Finding the most frequent value in an array and then choosing the lowest value if there is a tie

WoeIs :

I have the following program which looks up an array "a" and then output the value in the array which appears the most frequently. However one other condition that I would like to implement is that in the case of a tie where two different values appear an equal amount of times, the lowest value gets output.

So for the following code with an array:

int a[] = {34, 34, 20, 20, 15};

It outputs 34 however I want it to output 20 instead, since that is a lower value and appears just as many times in the array.

public class Arrays3 {
    public static void main(String[] args){
        int a[] = {34, 34, 20, 20, 15};
        mode(a);
    }
    public static int mode(int[] a) {
        int[] counts = new int[101];
        int maxCount = 0;
        int maxKey = 0;

        for(int i = 0; i < a.length; i++) {
            counts[a[i]]++;
            if(counts[a[i]] > maxCount) {
                maxCount = counts[a[i]];
                maxKey = a[i];
            }
        }
        System.out.println(maxKey);
        return maxKey;
    }
}

GBlodgett :

You could check against maxKey and then do something along the lines of:

 if(counts[a[i]] == maxCount && counts[a[i]] < maxKey) {
       maxKey = counts[a[i]];
 }

So that if there is ever a tie, the maxKey will be set to the smaller element. Then if count[a[i]] is ever greater than maxCount, maxKey will be overridden and become the element that occurs most often:

for(int i = 0; i < a.length; i++) {
     counts[a[i]]++;

     if(counts[a[i]] > maxCount) {
          maxCount = counts[a[i]];
          maxKey = a[i];
      }
      if(counts[a[i]] == maxCount && counts[a[i]] < maxKey) {
        maxKey = a[i];
      }
}
System.out.println(a[maxKey]);

Output

20