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Q: Count Symbols - Java Task (Maps)

+13 votes

Write a program that reads some text from the console and counts the occurrences of each character in it. Print the results in alphabetical (lexicographical) order.

Examples:

Input

Output

SoftUni rocks

 : 1 time/s (empty space)

S: 1 time/s

U: 1 time/s

c: 1 time/s

f: 1 time/s

i: 1 time/s

k: 1 time/s

n: 1 time/s

o: 2 time/s

r: 1 time/s

s: 1 time/s

t: 1 time/s

 

Input

Output

Did you know Math.Round rounds to the nearest even integer?

 : 9 time/s (empty space)

.: 1 time/s

?: 1 time/s

D: 1 time/s

M: 1 time/s

R: 1 time/s

a: 2 time/s

d: 3 time/s

e: 7 time/s

g: 1 time/s

h: 2 time/s

i: 2 time/s

k: 1 time/s

n: 6 time/s

o: 5 time/s

r: 3 time/s

s: 2 time/s

t: 5 time/s

u: 3 time/s

v: 1 time/s

w: 1 time/s

y: 1 time/s

 

 

asked in Java category by user eiorgert

1 Answer

+1 vote
 
Best answer

Here's my solution - the key row here is row number 15 (the highlighted one):

import java.util.*;

public class Pr_04_CountSymbols {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        String[] input = sc.nextLine().split("");//GETTING THE INPUT AS AN ARRAY OF STRINGS

        TreeMap<String, Integer> treeMap = new TreeMap<>();//USING THE TreeMap SO THE ELEMENTS WHEN PRINTING ARE IN ALPHABETICAL (LEXICOGRAPHICAL) ORDER

        for (int i = 0; i < input.length; i++) {//ITERATE THROUGH THE ARRAY input[]
            if (!treeMap.containsKey(input[i])) {//IF THERE ISN'T KEY input[i]IN THE treeMap...
                treeMap.put(input[i], 1);//...ADD IT THERE + ADD 1 AS A VALUE
            } else if (treeMap.containsKey(input[i])) {//IF THERE IS ALREADY KEY input[i] IN THE treeMap...
                treeMap.put(input[i], (treeMap.get(input[i]) + 1));//...ADD IT THERE + ADD TO THE CURRENT VALUE OF THIS KEY 1
            }
        }

        for (Map.Entry<String, Integer> entry : treeMap.entrySet()) {
            System.out.printf("%s: %s time/s\n", entry.getKey(),entry.getValue());
        }
    }
}

 

answered by user hues
selected by user golearnweb
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