Solution 2 for Scaler Topics Fortnightly Contest - 25
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This article is part of the Scaler Topics Fortnightly Contest - 25
Solution Approach
We’ll utilize two arrays: prefixR[i] to represent the count of ‘R’ beads in the prefix of length i and prefixB[i] to denote the count of ‘B’ beads in the prefix of length i.
Consider two positions, i and j, where 0 ≤ i ≤ j ≤ n. Our objective is to remove all ‘B’ beads from the prefix ending at position i, from the suffix starting at position j, and also all ‘R’ beads between positions i and j. The minimum number of beads needed to be removed can be calculated as prefixB[i] + (prefixB[N] - prefixB[j]) + (prefixR[j] - prefixR[i]).
To achieve this, we can employ a two-pointer approach using nested loops. Iterate through all possible positions of i and j and compute the answer by considering the given conditions.
Time Complexity: O(N * N) Space Complexity: O(2 * N)
C++ Implementation
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