What is the time complexity of Number of Subsequences That Satisfy the Given Sum Condition?

The optimal solution for Number of Subsequences That Satisfy the Given Sum Condition runs in O(n log n) time and uses O(n) space. Sorting the array takes O(n log n), and the two-pointer traversal takes O(n), making this approach much more efficient than brute force.

What is the brute force solution for Number of Subsequences That Satisfy the Given Sum Condition?

The brute force approach for Number of Subsequences That Satisfy the Given Sum Condition is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible subsequences and checks if their minimum and maximum sum is less than or equal to the target. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Number of Subsequences That Satisfy the Given Sum Condition?

Number of Subsequences That Satisfy the Given Sum Condition uses the following concepts: Array, Two Pointers, Binary Search, Sorting. The recommended patterns to study are: Two Pointers, Sorting.

What are the interview tips for Number of Subsequences That Satisfy the Given Sum Condition?

Always consider sorting the input when dealing with range-based problems. Be prepared to explain your thought process and why you chose a specific approach. Practice identifying patterns in problems, such as the two-pointer technique.

What are common mistakes when solving Number of Subsequences That Satisfy the Given Sum Condition?

Failing to account for duplicates in the array when counting subsequences. Not using modular arithmetic when counting large numbers of subsequences.

#1498

Number of Subsequences That Satisfy the Given Sum Condition

Medium

ArrayTwo PointersBinary SearchSortingTwo PointersSorting

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

By sorting the array and using a two-pointer technique, we can efficiently find valid subsequences without generating all of them. This method leverages the properties of sorted arrays to quickly find pairs that meet the criteria.

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Algorithm

3 steps

1Step 1: Sort the array nums.
2Step 2: Initialize two pointers: left at the start and right at the end of the array.
3Step 3: While left <= right, check if nums[left] + nums[right] <= target. If true, count all subsequences formed by nums[left] and nums[right] and move left pointer up. If false, move right pointer down.

solution.py17 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def numSubseq(nums, target):
5    nums.sort()
6    left, right = 0, len(nums) - 1
7    count = 0
8    power = [1] * (len(nums) + 1)
9    for i in range(1, len(nums) + 1):
10        power[i] = (power[i - 1] * 2) % MOD
11    while left <= right:
12        if nums[left] + nums[right] <= target:
13            count = (count + power[right - left]) % MOD
14            left += 1
15        else:
16            right -= 1
17    return count

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Complexity note: Sorting the array takes O(n log n), and the two-pointer traversal takes O(n), making this approach much more efficient than brute force.

1Sorting the array allows efficient pairing of minimum and maximum values.
2Using powers of 2 helps count the number of valid subsequences without generating them.

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