# 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 whole rely of distinct characters within the index vary [0, i] is a primary 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 rely of distict characters upto index 2 and 4 are 2 and three respectively which is prime.

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

Strategy: This downside will 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 verify for the required situations  at each character.

Comply with the under steps to resolve the issue:

• Create a precompute perform and calculate all prime components utilizing the sieve of Eratosthenes.
• Create a hashmap to retailer frequencies of characters which is able to assist us decide if the character is a reproduction or not.
• Iterate over the string from and when any even index is reached verify 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 ``utilizing` `namespace` `std;`` ` `int` `verify = 0;``int` `isPrime[100001];`` ` `void` `pre()``{``    ``verify = 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` `(!verify)``        ``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` `essential()``{``    ``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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