How do you solve Maximum Balanced Subsequence Sum on LeetCode?

Maximum Balanced Subsequence Sum (LeetCode #2926) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Transforming the balance condition allows for more efficient tracking of sums.

What is the brute force solution for Maximum Balanced Subsequence Sum?

The brute force approach for Maximum Balanced Subsequence Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences and checking if they are balanced. This method is straightforward but inefficient, as it examines every combination. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Maximum Balanced Subsequence Sum?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as arrays with all negative numbers or single elements. Explain your thought process clearly while coding to demonstrate your understanding.

What are common mistakes when solving Maximum Balanced Subsequence Sum?

Forgetting to check the balance condition correctly when updating sums. Not initializing the dp array properly, leading to incorrect results.

#2926

Maximum Balanced Subsequence Sum

Hard

ArrayBinary SearchDynamic ProgrammingBinary Indexed TreeSegment TreeDynamic ProgrammingArray Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal approach uses dynamic programming to build up the maximum sum of balanced subsequences efficiently. By transforming the balance condition, we can track the best sums in a single pass.

⚙️

Algorithm

4 steps

1Step 1: Create an array dp where dp[i] represents the maximum sum of a balanced subsequence ending at index i.
2Step 2: Initialize dp[i] with nums[i] for all i.
3Step 3: Iterate through the array, and for each index i, check all previous indices j to see if the balance condition holds and update dp[i] accordingly.
4Step 4: The result is the maximum value in the dp array.

solution.py14 lines

1# Full working Python code
2
3def max_balanced_subseq_sum(nums):
4    n = len(nums)
5    dp = [0] * n
6    for i in range(n):
7        dp[i] = nums[i]  # Start with the current element
8        for j in range(i):
9            if nums[i] - i >= nums[j] - j:  # Check balance condition
10                dp[i] = max(dp[i], dp[j] + nums[i])
11    return max(dp)
12
13# Example usage:
14print(max_balanced_subseq_sum([3, 3, 5, 6]))  # Output: 14

ℹ

Complexity note: The time complexity is O(n²) due to the nested loop for checking previous indices, but we only store results for each index, leading to linear space complexity.

1Transforming the balance condition allows for more efficient tracking of sums.
2Dynamic programming helps in building solutions incrementally, reducing redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.