Determine the maximum number of specific words in a string

Determine the maximum number of specific words in a string

LeetCode Brush problem Day 3
subject ： Give you a string text, You need to use text To piece together as many words as possible “balloon”（ balloon ）. character string text Each letter in can only be used once at most . Please return the maximum number of words you can piece together “balloon”.

Example ：
Input ：text = “nlaebolko”
Output ：1

Input ：text = “oollpqwa”
Output ：0

Topic link ：https://leetcode-cn.com/problems/maximum-number-of-balloons/

stay LeetCode Just submit the core function part of the code .

Ideas ：
"balloon" in , character l,o The number of is b,a,n Of 2 times . Statistics b,a,l,o,n After the number of , If it happens 5 individual b,5 individual a,10 individual l,10 individual o,5 individual n, be "balloon" The number of 2.

Short board effect , Take the minimum number of characters as the maximum number that can be output .
for example ：b,a,l,o,n There were , Yes 2 individual b,1 individual a,10 individual l,4 individual o,3 individual n.l,o,n Although the number of , but a Only 1 individual , So there's only one ”balloon“, The maximum number is 1.

C Language complete code ：

#include<stdio.h>
#include<string.h>
int maxNumberOfBalloons(char * text)
{
    
    int zimu[6]={
    0};  // Record text In the character a,b,l,o,n The number of times they appear separately , Initialize to 0
    int i;
    int length;
    int count=0;
    length=strlen(text);   // Calculation text The length of 
    for(i=0;i<length;i++)
    {
    
         if(text[i]=='a') // Statistical characters a Number of occurrences , Save in zimu[0] in , Similarly, record b,l,o,n The number of 
          zimu[0]++;
         if(text[i]=='b')
          zimu[1]++;
         if(text[i]=='l')
          zimu[2]++;
         if(text[i]=='o')
          zimu[3]++;
         if(text[i]=='n')
          zimu[4]++;
    }
// Statistics b,a,l,o,n After the number of , Short board effect , Take the minimum number of characters as the maximum number that can be output .
      zimu[2]=(zimu[2])/2;    //l,o The number of is half of the original 
      zimu[3]=(zimu[3])/2;
      int first=zimu[0];     // Make a The number of is equal to first
      for (int j = 0; j < 5; j++)
      {
    
          if(first<zimu[j])    // if a The number of characters is less than the number of other characters ,a Is the minimum 
            first=first;
          if(first>=zimu[j])// if a The number of characters is greater than the number of other characters , Make first=zimu[j], The number of other characters is the minimum 
            first=zimu[j];
      }
      return first;
}
int main()
{
    
    char number[1005];
    scanf("%s",&number);
    maxNumberOfBalloons(number);
    printf("%d\n", maxNumberOfBalloons(number));
}

Running results ：