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Problem solution · Java

Number of Subsequences with Odd Sum

Number of Subsequences with Odd Sum: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
24 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

Direct simulation

For Number of Subsequences with Odd Sum, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 24 lines of Java from the credited upstream file 3247.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 Subsequences with Odd Sum · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int subsequenceCount(int[] nums) {    final int MOD = 1_000_000_007;    int even = 0; // the number of subsequences with even sum    int odd = 0;  // the number of subsequences with odd sum     for (final int num : nums)      if (num % 2 == 0) {        // Appending an even number to a subsequence doesn't change the parity.        // The even number itself is also a valid subsequence.        even = (even + even + 1) % MOD;        odd = (odd + odd) % MOD;      } else {        // Appending an odd number to a subsequence changes the parity.        // The odd number itself is also a valid subsequence.        final int newEven = (even + odd) % MOD;        odd = (odd + even + 1) % MOD;        even = newEven;      }     return odd % MOD;  }}

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