Solution 3 for Scaler Topics Fortnightly Contest - 9

This article is part of the Scaler Topics Fortnightly Contest - 9

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Solution Approach

• Let's build the string character by character.
• While adding characters to end of the string we only care about the last character and length of the continuous occurrence of this character in the end of the string.
• Let the last character of the string be C and it's occurence in the end be CNT, if CNT is less than B then we can add C again to our string else we need to choose a different character.
• Doing the above process by brute force is not optimal, so here we can use DP.
• Let DP[i][j][k] denote the number of strings of length i, last character j and count of j in end of the string is k.

C++ Implementation

Java Implementation

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Python Implementation