Problem Description

Given an array nums, you need to find the total number of subsequences that have an odd sum. A subsequence is formed by deleting some (possibly zero) elements from the original array while maintaining the relative order of the remaining elements.

For example, if nums = [1, 2, 3], the subsequences with odd sums would be: [1], [3], [1, 2], [2, 3], and [1, 2, 3].

Since the number of such subsequences can be very large, return the answer modulo 10^9 + 7.

The solution uses dynamic programming to track two states:

• f[0]: Count of subsequences with even sum
• f[1]: Count of subsequences with odd sum

When processing each element x in the array:

If x is odd:

• New even-sum subsequences = previous even-sum subsequences + (previous odd-sum subsequences combined with x)
• New odd-sum subsequences = (previous even-sum subsequences combined with x) + previous odd-sum subsequences + subsequence containing only x

If x is even:

• New even-sum subsequences = previous even-sum subsequences + (previous even-sum subsequences combined with x) + subsequence containing only x
• New odd-sum subsequences = previous odd-sum subsequences + (previous odd-sum subsequences combined with x)

The final answer is f[1], representing the total count of subsequences with odd sums.

Intuition

The key insight is that when building subsequences, we need to track how adding each new element affects the parity (odd/even nature) of existing subsequence sums.

Think about what happens when we encounter a new element. For each existing subsequence, we have two choices: either include this element or skip it. If we skip it, the subsequence sum remains unchanged. If we include it, the parity of the sum depends on both the current element's parity and the existing subsequence's sum parity.

The parity rules are straightforward:

• Even + Even = Even
• Odd + Odd = Even
• Even + Odd = Odd
• Odd + Even = Odd

This suggests we should maintain two counters: one for subsequences with even sums and one for subsequences with odd sums. As we process each element, we update these counters based on how the new element would change the parity.

When we see an odd number:

• All previous even-sum subsequences become odd when we append this number
• All previous odd-sum subsequences become even when we append this number
• We also create one new odd-sum subsequence containing just this element

When we see an even number:

• All previous even-sum subsequences remain even when we append this number
• All previous odd-sum subsequences remain odd when we append this number
• We also create one new even-sum subsequence containing just this element

By tracking these two states throughout the array traversal, we can efficiently count all subsequences with odd sums without explicitly generating them. The beauty of this approach is that it reduces an exponential problem (generating all 2^n subsequences) to a linear time solution with just two state variables.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

We implement a dynamic programming solution using two state variables to track the count of subsequences by their sum parity.

Initialization:

• Create an array f with two elements: f[0] for even-sum subsequences count and f[1] for odd-sum subsequences count
• Initially, both are set to 0 since we haven't processed any elements yet
• Define mod = 10^9 + 7 for modulo operations

Processing Each Element:

For each number x in the array, we update our counters based on whether x is odd or even.

Case 1: When x is odd (x % 2 == 1):

The update formulas are:

f[0] = (f[0] + f[1]) % mod
f[1] = (f[0] + f[1] + 1) % mod

• New even-sum count: Previous even-sum subsequences (not including x) + previous odd-sum subsequences that become even when we add odd x
• New odd-sum count: Previous odd-sum subsequences (not including x) + previous even-sum subsequences that become odd when we add odd x + 1 (the subsequence containing only x)

Note: We use simultaneous assignment in Python to update both values using the old values of f[0] and f[1].

Case 2: When x is even (x % 2 == 0):

The update formulas are:

f[0] = (f[0] + f[0] + 1) % mod
f[1] = (f[1] + f[1]) % mod

• New even-sum count: Previous even-sum subsequences (not including x) + previous even-sum subsequences with x added (sum stays even) + 1 (the subsequence containing only x)
• New odd-sum count: Previous odd-sum subsequences (not including x) + previous odd-sum subsequences with x added (sum stays odd)

Final Result: After processing all elements, f[1] contains the total count of subsequences with odd sums, which is our answer.

The time complexity is O(n) where n is the length of the array, and the space complexity is O(1) as we only use two variables to maintain state.

Example Walkthrough

Let's walk through the solution with nums = [1, 2, 3].

Initial state:

• f[0] = 0 (even-sum subsequences)
• f[1] = 0 (odd-sum subsequences)

Step 1: Process element 1 (odd)

• Since 1 is odd, we apply the odd number rules
• New f[0] = (0 + 0) % mod = 0
• New f[1] = (0 + 0 + 1) % mod = 1
• Current subsequences: [1] with sum 1 (odd)
• State: f[0] = 0, f[1] = 1

Step 2: Process element 2 (even)

• Since 2 is even, we apply the even number rules
• New f[0] = (0 + 0 + 1) % mod = 1
• New f[1] = (1 + 1) % mod = 2
• Current subsequences:
  • Even sum: [2] (sum = 2)
  • Odd sum: [1] (sum = 1), [1,2] (sum = 3)
• State: f[0] = 1, f[1] = 2

Step 3: Process element 3 (odd)

• Since 3 is odd, we apply the odd number rules
• Using simultaneous assignment:
  • New f[0] = (1 + 2) % mod = 3
  • New f[1] = (1 + 2 + 1) % mod = 4
• Current subsequences:
  • Even sum: [2], [1,3], [1,2,3] (3 total)
  • Odd sum: [1], [3], [1,2], [2,3] (4 total)
• State: f[0] = 3, f[1] = 4

Result: The answer is f[1] = 4, representing 4 subsequences with odd sums.

Let's verify our answer:

• [1]: sum = 1 (odd) ✓
• [3]: sum = 3 (odd) ✓
• [1,2]: sum = 3 (odd) ✓
• [2,3]: sum = 5 (odd) ✓
• [1,2,3]: sum = 6 (even)
• [2]: sum = 2 (even)
• [1,3]: sum = 4 (even)

Indeed, we have exactly 4 subsequences with odd sums!

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the algorithm iterates through the array exactly once with a single for loop, and within each iteration, it performs only constant-time operations (modulo arithmetic and variable assignments).

The space complexity is O(1). The algorithm uses only a fixed amount of extra space regardless of the input size - specifically, it maintains an array f of size 2 and a constant mod variable. The space usage does not grow with the size of the input array nums.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect State Update Order

One of the most common mistakes is updating the states in the wrong order, causing the use of already-modified values instead of the original values from the previous iteration.

Incorrect Implementation:

if num % 2 == 1:
    even_sum_count = (even_sum_count + odd_sum_count) % MOD
    odd_sum_count = (even_sum_count + odd_sum_count + 1) % MOD  # Bug: uses modified even_sum_count

In this buggy code, when updating odd_sum_count, we're using the newly updated even_sum_count instead of the original value, leading to incorrect results.

Solution: Use temporary variables or simultaneous assignment to ensure both updates use the original values:

if num % 2 == 1:
    new_even = (even_sum_count + odd_sum_count) % MOD
    new_odd = (even_sum_count + odd_sum_count + 1) % MOD
    even_sum_count, odd_sum_count = new_even, new_odd

Or using Python's tuple unpacking:

if num % 2 == 1:
    even_sum_count, odd_sum_count = (even_sum_count + odd_sum_count) % MOD, \
                                    (even_sum_count + odd_sum_count + 1) % MOD

2. Misunderstanding the Problem Statement

Another pitfall is misinterpreting what constitutes an "odd sum subsequence." Some might mistakenly think it refers to:

• Subsequences with an odd number of elements
• Subsequences containing only odd numbers
• Subsequences where the count of odd numbers is odd

Clarification: The problem asks for subsequences where the sum of all elements is odd, regardless of how many elements are in the subsequence or whether they are odd or even.

3. Forgetting the Modulo Operation

Since the result can be very large, forgetting to apply modulo at each step can cause integer overflow in languages with fixed integer sizes, or incorrect results when the final modulo is applied to an already overflowed value.

Incorrect:

even_sum_count = even_sum_count + odd_sum_count  # Missing modulo

Correct:

even_sum_count = (even_sum_count + odd_sum_count) % MOD

4. Incorrect Initial State for Even Numbers

When processing an even number, some might forget to account for the subsequence containing only that even number (which has an even sum), missing the +1 in the even count update.

Incorrect:

if num % 2 == 0:
    even_sum_count = (even_sum_count * 2) % MOD  # Missing +1

Correct:

if num % 2 == 0:
    even_sum_count = (even_sum_count * 2 + 1) % MOD