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Problem solution · Java

Minimum Cost Good Caption

Minimum Cost Good Caption: a Java solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
78 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Minimum Cost Good Caption, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 78 lines of Java from the credited upstream file 3441.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 6 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Cost Good Caption · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String minCostGoodCaption(String caption) {    final int n = caption.length();    if (n < 3)      return "";     final int MAX_COST = 1_000_000_000;    int[][][] dp = new int[n][26][3];    Arrays.stream(dp).forEach(A -> Arrays.stream(A).forEach(B -> Arrays.fill(B, MAX_COST)));    // dp[i][j][k] := the minimum cost of caption[i..n - 1], where j is the last    // letter used, and k is the count of consecutive letters    for (char c = 'a'; c <= 'z'; ++c)      dp[n - 1][c - 'a'][0] = Math.abs(caption.charAt(n - 1) - c);     int minCost = MAX_COST;    for (int i = n - 2; i >= 0; --i) {      int newMinCost = MAX_COST;      for (char c = 'a'; c <= 'z'; ++c) {        final int j = c - 'a';        final int changeCost = Math.abs(caption.charAt(i) - c);        dp[i][j][0] = changeCost + minCost;        dp[i][j][1] = changeCost + dp[i + 1][j][0];        dp[i][j][2] = changeCost + Math.min(dp[i + 1][j][1], dp[i + 1][j][2]);        newMinCost = Math.min(newMinCost, dp[i][j][2]);      }      minCost = newMinCost;    }     // Reconstruct the string.    StringBuilder sb = new StringBuilder();    int cost = MAX_COST;    int letter = -1;     // Find the initial best letter.    for (int c = 25; c >= 0; --c)      if (dp[0][c][2] <= cost) {        letter = c;        cost = dp[0][c][2];      }     // Add the initial triplet.    cost -= appendLetter(caption, 0, (char) ('a' + letter), sb);    cost -= appendLetter(caption, 1, (char) ('a' + letter), sb);    cost -= appendLetter(caption, 2, (char) ('a' + letter), sb);     // Build the rest of the string.    for (int i = 3; i < n;) {      // Check if we should switch to a new letter.      final int nextLetter = getNextLetter(dp, i, cost);      if (nextLetter < letter || Arrays.stream(dp[i][letter]).min().getAsInt() > cost) {        letter = nextLetter;        cost -= appendLetter(caption, i, (char) ('a' + letter), sb);        cost -= appendLetter(caption, i + 1, (char) ('a' + letter), sb);        cost -= appendLetter(caption, i + 2, (char) ('a' + letter), sb);        i += 3;      } else {        cost -= appendLetter(caption, i, (char) ('a' + letter), sb);        i += 1;      }    }     return sb.toString();  }   private int getNextLetter(int[][][] dp, int i, int cost) {    int nextLetter = 26;    for (int c = 25; c >= 0; --c)      if (cost == dp[i][c][2])        nextLetter = c;    return nextLetter;  }   private int appendLetter(String caption, int i, char letter, StringBuilder sb) {    sb.append(letter);    return Math.abs(caption.charAt(i) - letter);  }}

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