Finding the Mode of Integers in an Array

arrays, java, mode

Solution

Here's a simpler way to solve this problem. Create an array called count of size 101. The indexes (0-100) represent the numbers you are counting. Traverse the input array and count the occurrences of each number. Finally, compare the counts to find the one that appears the most (tie goes to the lower number):

public static int mode(int[] input) {

    int[] count = new int[101];

    //count the occurrences
    for (int i=0; i < input.length; i++) {
        count[input[i]]++;
    }

    //go backwards and find the count with the most occurrences
    int index = count.length-1;
    for (int i=count.length-2; i >=0; i--) {
        if (count[i] >= count[index])
            index = i;
    }

    return index;
}

Problem

For this problem, I am to write a method called mode that returns the most frequently occurring element of an array of integers. Assume that the array has at least one element and that every element in the array has a value between 0 and 100 inclusive. Break ties by choosing the lower value. For example, if the array passed contains the values {27, 15, 15, 11, 27}, your method should return 15. (Hint: You may wish to look at the Tally program from earlier in this chapter to get an idea of how to solve this problem.) I am having a problem seeing what is going wrong for a specific input. For instance: mode({27, 15, 15, 27, 11, 11, 11, 14, 15, 15, 16, 19, 99, 100, 0, 27}) returns 15 which is correct, but mode({1, 1, 2, 3, 3}) returns 3 when it should be 1. Here is the code: ``` public static int mode(int[] input) { int returnVal = input[0]; // stores element to be returned int repeatCount = 0; // counts the record number of repeats int prevRepCnt = 0; // temporary count for repeats for (int i=0; i<input.length; i++) { // goes through each elem for (int j=i; j<input.length; j++) { // compares to each elem after the first elem if (i != j && input[i] == input[j]) { // if matching values repeatCount++; // gets the repeat count if (repeatCount>=prevRepCnt) { // a higher count of repeats than before returnVal=input[i]; // return that element } prevRepCnt = repeatCount; // Keeps the highest repeat record } repeatCount=0; // resets repeat Count for next comparison } } return returnVal; } ```

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