Software Development

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

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Given a string S of measurement 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 rely of distinct characters within the index vary [0, i] is a primary quantity. 


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

Enter: N = 2, S = “gg”
Output: 0
Rationalization: No invalid character

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

Concept: The thought is to precompute all of the prime numbers within the given vary of N after which simply test for the required circumstances  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 reproduction 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.

Beneath is the implementation of the above strategy.



#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 clear up(int N, string S)




    if (!test)


    int dis = 0;

    int ans = 0;



    unordered_map<char, int> f;

    for (int i = 0; i < N; i++)




        if (f[S[i]] == 0)








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




    return ans;


int predominant()

    int N = 6;

    string S = "aabagh";



    cout << clear up(N, S) << endl;


    return 0;

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

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