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

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Number of Subsequences That Satisfy the Given Sum Condition?

This problem uses Array, Two Pointers, Binary Search, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Subsequences That Satisfy the Given Sum Condition?

Insight 1: Sorting the input array `nums` allows us to efficiently find the minimum and maximum elements by considering elements from the beginning and the end of the sorted array. Insight 2: For a given minimum element `nums[left]`, the number of subsequences that can be formed with elements between `nums[left]` and the maximum element `nums[right]` (such that `nums[left] + nums[right] <= target`) is 2^(right - left). This is because each element in between can either be included or excluded in the subsequence.

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Number of Subsequences That Satisfy the Given Sum Condition - LeetCode 1498 Solution

Q: Which companies ask LeetCode 1498. Number of Subsequences That Satisfy the Given Sum Condition in interviews?

This problem is frequently asked at Turing.

Number of Subsequences That Satisfy the Given Sum Condition - Complete Solution Guide

Number of Subsequences That Satisfy the Given Sum Condition is LeetCode problem 1498, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of integers nums and an integer target . Return the number of non-empty subsequences of nums such that the sum of the minimum and maximum element on it is less or equal to target . Since the answer may be too large, return it modulo 10 9 + 7 . Example 1: Input: nums = [3,5,6,7], target = 9 Output: 4 Explanation: There are 4 subsequences that satisfy the condition. [3] -> Min value + max value <= target (3 + 3 <= 9) [3,5] -> (3 + 5 <= 9) [3,5,6] -> (3 + 6 <= 9) [3,6] -> (3

Detailed Explanation

The problem asks us to find the number of non-empty subsequences of a given array `nums` whose minimum and maximum element sum up to a value less than or equal to a given `target`. The result needs to be calculated modulo 10^9 + 7. A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, given `nums = [3, 5, 6, 7]` and `target = 9`, we want to find subsequences like `[3]`, `[3, 5]`, `[3, 6]` and `[3, 5, 6]` because the sum of their minimum and maximum values is at most 9.

Solution Approach

The solution first sorts the input array `nums`. Then, it uses a two-pointer approach with `left` and `right` pointers initialized to the beginning and the end of the sorted array, respectively. For each element at the `left` pointer (potential minimum value), it moves the `right` pointer to the largest element such that `nums[left] + nums[right] <= target`. The number of subsequences that can be formed using the elements between `left` and `right` is 2^(right - left). These counts are accumulated (modulo 10^9 + 7) to produce the final result. Powers of 2 are pre-calculated to optimize calculations within the main loop.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize `left` to 0 and `right` to `n - 1`, where `n` is the length of `nums`.
Step 3: Precalculate powers of 2 modulo 10^9 + 7 and store them in an array `powers` (powers[i] = 2^i % mod).
Step 4: Initialize `ans` to 0, which will store the count of valid subsequences.
Step 5: While `left <= right`:
Step 6: If `nums[left] + nums[right] <= target`, it means all elements between `left` and `right` can be part of a valid subsequence with `nums[left]` as the minimum. Add `powers[right - left]` to `ans` modulo 10^9 + 7. Increment `left`.
Step 7: Else, `nums[left] + nums[right] > target`, which means `nums[right]` is too large to form a valid subsequence with `nums[left]` as the minimum. Decrement `right`.
Step 8: Return `ans`.

Key Insights

Insight 1: Sorting the input array `nums` allows us to efficiently find the minimum and maximum elements by considering elements from the beginning and the end of the sorted array.
Insight 2: For a given minimum element `nums[left]`, the number of subsequences that can be formed with elements between `nums[left]` and the maximum element `nums[right]` (such that `nums[left] + nums[right] <= target`) is 2^(right - left). This is because each element in between can either be included or excluded in the subsequence.
Insight 3: Precalculating powers of 2 modulo 10^9 + 7 avoids repeated calculations and ensures the result doesn't overflow.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Turing.

Number of Subsequences That Satisfy the Given Sum Condition | LeetCode 1498

Medium

Array Two Pointers Binary Search Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int numSubseq(int[] nums, int target) {
        int n = nums.length;
        Arrays.sort(nums);
        int mod = (int)1e9 + 7;
        
        int[] powers = new int[n];
        powers[0] = 1;
        for (int i = 1; i < n; i++) {
            powers[i] = (powers[i-1] * 2) % mod;
        }
        
        int ans = 0;
        int left = 0, right = n - 1;
        
        while (left <= right) {
            if (nums[left] + nums[right] <= target) {
                ans = (ans + powers[right - left]) % mod;
                left += 1;
            } else {
                right -= 1;
            }
        }
        
        return ans;
    }
}

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Problem Info

Description

Problem 1498: Number of Subsequences That Satisfy the Given Sum Condition - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

Turing

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1498

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Sort the input array `nums` in ascending order.
2
Step 2: Initialize `left` to 0 and `right` to `n - 1`, where `n` is the length of `nums`.
3
Step 3: Precalculate powers of 2 modulo 10^9 + 7 and store them in an array `powers` (powers[i] = 2^i % mod).
4
Step 4: Initialize `ans` to 0, which will store the count of valid subsequences.
5
Step 5: While `left <= right`:
6
Step 6: If `nums[left] + nums[right] <= target`, it means all elements between `left` and `right` can be part of a valid subsequence with `nums[left]` as the minimum. Add `powers[right - left]` to `ans` modulo 10^9 + 7. Increment `left`.
7
Step 7: Else, `nums[left] + nums[right] > target`, which means `nums[right]` is too large to form a valid subsequence with `nums[left]` as the minimum. Decrement `right`.
8
Step 8: Return `ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting the array takes O(n log n) time. The two-pointer approach takes O(n) time. The precalculation of powers of 2 takes O(n) time. Therefore, the overall time complexity is dominated by the sorting step, O(n log n). The space complexity is O(n) due to the `powers` array used to store powers of 2.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Generate all possible subsequences of the array and check if the sum of the minimum and maximum element in each subsequence is less than or equal to the target. This would involve generating 2^n subsequences.

Time: O(n * 2^n)Space: O(n)

Trade-offs: This approach is extremely inefficient and will lead to a time limit exceeded (TLE) error for large input arrays. It's only useful for understanding the problem but not for solving it efficiently.

✨

Key Insights

1Insight 1: Sorting the input array `nums` allows us to efficiently find the minimum and maximum elements by considering elements from the beginning and the end of the sorted array.
2Insight 2: For a given minimum element `nums[left]`, the number of subsequences that can be formed with elements between `nums[left]` and the maximum element `nums[right]` (such that `nums[left] + nums[right] <= target`) is 2^(right - left). This is because each element in between can either be included or excluded in the subsequence.
3Insight 3: Precalculating powers of 2 modulo 10^9 + 7 avoids repeated calculations and ensures the result doesn't overflow.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: An empty input array is not possible based on the constraints (1 <= nums.length <= 10^5).
Edge case 2: When a single element satisfies the target condition, the subsequence formed will be the element itself. The logic correctly handles this situation because `right - left` will be 0 and `powers[0]` will be 1, contributing 1 to the answer, which is the valid subsequence with just the element.
Edge case 3: When all elements can be combined to produce a valid subsequence. The two pointer will eventually compare each pair in the sorted array until the correct count is calculated.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the input array. Without sorting, it's difficult to efficiently find the minimum and maximum elements within the required time complexity.

✗Common mistake 2: Not taking the modulo at each step. Failing to take the modulo during the calculation of powers of 2 and the accumulation of the answer can lead to integer overflow, resulting in incorrect results.

✗Common mistake 3: Incorrectly calculating the number of subsequences. It's important to understand why 2^(right-left) represents the number of possible subsequences with `nums[left]` as the minimum, up to and including `nums[right]`.

✗Common mistake 4: Assuming the subsequences need to be contiguous. Subsequences do not need to be contiguous, this is a common misunderstanding.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimization problems involving resource allocation. Imagine having a list of tasks with varying costs and benefits. You want to find the number of task combinations that satisfy a budget constraint (similar to the target).

🏦Real-world scenario 2: Data analysis where you want to identify patterns or combinations of data points that meet certain criteria. For example, finding the number of customer segments whose spending habits fall within a specific range.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Subsequences That Satisfy the Given Sum Condition"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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