Trading Space for Time: Boosting Algorithm Efficiency

When solving algorithmic problems, efficiency is often a balancing act between time complexity and space complexity. One powerful strategy is trading space for time. Let’s explore how this concept can help optimize a brute-force solution for counting "bad pairs" in a list. You can find this problem here.

The Brute-Force Approach

The brute-force solution is quite simple but becomes insufficient for large input sizes due to its time complexity:

class Solution:
    def countBadPairs(self, nums: List[int]) -> int:
        n = len(nums)
        bad_pairs = 0

        for i in range(n):
            for j in range(i + 1, n):
                if i < j and (j - i != nums[j] - nums[i]):
                    bad_pairs += 1

        return bad_pairs

This approach has a time complexity of O(n²) because of the nested loops. For small lists, this might work fine, but as n grows, it quickly becomes impractical.

Can We Make It More Efficient?

Yes! Instead of counting bad pairs directly, it’s easier to count good pairs (i.e., pairs that satisfy the condition j - i == nums[j] - nums[i]) and subtract that count from the total number of possible pairs.

Observing the Pattern

Notice that the condition (j - i != nums[j] - nums[i]) is equivalent to:

nums[i] - i != nums[j] - j

Thus, we can keep track of nums[i] - i using a HashMap (dictionary in Python) to count how often each difference appears.

Calculating Total Pairs

The total number of pairs in any list of length n is:

total_pairs = n * (n - 1) // 2

The Optimized Approach

1. Use a dictionary diff_count to track how many times each difference diff = nums[i] - i has appeared.
2. For each element in the list, if diff has been seen before, it means there exists some j < i such that nums[i] - i == nums[j] - j. This contributes to the count of good pairs.
3. After checking, update the dictionary with the current diff.
4. Finally, calculate the number of bad pairs by subtracting the count of good pairs from the total number of pairs:

from collections import defaultdict

class Solution:
    def countBadPairs(self, nums: List[int]) -> int:
        n = len(nums)

        # Total possible pairs
        total_pairs = n * (n - 1) // 2

        # Map to store the frequency of (nums[i] - i)
        diff_count = defaultdict(int)

        # Count good pairs
        good_pairs = 0

        for i in range(n):
            # Compute the difference
            diff = nums[i] - i

            # If the difference has been seen before, it contributes to good pairs
            if diff in diff_count:
                good_pairs += diff_count[diff]

            # Update the count of this difference
            diff_count[diff] += 1

        # Bad pairs are total pairs minus good pairs
        num_bad_pairs = total_pairs - good_pairs

        return num_bad_pairs

Complexity Analysis

• Time Complexity: O(n) – We iterate through the list once, and each dictionary operation takes constant time on average.
• Space Complexity: O(n) – We store at most n unique differences in the dictionary.

Trading Space for Time

In the brute-force approach, we had:

• Time Complexity: O(n²)
• Space Complexity: O(1)

In the optimized approach, we traded additional space for a dictionary (O(n) space complexity) to reduce the time complexity to O(n). This trade-off is worth it when the input size grows large because the gain in time efficiency is significant.

Conclusion

This example illustrates how trading space for time can drastically improve algorithm performance. By using a HashMap to store intermediate results, we reduced the time complexity from O(n²) to O(n), making our solution more scalable and practical for large inputs.

In algorithm design, always ask yourself: "Can I use extra space to reduce the time complexity?" Often, the answer leads to a more efficient and elegant solution.