What is the time complexity of Count of Range Sum?

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

What data structures are used to solve Count of Range Sum?

This problem uses Array, Binary Search, Divide and Conquer, Binary Indexed Tree, Segment Tree, Merge Sort, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count of Range Sum?

Insight 1: Directly calculating all possible range sums (O(n^2) ranges, O(n) to compute each sum) leads to an O(n^3) solution which is too slow for the given constraints (n <= 10^5). Insight 2: Using prefix sums allows for calculating the sum of any range S(i, j) in O(1) time: S(i, j) = prefix_sum[j+1] - prefix_sum[i].

Which companies ask LeetCode 327. Count of Range Sum in interviews?

This problem is frequently asked at Morgan Stanley.

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Count of Range Sum - LeetCode 327 Solution

Q: How to solve Count of Range Sum?

The provided solutions employ a Divide and Conquer strategy using a modified Merge Sort algorithm to count the range sums. First, prefix sums are calculated to enable O(1) range sum computation. The core of the solution lies within the merge sort function. During the merge step, we iterate through the right half of the array and, for each element, use two pointers in the left half to find the number of elements within the desired range (val_j - upper, val_j - lower). This process efficiently counts the number of range sums without explicitly calculating them all. Finally, the two halves are merged in sorted order to ensure the correctness of the merge sort algorithm.

Q: Which companies ask LeetCode 327. Count of Range Sum in interviews?

This problem is frequently asked at Morgan Stanley.

Count of Range Sum - Complete Solution Guide

Count of Range Sum is LeetCode problem 327, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer array nums and two integers lower and upper , return the number of range sums that lie in [lower, upper] inclusive . Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j inclusive, where i <= j . Example 1: Input: nums = [-2,5,-1], lower = -2, upper = 2 Output: 3 Explanation: The three ranges are: [0,0], [2,2], and [0,2] and their respective sums are: -2, -1, 2. Example 2: Input: nums = [0], lower = 0, upper = 0 Output: 1 Constraints: 1 <= nums

Detailed Explanation

The problem asks us to find the number of subarray sums (range sums) within a given array `nums` that fall between a specified `lower` and `upper` bound (inclusive). A range sum S(i, j) is the sum of elements from index `i` to `j` in `nums`, where `i <= j`. The challenge lies in efficiently counting these valid range sums given the size constraints of the input array.

Solution Approach

The provided solutions employ a Divide and Conquer strategy using a modified Merge Sort algorithm to count the range sums. First, prefix sums are calculated to enable O(1) range sum computation. The core of the solution lies within the merge sort function. During the merge step, we iterate through the right half of the array and, for each element, use two pointers in the left half to find the number of elements within the desired range (val_j - upper, val_j - lower). This process efficiently counts the number of range sums without explicitly calculating them all. Finally, the two halves are merged in sorted order to ensure the correctness of the merge sort algorithm.

Step-by-Step Algorithm

Step 1: Calculate the prefix sums of the input array `nums`. `prefix_sums[i]` stores the sum of elements from `nums[0]` to `nums[i-1]`.
Step 2: Implement a recursive merge sort function `merge_sort_and_count(l, r)` that sorts the `prefix_sums` array within the range `[l, r)` and counts the number of range sums that fall within the `[lower, upper]` bounds.
Step 3: Within the `merge_sort_and_count` function, divide the array into two halves: `[l, mid)` and `[mid, r)`.
Step 4: Recursively call `merge_sort_and_count` on both halves to sort them and count the range sums within each half.
Step 5: During the merge process, iterate through the right half of the array (`[mid, r)`). For each element `prefix_sums[j]` in the right half:
Step 6: Calculate `target_lower = prefix_sums[j] - upper` and `target_upper = prefix_sums[j] - lower`. These represent the range the prefix sum in the left half has to fall within.
Step 7: Use two pointers, `k1` and `k2`, to iterate through the left half of the array (`[l, mid)`) to find the number of elements `prefix_sums[k]` such that `target_lower <= prefix_sums[k] <= target_upper`. `k1` points to the first element >= `target_lower`, and `k2` points to the first element > `target_upper`. The count of valid range sums is `k2 - k1`. This avoids recalculating the range each time by only incrementing the pointers until the next valid value.
Step 8: Merge the two sorted halves (`[l, mid)` and `[mid, r)`) into a single sorted array using a temporary array `temp`.
Step 9: Copy the sorted elements from the `temp` array back into the `prefix_sums` array within the range `[l, r)`.
Step 10: Return the total count of range sums found.
Step 11: Call `merge_sort_and_count(0, n + 1)` to sort the entire `prefix_sums` array and get the final count of range sums.

Key Insights

Insight 1: Directly calculating all possible range sums (O(n^2) ranges, O(n) to compute each sum) leads to an O(n^3) solution which is too slow for the given constraints (n <= 10^5).
Insight 2: Using prefix sums allows for calculating the sum of any range S(i, j) in O(1) time: S(i, j) = prefix_sum[j+1] - prefix_sum[i].
Insight 3: The problem can be efficiently solved using a modified Merge Sort algorithm. During the merge process, we can efficiently count the number of valid range sums by iterating through one half of the sorted prefix sums and using two pointers to identify the indices in the other half that satisfy the range condition.
Insight 4: Since integer overflow could occur, the use of long data type is important.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Divide and Conquer, Binary Indexed Tree, Segment Tree, Merge Sort, Ordered Set.

Companies

Asked at: Morgan Stanley.

Count of Range Sum | LeetCode 327

Hard

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

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

Solution Code

Available in:

class Solution {
    public int countRangeSum(int[] nums, int lower, int upper) {
        int n = nums.length;
        long[] prefixSums = new long[n + 1];
        for (int i = 0; i < n; i++) {
            prefixSums[i + 1] = prefixSums[i] + nums[i];
        }

        return mergeSortAndCount(prefixSums, 0, n + 1, lower, upper);
    }

    private int mergeSortAndCount(long[] prefixSums, int l, int r, int lower, int upper) {
        if (r - l <= 1) {
            return 0;
        }

        int mid = l + (r - l) / 2;

        int count = mergeSortAndCount(prefixSums, l, mid, lower, upper) + mergeSortAndCount(prefixSums, mid, r, lower, upper);

        int k1 = l, k2 = l;
        for (int j = mid; j < r; j++) {
            long val_j = prefixSums[j];

            long target_lower = val_j - upper;
            long target_upper = val_j - lower;

            while (k1 < mid && prefixSums[k1] < target_lower) {
                k1++;
            }

            while (k2 < mid && prefixSums[k2] <= target_upper) {
                k2++;
            }

            count += (k2 - k1);
        }

        // Merge the two sorted halves
        long[] temp = new long[r - l];
        int p1 = l, p2 = mid, idx = 0;
        while (p1 < mid && p2 < r) {
            if (prefixSums[p1] <= prefixSums[p2]) {
                temp[idx++] = prefixSums[p1++];
            } else {
                temp[idx++] = prefixSums[p2++];
            }
        }

        while (p1 < mid) {
            temp[idx++] = prefixSums[p1++];
        }
        while (p2 < r) {
            temp[idx++] = prefixSums[p2++];
        }

        for(int i = 0; i < temp.length; i++){
            prefixSums[l + i] = temp[i];
        }


        return count;
    }
}

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Problem Info

Description

Problem 327: Count of Range Sum - Detailed explanation and solution approaches.

Topics

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

Companies

Morgan Stanley

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:327

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

11 steps

1
Step 1: Calculate the prefix sums of the input array `nums`. `prefix_sums[i]` stores the sum of elements from `nums[0]` to `nums[i-1]`.
2
Step 2: Implement a recursive merge sort function `merge_sort_and_count(l, r)` that sorts the `prefix_sums` array within the range `[l, r)` and counts the number of range sums that fall within the `[lower, upper]` bounds.
3
Step 3: Within the `merge_sort_and_count` function, divide the array into two halves: `[l, mid)` and `[mid, r)`.
4
Step 4: Recursively call `merge_sort_and_count` on both halves to sort them and count the range sums within each half.
5
Step 5: During the merge process, iterate through the right half of the array (`[mid, r)`). For each element `prefix_sums[j]` in the right half:
6
Step 6: Calculate `target_lower = prefix_sums[j] - upper` and `target_upper = prefix_sums[j] - lower`. These represent the range the prefix sum in the left half has to fall within.
7
Step 7: Use two pointers, `k1` and `k2`, to iterate through the left half of the array (`[l, mid)`) to find the number of elements `prefix_sums[k]` such that `target_lower <= prefix_sums[k] <= target_upper`. `k1` points to the first element >= `target_lower`, and `k2` points to the first element > `target_upper`. The count of valid range sums is `k2 - k1`. This avoids recalculating the range each time by only incrementing the pointers until the next valid value.
8
Step 8: Merge the two sorted halves (`[l, mid)` and `[mid, r)`) into a single sorted array using a temporary array `temp`.
9
Step 9: Copy the sorted elements from the `temp` array back into the `prefix_sums` array within the range `[l, r)`.
10
Step 10: Return the total count of range sums found.
11
Step 11: Call `merge_sort_and_count(0, n + 1)` to sort the entire `prefix_sums` array and get the final count of range sums.

💡

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?

The time complexity is dominated by the merge sort algorithm, which has a time complexity of O(n log n). The counting of range sums during the merge step takes O(n) time, where n is the size of the input array. The space complexity is O(n) because we use an auxiliary array of size n to store the prefix sums and a temporary array of size n during the merge process.

📖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

Binary Indexed Tree (BIT) / Fenwick Tree

This approach involves iterating through the prefix sums and for each sum, calculating the range [sum - upper, sum - lower] and querying the BIT to count how many previous prefix sums fall in that range.

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

Trade-offs: BIT requires coordinate compression and might be harder to implement and understand compared to the merge sort approach. However, it can be slightly faster in practice for certain input distributions.

Segment Tree

Similar to BIT, this approach maintains a segment tree of prefix sums. For each prefix sum, we query the segment tree to find how many previous prefix sums fall within the range [sum - upper, sum - lower].

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

Trade-offs: Segment trees offer more flexibility than BITs but are generally more complex to implement and have slightly higher overhead. Not usually preferred for this specific problem.

✨

Key Insights

1Insight 1: Directly calculating all possible range sums (O(n^2) ranges, O(n) to compute each sum) leads to an O(n^3) solution which is too slow for the given constraints (n <= 10^5).
2Insight 2: Using prefix sums allows for calculating the sum of any range S(i, j) in O(1) time: S(i, j) = prefix_sum[j+1] - prefix_sum[i].
3Insight 3: The problem can be efficiently solved using a modified Merge Sort algorithm. During the merge process, we can efficiently count the number of valid range sums by iterating through one half of the sorted prefix sums and using two pointers to identify the indices in the other half that satisfy the range condition.
4Insight 4: Since integer overflow could occur, the use of long data type is important.

📚

Key Terms Explained

For Students

Array

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

Binary Search

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

Edge Cases

Edge case 1: Empty input array (nums.length == 0). The code correctly handles this case because the loop that calculates the prefix sums won't execute, and the merge sort will return 0, which is the correct answer.
Edge case 2: All elements in `nums` are 0. The algorithm handles this case correctly because the prefix sums will also be 0. The counting process will accurately count the number of ranges that fall between `lower` and `upper`.
Edge case 3: `lower` and `upper` are both 0. The algorithm counts the number of ranges that sum to zero. It handles this correctly as well.
Edge case 4: Integer overflow. Since the problem constraints involve integers that can be close to the limits of a 32-bit integer, using `long` for `prefix_sums` is crucial to avoid potential overflows during the calculation of range sums. This avoids incorrect range sum calculations due to wrapping around the integer limits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using integers instead of long for prefix sums. This can lead to integer overflows when the sum of a range exceeds the maximum value of an integer, resulting in incorrect counts.

✗Common mistake 2: Directly calculating all possible range sums without using prefix sums or an optimized approach like merge sort. This leads to a time complexity of O(n^3), which will result in a Time Limit Exceeded (TLE) error.

✗Common mistake 3: Incorrectly implementing the merge step in the merge sort algorithm. Specifically, failing to merge the sorted halves correctly can lead to inaccurate range sum counts because the prefix sums are no longer sorted.

✗Common mistake 4: Not handling the edge cases where the input array is empty or when lower and upper are both zero.

✅

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: Financial analysis. Analyzing stock price trends and identifying periods where cumulative returns fall within certain target ranges.

🏦Real-world scenario 2: Signal processing. Detecting segments of a signal whose integrated values (sums) fall within pre-defined thresholds. This might be used for anomaly detection or pattern recognition in sensor data.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: October 26, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count of Range Sum"?

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

🤖 Need help with this problem?