Leetcode Problem 1883. Minimum Skips to Arrive at Meeting On Time

1883. Minimum Skips to Arrive at Meeting On Time

Leetcode Solutions

Dynamic Programming Approach

  1. Initialize a 2D DP array dp with dimensions (n+1) x (n+1) and fill it with a large value (e.g., 1e15) to represent infinity.
  2. Set the base case dp[0][0] to 0.
  3. Iterate over each road i from 1 to n: a. Calculate the time needed to reach the end of the i-th road without any skips: dp[i][0] = dist[i-1] + ((dp[i-1][0] + speed - 1) / speed) * speed. b. For each possible number of skips j from 1 to i, calculate the minimum time needed with j skips: dp[i][j] = dist[i-1] + min(dp[i-1][j-1], ((dp[i-1][j] + speed - 1) / speed) * speed).
  4. Iterate over the DP array to find the minimum number of skips needed to arrive on time: Check if dp[n][i] <= speed * hoursBefore for each i from 0 to n.
  5. Return the minimum number of skips if possible, otherwise return -1.
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Greedy Approach with Priority Queue

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