[FIXED] How can i show only the largest number from my output?

Issue

In this program I am trying to show the longest consecutive numbers. (For example: input: 1 1 1 2 2 2 2 3 3 3 3 3, output: 3 4 5)

My question is how can I show only the largest number from my output. In this example 5.

We are not allowed to use Arrays, is there any other way to solve this problem?

public static void main(String args[]) {
        
    Scanner sc = new Scanner(System.in);
        
    int zahl = 0;
    int anzahl = -1;
    String maxString = "";
        
    while (sc.hasNextInt()) { 
              
        int i = sc.nextInt();
 
        if (anzahl == i | anzahl == -1) {
            zahl++;
        } else if(anzahl != i) {
            maxString += zahl + "\n" ;
            zahl = 1;
        }
              
        anzahl = i;
    }
          
    sc.close();
    System.out.println(maxString + (zahl + 1));
}

Solution

Just create a variable to track the maximum number outside the loop, and change it when new subsequence is detected and after the loop to check the length of the last subsequence:

Scanner sc = new Scanner("1 1 1 2 2 2 2 2 3 3 3 4 4 4 4 4 4 5 5 5");

int zahl = 0;
int anzahl = -1;
String maxString = "";
int maxFreq = 0; // maximal frequency of numbers

while (sc.hasNextInt()) { 
  
  int i = sc.nextInt();

  if(anzahl == i || anzahl == -1) {
      zahl++;
  } else {
      maxFreq = Math.max(maxFreq, zahl); // update maxFreq
      maxString += zahl + "\n" ;
      zahl = 1;
  }
  
  anzahl = i;
}
maxFreq = Math.max(maxFreq, zahl); // check the tail

sc.close();
//System.out.println(maxString + (zahl + 1) );
System.out.println("maxFreq=" + maxFreq);

prints maxFreq=6 (the count of 4s).

For this input Scanner sc = new Scanner("1 1 1 2 2 2 2 2 2 2"); the code prints maxFreq=7

Answered By - Alex Rudenko
Answer Checked By - Clifford M. (JavaFixing Volunteer)