Solution 1 - Count Letters in "balloon"

Answer is min(b, a, l/2, o/2, n).

TimeComplexity: O(n)

SpaceComplexity: O(1)

""" 1189.1 - Maximum Number of Balloons - Solution 1 - Count Letters """

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        c = {}
        for ch in text:
            c[ch] = c.get(ch, 0) + 1
        return min(
            c.get("b", 0),
            c.get("a", 0),
            c.get("l", 0) // 2,
            c.get("o", 0) // 2,
            c.get("n", 0),
        )

function maxNumberOfBalloons(text:string):number{
 const c:Record<string,number>={};
 for(const ch of text) c[ch]=(c[ch]??0)+1;
 return Math.min(c["b"]??0, c["a"]??0, Math.floor((c["l"]??0)/2), Math.floor((c["o"]??0)/2), c["n"]??0);
}

package main

func maxNumberOfBalloons(text string) int {
	cnt := map[rune]int{}
	for _, ch := range text {
		cnt[ch]++
	}
	b := cnt['b']
	a := cnt['a']
	l := cnt['l'] / 2
	o := cnt['o'] / 2
	n := cnt['n']
	res := b
	for _, v := range []int{a, l, o, n} {
		if v < res { res = v }
	}
	return res
}

func main() {}

use std::collections::HashMap;

fn max_number_of_balloons(text: &str) -> i32 {
	let mut cnt: HashMap<char, i32> = HashMap::new();
	for ch in text.chars() { *cnt.entry(ch).or_insert(0) += 1; }
	let b = *cnt.get(&'b').unwrap_or(&0);
	let a = *cnt.get(&'a').unwrap_or(&0);
	let l = *cnt.get(&'l').unwrap_or(&0) / 2;
	let o = *cnt.get(&'o').unwrap_or(&0) / 2;
	let n = *cnt.get(&'n').unwrap_or(&0);
	b.min(a).min(l).min(o).min(n)
}

fn main() {}

1189 - Maximum Number of Balloons

Solution 1 - Count Letters in "balloon"