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Yes, it is possible to apply similar ideas. The "pre-computation + prefix sum" technique involves computing cumulative sums to answer range queries efficiently. A standard prefix sum array `P` where `P[k]` stores the sum of `A[0...k-1]` allows `sum(A[i...j])` to be computed as `P[j+1] - P[i]` in O(1) time after O(N) pre-computation. The idea suggested in the question, "pre-compute the sum for each possible interval length," can be implemented using a 2D table. Let's define `T[i][L]` as the sum of the sub-array starting at index `i` with length `L`. That is, `T[i][L] = A[i] + A[i+1] + ... + A[i+L-1]`. Here's how this approach works: 1. **Pre-computation:** * We can build this `N x N` table `T` using dynamic programming. * For `L = 1`: `T[i][1] = A[i]` for all valid `i`. * For `L > 1`: `T[i][L] = T[i][L-1] + A[i+L-1]` for all valid `i`. * This pre-computation will take `O(N^2)` time (N starting positions, N possible lengths) and `O(N^2)` space to store the table. Example: If `A = [1, 2, 3, 4]` (N=4) `T` table (sums `A[i...i+L-1]`) `L=1`: `T[0][1] = A[0] = 1` `T[1][1] = A[1] = 2` `T[2][1] = A[2] = 3` `T[3][1] = A[3] = 4` `L=2`: `T[0][2] = T[0][1] + A[1] = 1 + 2 = 3` `T[1][2] = T[1][1] + A[2] = 2 + 3 = 5` `T[2][2] = T[2][1] + A[3] = 3 + 4 = 7` `L=3`: `T[0][3] = T[0][2] + A[2] = 3 + 3 = 6` `T[1][3] = T[1][2] + A[3] = 5 + 4 = 9` `L=4`: `T[0][4] = T[0][3] + A[3] = 6 + 4 = 10` 2. **Querying:** * To find the sum of a given range `A[i...j]`: * Calculate the length of the interval: `L = j - i + 1`. * Look up the pre-computed sum directly: `T[i][L]`. * This query takes `O(1)` time. This approach indeed pre-computes the sum for every possible interval (each defined by its starting position `i` and length `L`) and allows for `O(1)` query time. It is "similar" to prefix sums in that it uses pre-computation and dynamic programming to speed up range sum queries, even though it's less space and time efficient than a standard prefix sum array for this particular problem.

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