/*
 * CSE 142
 * Lecture - 5/14/2012
 * Arrays For Tallying
 * Digit Frequency
 *
 * This program tests our implementation of mostFrequentDigits.
 * Our goal is to calculate which digit
 * occurs most frequently in several numbers.
 */
public class Frequency {
	public static final int BASE = 10;

	public static void main(String[] args) {
		int[] testDigits = {123121, 44222, 98799, 5678};
		for (int i = 0; i < testDigits.length; i++) {
			int currentTest = testDigits[i];
			System.out.println(currentTest + " -> " +
			        mostFrequentDigit(currentTest));
		}
	}
	
	// Returns the most frequently occuring digit within a number n.
	// If there is a tie, returns the lower digit.
	// Uses the count and max pattern we discussed.
	public static int mostFrequentDigit(int n) {
		// make an array with BASE elements (1 for each counter)
		int[] counts = new int[BASE];
		// loop over input using mod by BASE trick
		while (n > 0) {
			int digit = n % BASE;
		   // update correct counter (element 0 is count of 0s, elemnt 1 is count of 1s etc)
			counts[digit]++;
			n /= BASE;
		}
		// Traverse counts to find index of highest valued element.
		// Slightly different than other max problems we've solved.
		int highestIndex = 0;
		for (int i = 0; i < BASE; i++) {
			if (counts[i] > counts[highestIndex]) {
				highestIndex = i;
			}
		}
		return highestIndex;
	}
	
	// This is the implementation we looked at that doesn't use arrays.
	// It is super redundant, and doesn't handle bases other than 10 well.
	public static int terribleMostFrequentDigit(int n) {
		int count0 = 0;
		int count1 = 0;
		int count2 = 0;
		int count3 = 0;
		int count4 = 0;
		int count5 = 0;
		int count6 = 0;
		int count7 = 0;
		int count8 = 0;
		int count9 = 0;
		while (n > 0) {
			int digit = n % 10;
			if (digit == 0) {
				count0++;
			} else if (digit == 1) {
				count1++;
			} else if (digit == 2) {
				count2++;
			} else if (digit == 3) {
				count3++;
			} else if (digit == 4) {
				count4++;
			} else if (digit == 5) {
				count5++;
			} else if (digit == 6) {
				count6++;
			} else if (digit == 7) {
				count7++;
			} else if (digit == 8) {
				count8++;
			} else { // digit == 9
				count9++;
			}
			n /= 10;
		}
		
		int best = 0;
		int bestCount = count0;
		if (count1 > bestCount) {
			bestCount = count1;
			best = 1;
		}
		if (count2 > bestCount) {
			bestCount = count2;
			best = 2;
		}
		if (count3 > bestCount) {
			bestCount = count3;
			best = 3;
		}
		if (count4 > bestCount) {
			bestCount = count4;
			best = 4;
		}
		if (count5 > bestCount) {
			bestCount = count5;
			best = 5;
		}
		if (count6 > bestCount) {
			bestCount = count6;
			best = 6;
		}
		if (count7 > bestCount) {
			bestCount = count7;
			best = 7;
		}
		if (count8 > bestCount) {
			bestCount = count8;
			best = 8;
		}
		if (count9 > bestCount) {
			bestCount = count9;
			best = 9;
		}
		return best;
	}
}