LeetCode #327 — HARD

Count of Range Sum

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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.length <= 10⁵
-2³¹ <= nums[i] <= 2³¹ - 1
-10⁵ <= lower <= upper <= 10⁵
The answer is guaranteed to fit in a 32-bit integer.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[-2,5,-1]
-2
2

Example 2

[0]
0
0

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #327: Count of Range Sum
class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[n + 1];
    }

    public void update(int x, int v) {
        while (x <= n) {
            c[x] += v;
            x += x & -x;
        }
    }

    public int query(int x) {
        int s = 0;
        while (x != 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class Solution {
    public int countRangeSum(int[] nums, int lower, int upper) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        long[] arr = new long[n * 3 + 3];
        for (int i = 0, j = 0; i <= n; ++i, j += 3) {
            arr[j] = s[i];
            arr[j + 1] = s[i] - lower;
            arr[j + 2] = s[i] - upper;
        }
        Arrays.sort(arr);
        int m = 0;
        for (int i = 0; i < arr.length; ++i) {
            if (i == 0 || arr[i] != arr[i - 1]) {
                arr[m++] = arr[i];
            }
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m);
        int ans = 0;
        for (long x : s) {
            int l = search(arr, m, x - upper);
            int r = search(arr, m, x - lower);
            ans += tree.query(r) - tree.query(l - 1);
            tree.update(search(arr, m, x), 1);
        }
        return ans;
    }

    private int search(long[] nums, int r, long x) {
        int l = 0;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l + 1;
    }
}

// Accepted solution for LeetCode #327: Count of Range Sum
type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += x & -x
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= x & -x
	}
	return s
}

func countRangeSum(nums []int, lower int, upper int) (ans int) {
	n := len(nums)
	s := make([]int, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	arr := make([]int, (n+1)*3)
	for i, j := 0, 0; i <= n; i, j = i+1, j+3 {
		arr[j] = s[i]
		arr[j+1] = s[i] - lower
		arr[j+2] = s[i] - upper
	}
	sort.Ints(arr)
	m := 0
	for i := range arr {
		if i == 0 || arr[i] != arr[i-1] {
			arr[m] = arr[i]
			m++
		}
	}
	arr = arr[:m]
	tree := newBinaryIndexedTree(m)
	for _, x := range s {
		l := sort.SearchInts(arr, x-upper) + 1
		r := sort.SearchInts(arr, x-lower) + 1
		ans += tree.query(r) - tree.query(l-1)
		tree.update(sort.SearchInts(arr, x)+1, 1)
	}
	return
}

# Accepted solution for LeetCode #327: Count of Range Sum
class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x, v):
        while x <= self.n:
            self.c[x] += v
            x += x & -x

    def query(self, x):
        s = 0
        while x > 0:
            s += self.c[x]
            x -= x & -x
        return s


class Solution:
    def countRangeSum(self, nums: List[int], lower: int, upper: int) -> int:
        s = list(accumulate(nums, initial=0))
        arr = sorted(set(v for x in s for v in (x, x - lower, x - upper)))
        tree = BinaryIndexedTree(len(arr))
        ans = 0
        for x in s:
            l = bisect_left(arr, x - upper) + 1
            r = bisect_left(arr, x - lower) + 1
            ans += tree.query(r) - tree.query(l - 1)
            tree.update(bisect_left(arr, x) + 1, 1)
        return ans

// Accepted solution for LeetCode #327: Count of Range Sum
/**
 * [327] Count of Range Sum
 *
 * 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.length <= 10^5
 * 	-2^31 <= nums[i] <= 2^31 - 1
 * 	-10^5 <= lower <= upper <= 10^5
 * 	The answer is guaranteed to fit in a 32-bit integer.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-of-range-sum/
// discuss: https://leetcode.com/problems/count-of-range-sum/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/count-of-range-sum/discuss/1378661/Rust-Divide-and-Conquer-using-prefix-sum
    pub fn count_range_sum(nums: Vec<i32>, lower: i32, upper: i32) -> i32 {
        Self::count_sub_range_sum(
            std::rc::Rc::new(std::cell::RefCell::new(
                Some(0 as i64)
                    .iter()
                    .chain(nums.iter().map(|x| *x as i64).collect::<Vec<i64>>().iter())
                    .scan(0, |state, x| {
                        *state += x;
                        Some(*state as i64)
                    })
                    .collect::<Vec<i64>>(),
            )),
            0,
            nums.len(),
            lower as i64,
            upper as i64,
        )
    }

    fn count_sub_range_sum(
        prefix: std::rc::Rc<std::cell::RefCell<Vec<i64>>>,
        start: usize,
        end: usize,
        lower: i64,
        upper: i64,
    ) -> i32 {
        if start >= end {
            return 0;
        }

        // Pre-compute the two halves, thus assuming both sorted
        let mid: usize = (start + end) / 2;
        let mut count: i32 =
            Self::count_sub_range_sum(std::rc::Rc::clone(&prefix), start, mid, lower, upper)
                + Self::count_sub_range_sum(
                    std::rc::Rc::clone(&prefix),
                    mid + 1,
                    end,
                    lower,
                    upper,
                );

        // loop over every pos in left and use two pointers for lower & upper
        let (mut i, mut j) = (mid + 1, mid + 1);
        for pos in start..=mid {
            // compute the range of acceptable values
            while i <= end && (*prefix.borrow())[i] - (*prefix.borrow())[pos] < lower {
                i += 1;
            }
            while j <= end && (*prefix.borrow())[j] - (*prefix.borrow())[pos] <= upper {
                j += 1;
            }
            count += (j - i) as i32;
        }

        // partially sort the vector
        let mut replacement = (*prefix.borrow())[start..=end].to_vec();
        replacement.sort();
        (*prefix.borrow_mut())[start..=end].copy_from_slice(&replacement[..]);

        count
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0327_example_1() {
        let nums = vec![-2, 5, -1];
        let lower = -2;
        let upper = 2;

        let result = 3;

        assert_eq!(Solution::count_range_sum(nums, lower, upper), result);
    }

    #[test]
    fn test_0327_example_2() {
        let nums = vec![0];
        let lower = 0;
        let upper = 0;

        let result = 1;

        assert_eq!(Solution::count_range_sum(nums, lower, upper), result);
    }
}

// Accepted solution for LeetCode #327: Count of Range Sum
class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, v: number) {
        while (x <= this.n) {
            this.c[x] += v;
            x += x & -x;
        }
    }

    query(x: number): number {
        let s = 0;
        while (x > 0) {
            s += this.c[x];
            x -= x & -x;
        }
        return s;
    }
}

function countRangeSum(nums: number[], lower: number, upper: number): number {
    const n = nums.length;
    const s = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    let arr: number[] = Array((n + 1) * 3);
    for (let i = 0, j = 0; i <= n; ++i, j += 3) {
        arr[j] = s[i];
        arr[j + 1] = s[i] - lower;
        arr[j + 2] = s[i] - upper;
    }
    arr.sort((a, b) => a - b);
    let m = 0;
    for (let i = 0; i < arr.length; ++i) {
        if (i === 0 || arr[i] !== arr[i - 1]) {
            arr[m++] = arr[i];
        }
    }
    arr = arr.slice(0, m);
    const tree = new BinaryIndexedTree(m);
    let ans = 0;
    for (const x of s) {
        const l = search(arr, m, x - upper);
        const r = search(arr, m, x - lower);
        ans += tree.query(r) - tree.query(l - 1);
        tree.update(search(arr, m, x), 1);
    }
    return ans;
}

function search(nums: number[], r: number, x: number): number {
    let l = 0;
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] >= x) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l + 1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.