 0) return -1; o--; a++; break; case 'k': if(a == 0) return -1; a--; k++; frogs--; break; } } if(c == 0 && r == 0 && o == 0 && a == 0 && k == 0) return maxFrogs; else return -1; }