Aisha is playing a game where she controls a character to catch pancakes falling onto a stage

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Aisha is playing a game where she controls a character to catch pancakes falling onto a stage. • The character can move left and right on a stage of length n. The character’s location : is in the range {1, 2, ….., n}. Let or by the location of the character at time t. At each second, Aisha can move the character to the left (C++1 = 2t – 1) one step, to the right 2+1 = 2t +1) one step, or stay in the same spot (3t+1 = x+). (Exceptions: can’t move left when current 2 = 1 and can’t move right when I = n.)

  • The game lasts m seconds.
  • Aisha knows the game really well, so she knows when and where pancakes will fall, i.e. pairs (t,p), which means at t seconds, a pancake will be at position p. You are given this, and assume lookup of this information in constant time.

The character starts at location 1 at time 0. Please design an algorithm to find out the most pancakes Aisha can catch. It should run in Omn) time – please also justify why.

Example: (t, p) sequence: (1, 1), (2, 2), (3, 4), (3, 5), (4, 3), (4,4), (5, 2)). Optimal solu- tion: 4 pancakes. Aisha can stay still for first step (21 = 1) and catch pancake, move right (22 = 2) and catch pancake, stay still (23 = 2) then move right (14 = 3) and catch pancake, and move left at (35 = 2) and catch pancake.

The correct answer and explanation is:

Algorithm

The task can be solved using dynamic programming (DP). Let dp[t][x] represent the maximum number of pancakes Aisha can catch if she is at position x at time t. We can build this DP table based on the following rules:

  1. Initialization: At t = 0, dp[0][1] = 0 (Aisha starts at position 1 and has not caught any pancakes).
  2. Transition: For each t and x, the value of dp[t][x] depends on:
    • Staying at the same position (x): dp[t][x] = max(dp[t-1][x], dp[t][x]).
    • Moving left (x-1), if x > 1: dp[t][x] = max(dp[t-1][x-1], dp[t][x]).
    • Moving right (x+1), if x < n: dp[t][x] = max(dp[t-1][x+1], dp[t][x]).
  3. Catch pancakes: If a pancake falls at position x at time t, increment dp[t][x] by 1.
  4. Final result: The maximum value in the last row of the DP table (dp[m][*]) gives the maximum number of pancakes Aisha can catch.

Implementation Steps

  1. Initialize a DP table of size m+1 x n+1 with zeros.
  2. Preprocess the pancake positions for constant-time lookup using a hash table.
  3. Fill the DP table row by row (time step by time step).
  4. Return the maximum value in the last row.

Why O(mn)O(mn)?

  • We iterate over m time steps.
  • For each time step, we update n positions.
  • Each update involves constant-time operations (checking moves and adding pancake values). Thus, the time complexity is O(mn)O(mn).

Example Walkthrough

Given n = 5, m = 5, and pancakes at (1,1), (2,2), (3,4), (3,5), (4,3), (4,4), (5,2):

  1. Start at (1, 1), catch 1 pancake.
  2. Move to (2, 2) in the second second, catch another pancake.
  3. Move to (3, 3) (not catching at (3,4) or (3,5)), then move to (4, 3) for a pancake.
  4. Return to (5, 2) for a total of 4 pancakes.

This algorithm ensures correctness and efficiency by systematically exploring all possible movements while avoiding redundant computations.

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