Split String with Same Distinct Counts - Google Top Interview Questions


Problem Statement :


You are given a lowercase alphabet string s. 

Return the number of ways to split the string into two strings such that the number of distinct characters in each string is the same.

Constraints

1 ≤ n ≤ 100,000 where n is the length of s

Example 1

Input

s = "abaab"

Output

2] = hs.size();
        }

Explanation

We can split it by "ab" + "aab" and "aba" + "ab"


Solution :



title-img 




                        Solution in C++ :

int solve(string s) {
    vector<int> r(s.size(), 0);

    unordered_map<char, int> fre;

    for (int j = s.size() - 1; j >= 0; j--) fre[s[j]]++, r[j] = fre.size();

    fre.clear();
    int res = 0;

    for (int i = 0; i + 1 < s.size(); i++) {
        fre[s[i]]++;
        if (fre.size() == r[i + 1]) res++;
    }

    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(String s) {
        int n = s.length();
        int[] prefix = new int[n];
        int[] suffix = new int[n];
        HashSet<Character> hs = new HashSet();

        for (int i = 0; i < n; i++) {
            hs.add(s.charAt(i));
            prefix[i] = hs.size();
        }

        hs = new HashSet();
        for (int i = n - 1; i >= 0; i--) {
            hs.add(s.charAt(i));
            suffix[i
        int res = 0;
        for (int i = 0; i < n - 1; i++) {
            if (prefix[i] == suffix[i + 1])
                res++;
        }
        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s):

        forward_cnt = [0] * len(s)
        seen = set()
        temp = 0
        for i in range(len(s)):
            if s[i] not in seen:
                temp += 1
                forward_cnt[i] = temp
                seen.add(s[i])
            else:
                forward_cnt[i] = forward_cnt[i - 1]

        backward_cnt = [0] * len(s)
        seen = set()
        temp = 0
        for i in range(len(s) - 1, -1, -1):
            if s[i] not in seen:
                temp += 1
                backward_cnt[i] = temp
                seen.add(s[i])
            else:
                backward_cnt[i] = backward_cnt[i + 1]

        ans = 0
        for i in range(len(s) - 1):
            if forward_cnt[i] == backward_cnt[i + 1]:
                ans += 1
        return ans


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