[Solved] Jay has a chocolate box that contains different types of chocolate

Jay has a chocolate box that contains different types of chocolate represented by a lowercase character. Chocolates are arranged in a row. Jay wants to divide the row into pieces such that both the pieces have an equal number of unique chocolate types (denotes by lowercase character).

Can you help jay to determine whether it is possible to divide the chocolate row according to the given condition? And if possible then print how many ways are there to which he can divide the chocolate row.

Input Format

The first line contains T’s value, the number of test cases

The second line contains the length of the chocolate row

Constraints

1<=T<=1000

1<=|S|<=100000

Output Format

Print the number of ways to divide the row according to the given condition in a new line for each test case.

Sample Input

3

ababa

aaaa

d

Sample Output

2

3

0

Explanation:

for test case 1: position pieces are – [“ab”,”aba”], [“aba”,”ba”].

for test case 2: position pieces are – [“a”,”aaa”], [“aa”,”aa”], [“aaa”,”a”].

for test case 3: there is no possible division.

Solution

1. Take input for the number of test cases.
2. Take input for the string.
3. Take two variables count and c1 and c2.
4. Take a for loop from 0 to length of string – 1.
5. Take two substrings s1 and s2.
6. Take two variables c1 and c2.
7. Take a for loop from 0 to length of s1.
8. If the character at index j is not equal to the character at index 0, then increment c1.
9. Take a for loop from 0 to length of s2.
10. If the character at index j is not equal to the character at index 0, then increment c2.
11. If c1 is equal to c2, then increment count.
12. Print count.

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        in.nextLine();
        while (t-- > 0) {
            String s = in.nextLine();
            int count = 0;
            for (int i = 0; i < s.length() - 1; i++) {
                // https://github.com/sauravhathi
                String s1 = s.substring(0, i + 1);
                String s2 = s.substring(i + 1);
                int c1 = 0;
                int c2 = 0;
                for (int j = 0; j < s1.length(); j++) {
                    if (s1.charAt(j) != s1.charAt(0)) {
                        c1++;
                    }
                }
                for (int j = 0; j < s2.length(); j++) {
                    if (s2.charAt(j) != s2.charAt(0)) {
                        c2++;
                    }
                }
                if (c1 == c2) {
                    count++;
                }
            }
            System.out.println(count);
        }
    }
}

#include <iostream>
#include <string>

using namespace std;

int main() {
    int t;
    cin >> t;
    while (t-- > 0) {
        string s;
        cin >> s;
        int count = 0;
        for (int i = 0; i < s.length() - 1; i++) {
            // https://github.com/sauravhathi
            string s1 = s.substr(0, i + 1);
            string s2 = s.substr(i + 1);
            int c1 = 0;
            int c2 = 0;
            for (int j = 0; j < s1.length(); j++) {
                if (s1[j] != s1[0]) {
                    c1++;
                }
            }
            for (int j = 0; j < s2.length(); j++) {
                if (s2[j] != s2[0]) {
                    c2++;
                }
            }
            if (c1 == c2) {
                count++;
            }
        }
        cout << count << endl;
    }
}

t = int(input())
for _ in range(t):
    s = input()
    count = 0
    for i in range(len(s) - 1):
    # https://github.com/sauravhathi
        s1 = s[:i + 1]
        s2 = s[i + 1:]
        c1 = 0
        c2 = 0
        for j in range(len(s1)):
            if s1[j] != s1[0]:
                c1 += 1
        for j in range(len(s2)):
            if s2[j] != s2[0]:
                c2 += 1
        if c1 == c2:
            count += 1
    print(count)

Happy Learning – If you require any further information, feel free to contact me.