CPMagical CP Library

Problem solution · Java

Number of Good Binary Strings

Number of Good Binary Strings: 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
32 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 Number of Good Binary Strings, 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

  • 32 lines of Java from the credited upstream file 2533.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 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 codeNumber of Good Binary Strings · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int goodBinaryStrings(int minLength, int maxLength, int oneGroup, int zeroGroup) {    final int MOD = 1_000_000_007;    // dp[i] := the number of good binary strings with length i    int[] dp = new int[maxLength + 1];    dp[0] = 1; // ""     for (int i = 0; i <= maxLength; ++i)      // There are good binary strings with length i, so we can append      // consecutive 0s or 1s after it.      if (dp[i] > 0) {        final int appendZeros = i + zeroGroup;        if (appendZeros <= maxLength) {          dp[appendZeros] += dp[i];          dp[appendZeros] %= MOD;        }        final int appendOnes = i + oneGroup;        if (appendOnes <= maxLength) {          dp[appendOnes] += dp[i];          dp[appendOnes] %= MOD;        }      }     int ans = 0;    for (int i = minLength; i <= maxLength; ++i) {      ans += dp[i];      ans %= MOD;    }    return ans;  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗