Get the biggest number "n" in an array that occurs "n" times

Question

Title sums it up. For example {2,3,3,3,1} would return 3, {2,2,4,4,4} would return 2; In my code I only get the biggest number that occurs n times. If 3 would occur 2 times and 2 would occur 2 times the program should return 2 but in my code it return 0.

public class arrays {

   public static int NmalN(int[] arr) {
       int max = 0;
       int counter = 1;
       for (int i=0; i<arr.length;) {
           if (arr[i]>max) {
               max = arr[i];
               i++;
           }
           else i++;
       }
       for(int j = 0; j<arr.length; j++) {
           if (arr[j]==max) {
               counter++;
               j++;
           }
           else {
               j++;
           }
       }

       if(max == counter) { 
           return counter;
       }
       else return 0; 
   }


   public static void main(String[] args) {
       int [] arr = {1,2,3,3,2,};
       System.out.println("answer: "+ (NmalN(arr)));

   }

}

Answer 1

As I read your code, you find the biggest number in the array, max . If that number occurs exactly max times, no more no less, you return it, otherwise you return 0. When 3 occurs only twice, you should probably try with 2, then with 1, etc. rather than just returning 0. Also think: is it OK if the number occurs more than max times?

Your loop for finding max looks fine (though convention would place i++ inside for (int i=0; i<arr.length; i++) and not inside the loop). After that you need a for loop to count down from max to 0, for each number n in that range checking if it occurs at least n times. You way of counting is almost correct, but as Vasan said in a comment, you are doing j++ both in for(int j = 0; j<arr.length; j++) and in the loop, that is, twice, so remove the one inside the loop. Whenever you find that counter >= n , return n . If not earlier, it will happen when n is 0, since 0 is always present 0 times.

Answer 2

    import java.util.Arrays;

    public class find_Biggest_Num_N_In_Array_Occures_N_times {
     public static int Solution(int[] ar){
     int[] counter= new int[ar.length];
     Arrays.fill(counter, 0);

     for(int i=0; i<ar.length;i++){
     int pos=ar[i];
     counter[pos]++;
     }

    for(int j=ar.length-1;j>=1;j--){
      if(counter[j]==j){
        return j;
      }
    }
    return -1;
 }

public static void main(String args[]){
    System.out.println(Solution(new int[]{1,2,2,3,4,4,4,4,3,5,5,5,5,5,1}));
}
}

Answer 3

public static int solution(Integer[] intArray) {
       int result = 0;
       Arrays.sort(intArray, Collections.reverseOrder());
       for (int i=0; i<intArray.length;i++) {
           int count = 0;
           for(int j=0;j<intArray.length;j++) {
               if(intArray[i]==intArray[j]) {
                   count++;
               }
           }
           if(count == intArray[i]) {
               result = intArray[i];
               break;
           }
       }
       return result; 
   }

Answer 4

In javascript, it'll be like this

function getLargestFrequency(arr) {
    let frequency = {};

    // Count frequency of each number
    for (let i = 0; i < arr.length; i++) {
        let num = arr[i];
        if (frequency[num]) {
            frequency[num]++;
        } else {
            frequency[num] = 1;
        }
    }

    let maxFrequency = 0;
    let maxNum;
    for (let num in frequency) {
        if (frequency[num] > maxFrequency) {
            maxFrequency = frequency[num];
            maxNum = num;
        }
    }
    // Initialize the maxFrequencyCount with maxFrequency
    let maxFrequencyCount = maxFrequency;
    // Iterate over the frequency object
    for (let num in frequency) {
        if (frequency[num] === maxFrequencyCount) {
            // If the current number's frequency count is equal to maxFrequencyCount, return it
            return num;
        }
    }
    // If no number with the highest frequency count is found, return 0
    return 0;
}

and you can see the output

let array1 = [2,3,3,3,1];
let result1 = getLargestFrequency(array1);
console.log(result1); // should return 3

let array2 = [2,2,4,4,4];
let result2 = getLargestFrequency(array2);
console.log(result2); // should return 2

Answer 5

I think this should work ` import java.util.Arrays;

public class find_Biggest_Num_N_In_Array_Occures_N_times {
 public static int Solution(int[] ar){
 Arrays.sort(ar);
 int maxElement = ar[ar.length -1];
 int[] counter= new int[maxElement + 1];
 Arrays.fill(counter, 0);

 for(int i=0; i<ar.length;i++){
 int pos=ar[i];
 counter[pos]++;
 }

for(int j=ar.length-1;j>=1;j--){
  if(counter[j]==j){
    return j;
  }
}
return -1;

} }`