Rely even indices of String whose prefix has prime variety of distinct Characters


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Given a string S of dimension N. The duty is to seek out the variety of Invalid characters. Index i (0 ≤ i < N) is invalid if i is even and the full depend of distinct characters within the index vary [0, i] is a first-rate quantity. 

Examples:

Enter: N = 6, S = “aabagh”
Output:  2
Clarification: Characters at index 2 and 4 are invalid as 2 and 4 each are even and depend of distict characters upto index 2 and 4 are 2 and three respectively which is prime.

Enter: N = 2, S = “gg”
Output: 
Clarification: No invalid character

Strategy: This drawback may be solved utilizing the prefix array idea.

Thought: The concept is to precompute all of the prime numbers within the given vary of N after which simply test for the required situations  at each character.

Observe the beneath steps to unravel the issue:

  • Create a precompute perform and calculate all prime elements utilizing the sieve of Eratosthenes.
  • Create a hashmap to retailer frequencies of characters which can assist us decide if the character is a replica or not.
  • Iterate over the string from and when any even index is reached test the next:
    • The variety of distinct characters within the prefix is prime
    • Whether it is true, then incremented the ans by 1.

Under is the implementation of the above strategy.

C++

  

#embody <bits/stdc++.h>

utilizing namespace std;

  

int test = 0;

int isPrime[100001];

  

void pre()

{

    test = 1;

    memset(isPrime, 1, sizeof isPrime);

    isPrime[0] = isPrime[1] = 0;

    for (int i = 4; i <= 100000; i += 2)

        isPrime[i] = 0;

  

    for (int i = 3; i * i <= 100000; i += 2) {

        if (isPrime[i]) {

            for (int j = 3; j * i <= 100000; j += 2)

                isPrime[i * j] = 0;

        }

    }

}

  

int resolve(int N, string S)

{

    

    

    

    if (!test)

        pre();

    int dis = 0;

    int ans = 0;

  

    

    unordered_map<char, int> f;

    for (int i = 0; i < N; i++) {

  

        

        

        if (f[S[i]] == 0) {

            f[S[i]]++;

            dis++;

        }

  

        

        

        

        if (((i % 2) == 0) and isPrime[dis]) {

            ans++;

        }

    }

    return ans;

}

  

int fundamental()

{

    int N = 6;

    string S = "aabagh";

  

    

    cout << resolve(N, S) << endl;

  

    return 0;

}

Time Complexity: O(N√N)
Auxiliary House: O(N)

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