Learning Algorithms with JS, Python and Java 4: MaxChar

This is the fourth article of my attempts to follow Stephen Grider's Udemy course in three different languages.

First I quote todays's question:

--- Directions
Given a string, return the character that is most
commonly used in the string.
--- Examples
maxChar("abcccccccd") === "c"
maxChar("apple 1231111") === "1"

JavaScript:

function maxChar(str) {
    const charMap = {};
    let max = 0;
    let maxChar = '';

    for (let char of str) {
        if (charMap[char]) {
            charMap[char]++
        } else {
            charMap[char] = 1;
        }
    }

    for (let char in charMap) {
        if (charMap[char] > max) {
            max = charMap[char];
            maxChar = char;
        }
    }

    return maxChar;
}

Stephen rewrites the first for loop as:

    for (let char of str) {
        charMap[char] = charMap[char] + 1 || 1;
    }

This is possible because anObject['nonexistentKey'] returns undefined, undefined + 1 return NaN, which is falsy.

Python:

def max_char(str):
    char_map = {}
    max = 0
    max_char = ''

    for char in str:
        char_map.setdefault(char, 0)
        char_map[char] += 1

    for char in char_map:
        if char_map[char] > max:
            max = char_map[char]
            max_char = char

    return max_char

In Python, you can't just try char_map[char] += 1 as in JS, because it would throw a KeyError if the key doesn't exist yet.

More concisely:

def max_char(str):
    char_map = {}

    for char in str:
        char_map.setdefault(char, 0)
        char_map[char] += 1

    return max(char_map, key=char_map.get)

Source: https://stackoverflow.com/questions/268272/getting-key-with-maximum-value-in-dictionary
max(char_map, key=char_map.get) is the same as max(char_map.keys(), key=char_map.get).
key=char_map.get is needed to find the (key, value) tuple pair in the dictionary whose value is the largest. max({'a': 2, 'b': 1}) returns 'b'.

Java:

import java.util.HashMap;
import java.util.Map;

public static char maxChar2(String str) {
    Map<Character, Integer> charMap = new HashMap<>();

    for (char chr : str.toCharArray()) {
        Integer chrCount = charMap.get(chr);
        if (chrCount == null) {
            chrCount = 0;
        }
        charMap.put(chr, chrCount + 1);
    }

    int max = 0;
    char maxChar = '\0';

    for (Map.Entry<Character, Integer> entry : charMap.entrySet()) {
        int value = entry.getValue();
        if (value > max) {
            max = value;
            maxChar = entry.getKey();
        }
    }

    return maxChar;
}

Using Stream:

public static char maxChar(String str) {
    Map<Character, Long> charMap = str.chars()
            .mapToObj(i -> (char) i)
            .collect(Collectors.groupingBy(
                    c -> c, Collectors.counting()));

    return charMap.entrySet().stream()
            .collect(Collectors.groupingBy(
                    Map.Entry::getValue,
                    TreeMap::new,
                    Collectors.mapping(
                            Map.Entry::getKey, Collectors.toList())))
            .lastEntry()
            .getValue()
            .get(0);
}

The return statement can be made simpler:

    return charMap.entrySet().stream()
            .max(Map.Entry.comparingByValue())
            .get()
            .getKey();

References:
https://codereview.stackexchange.com/questions/133058/print-max-occurring-character-in-a-string
https://stackoverflow.com/questions/48280361/how-to-count-occurrences-for-each-value-in-multimap-java
https://stackoverflow.com/questions/42060294/using-java8-stream-to-find-the-highest-values-from-map

Comments

[Karolis Ryselis]

Python could do it even better.

import operator
from collections import Counter

def max_char(s):
    c = Counter(s)
    try:
        return max(c.items(), key=operator.itemgetter(1))[0]
    except ValueError:
        return ""

[Pierre Bouillon]

I would have done this that way and using the built in 'max' method for the Counter object

from collections import Counter

def max_char(s):
    return Counter(s).most_common(1)[0][0] if s else ''

To explain a little bit to non Python dev:
if s is a valid chain, I build a Counter object with it, get the list of the n (letter, count) elements, getting the first one and returning only the letter.

Step by step:

s = 'abcccccccd'
Counter(s).most_common(1)        # [('c', 7)]
Counter(s).most_common(1)[0]     # ('c', 7)
Counter(s).most_common(1)[0][0]  # 'c

[tommy-3]

Wow, this looked enigmatic at first, but I understood how it works. Thank you!

[tommy-3]

Thank you for the comment!

I learned quite a lot from this code. I really appreciate it. Below is just for my understanding:

import operator
from collections import Counter

c = Counter('abbb')
print(c)    # prints: Counter({'b': 3, 'a': 1})
print(c.items())    # prints: dict_items([('a', 1), ('b', 3)])

item = list(c.items())[0]
print(item)    # prints: ('a', 1)
print(item[0])    # prints: a
print(item[1])    # prints: 1

itemgetter_1 = operator.itemgetter(1)
print(itemgetter_1(['x', 'y', 'z']))    # prints: y
print(itemgetter_1(item))    # prints: 1

print(max(c.items(), key=operator.itemgetter(1)))    # prints: ('b', 3)
print(max(c.items(), key=operator.itemgetter(1))[0])    # prints: b

Actually we can make it more concise with the way I used above for the max function:

import operator
from collections import Counter

def max_char(s):
    c = Counter(s)
    try:
        return max(c, key=c.get)
    except ValueError:
        return ""

[Rattanak Chea]

I like how we can write python like psuedocode with few lines.

count element in an element

=> Counter(s)

Though it can be hard to understand and remember all the shortcuts.