algoadvance

Given an integer array nums of length n, you are tasked with splitting it into one or more non-overlapping subarrays. Each subarray should contain the maximum possible elements but must adhere to the following conditions:

• Each subarray is a consecutive sequence from nums.
• The sum of any subarray must be an odd number.

Return the minimum number of subarrays required to achieve this goal.

Clarifying Questions:

1. Input Size and Values:
   • What are the constraints on the size of the nums array?
   • Is there a range for the values within nums?
2. Edge Cases:
   • What should be returned if the input nums is empty?
   • How to handle cases where all elements are even? Should we split each element individually?
3. Properties of Odd Sum:
   • What is the definition of an odd sum in the context of this problem?

Code:

No Python code has been provided yet. Let’s proceed to the strategy.

Strategy:

1. Understanding “Odd Sum”:
   • A sequence has an odd sum if the sum of its elements is an odd number.
   • This means if we are to split the array into subarrays, each subarray must have an odd-summation.
2. Breaking it Down:
   • To ensure a subarray has an odd sum, the sum of its elements must be odd.
   • We know that adding any even number does not change the parity (odd/even nature) of the sum. Hence, an odd sum is obtained if there is an odd number of odd integers.
3. Approach:
   • Traverse the array while maintaining the cumulative sum.
   • Whenever the cumulative sum becomes odd, consider this as a valid subarray.
   • If the sum is even and cannot be made odd by extending, start a new subarray to satisfy the condition.
4. Algorithm Steps:
   • Initialize variables to keep track of the minimum number of subarrays.
   • Traverse the array and keep a running sum.
   • When the running sum is odd, consider it as a subarray and reset the running sum. Increment the count of subarrays.
   • If the running sum stays even, continue to the next element.
   • Finalize the subarray count and return it.

Time Complexity:

• The time complexity of this approach is O(n) since we traverse the array once.

Implementation:

def min_subarray_to_make_sum_odd(nums):
    subarray_count = 0
    current_sum = 0
    
    for num in nums:
        current_sum += num
        if current_sum % 2 != 0:  # Check if current sum is odd
            subarray_count += 1    # A subarray with an odd sum is found
            current_sum = 0        # Reset current sum for the next subarray
            
    # If there's any leftover sum that's not zero, it means there's one more subarray
    if current_sum != 0:
        subarray_count += 1

    return subarray_count

# Example usage
nums = [1, 2, 3, 4, 5]
print(min_subarray_to_make_sum_odd(nums))  # Output: expected number of subarrays

In this implementation:

• We loop through each element of the array nums.
• We maintain the cumulative sum of elements seen so far.
• If this sum becomes odd, we treat this as the end of a valid subarray and reset the sum.
• The overall complexity remains linear, O(n), as desired.

This solution adheres to the problem requirements and efficiently determines the minimum number of subarrays with an odd sum.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?