Problem Description

Given an array nums, return the number of subsequences with an odd sum of elements. Since the answer may be very large, return it modulo (10^9 + 7).

Intuition

To solve the problem of counting subsequences with an odd sum, we can use a dynamic programming approach.

We keep track of two numbers: f[0] for the number of subsequences with an even sum and f[1] for the number of subsequences with an odd sum. Initially, both are set to 0 because there are no subsequences yet.

As we traverse through each element of the input array nums, we need to update these two counters based on whether the current element is odd or even:

1. If the current element x is odd:

   • Adding x to any existing subsequence with an even sum will result in an odd sum. Thus, the new number of subsequences with an odd sum (f[1]) becomes the old count of even-sum subsequences plus the current count of odd-sum subsequences, plus one for the new subsequence consisting only of this element.
   • Similarly, adding x to any existing subsequence with an odd sum will result in an even sum, hence the new number of even-sum subsequences (f[0]) becomes the old count of odd-sum subsequences plus the current count of even-sum subsequences.

2. If the current element x is even:

   • Adding x to any existing subsequence with an even sum will still result in an even sum, so f[0] is updated to include twice the number of existing even-sum subsequences plus one for the new subsequence consisting only of this element.
   • Adding x to any existing subsequence with an odd sum remains odd, so f[1] is updated to twice its current count.

Finally, return f[1] as it represents the count of subsequences with an odd sum.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

To achieve the goal of counting subsequences with an odd sum efficiently, we utilize a dynamic programming approach as outlined in the reference solution. Here's a step-by-step explanation:

1. Variables Initialization:

   • We initialize f as a list of two integers: f[0] and f[1], both initially set to 0. f[0] will track the count of subsequences with an even sum, while f[1] will count those with an odd sum.

2. Iterate through the Array:

   • For each element x in the array nums, we need to update our counts depending on whether x is odd or even.

3. If x is Odd:

   • Update f[0] to the sum of the previous f[0] and f[1], as adding an odd number to an even sum makes it odd, and adding it to an odd sum makes it even:

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

   • Update f[1] by adding to it the sum of the old f[0] plus f[1] plus one to account for subsequences containing just the number x:

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

4. If x is Even:

   • Update f[0] by doubling the previous f[0] and adding one, as adding an even number retains the even nature of sums:

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

   • Update f[1] by doubling the previous f[1], since adding an even number to an odd sum retains its odd nature:

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

5. Modulo Operation:

   • Since the result can be very large, every update to f[0] and f[1] is followed by taking modulo 10^9 + 7 to prevent overflow and maintain compliance with the problem constraints.

6. Return the Result:

   • After processing all elements in nums, the variable f[1] holds the number of subsequences with an odd sum. Return this value as the final result:

     return f[1]

This dynamic programming solution ensures we efficiently calculate the number of subsequences with odd sums in linear time complexity relative to the size of the input array, nums.

Example Walkthrough

Let's walk through the solution approach with a small example.

Consider the array nums = [1, 2, 3].

1. Initialization:

   • Begin with f = [0, 0], where f[0] is the count of subsequences with an even sum, and f[1] is the count with an odd sum.

2. Process Element 1 (odd):

   • Before processing, f = [0, 0].
   • Add 1 to existing even sum sequences (f[0]): No impact since no even sequences exist yet.
   • Add 1 to existing odd sum sequences (f[1]): No impact since no odd sequences exist yet.
   • Add sequence [1]: Increase odd sum sequences (f[1]) by 1.
   • After processing, f = [0, 1].

3. Process Element 2 (even):

   • Before processing, f = [0, 1].
   • Add 2 to existing even sum sequences (f[0]): Still results in even sums, double f[0] and add 1 (new sequence [2]).
   • Add 2 to existing odd sum sequences (f[1]): Results in odd sums, double f[1].
   • Update: f[0] becomes 0*2 + 1 = 1; f[1] becomes 1*2 = 2.
   • After processing, f = [1, 2].

4. Process Element 3 (odd):

   • Before processing, f = [1, 2].
   • Add 3 to existing even sum sequences (f[0]): Results in odd sums, use old f[0] + f[1].
   • Add 3 to existing odd sum sequences (f[1]): Results in even sums, double f[1].
   • Add sequence [3]: Increase odd sum (f[1]) by 1.
   • Calculation:
     • New f[0] = 1 + 2 = 3.
     • New f[1] = 1 + 2 + 1 = 4.
   • After processing, f = [3, 4].

5. Final result:

   • The number of subsequences with an odd sum is held in f[1], which equals 4.

The steps above illustrate how the dynamic programming solution processes each element in nums by updating the count of subsequences with even and odd sums.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the array nums. This is because the code iterates through each element of the input array exactly once.

The space complexity of the code is O(1), as it uses a fixed amount of additional space (the list f with two elements) regardless of the input size.

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