Question

327. 区间和的个数

English Version

题目描述

给你一个整数数组 nums 以及两个整数 lower 和 upper 。求数组中，值位于范围 [lower, upper] （包含 lower 和 upper）之内的 区间和的个数 。

区间和 S(i, j) 表示在 nums 中，位置从 i 到 j 的元素之和，包含 i 和 j (i ≤ j)。

 

示例 1：

输入：nums = [-2,5,-1], lower = -2, upper = 2
输出：3
解释：存在三个区间：[0,0]、[2,2] 和 [0,2] ，对应的区间和分别是：-2 、-1 、2 。

示例 2：

输入：nums = [0], lower = 0, upper = 0
输出：1

 

提示：

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• -105 <= lower <= upper <= 105
• 题目数据保证答案是一个 32 位 的整数

解法

方法一：前缀和 + 树状数组

题目要求区间和，因此我们可以先求出前缀和数组 $s$，其中 $s[i]$ 表示 $nums$ 中前 $i$ 个元素的和。那么对于任意的 $i \lt j$，$s[j+1] - s[i]$ 就是 $nums$ 中下标在 $[i, j]$ 的元素之和。

而 $lower \leq s[j+1] - s[i] \leq upper$，可以转换为 $s[j+1] - upper \leq s[i] \leq s[j+1] - lower$，也就是说，对于当前前缀和 $s[j+1]$，我们需要统计 $s$ 中有多少个下标 $i$ 满足 $s[j+1] - upper \leq s[i] \leq s[j+1] - lower$。

我们可以用树状数组来维护每个前缀和出现的次数，这样对于每个前缀和 $s[j+1]$，我们只需要查询树状数组中有多少个前缀和 $s[i]$ 满足 $s[j+1] - upper \leq s[i] \leq s[j+1] - lower$ 即可。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组长度。

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

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;
  }
}

class BinaryIndexedTree {
public:
  BinaryIndexedTree(int _n)
    : n(_n)
    , c(_n + 1) {}

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

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

private:
  int n;
  vector<int> c;
};

class Solution {
public:
  int countRangeSum(vector<int>& nums, int lower, int upper) {
    using ll = long long;
    int n = nums.size();
    ll s[n + 1];
    s[0] = 0;
    for (int i = 0; i < n; ++i) {
      s[i + 1] = s[i] + nums[i];
    }
    ll arr[(n + 1) * 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;
    }
    sort(arr, arr + (n + 1) * 3);
    int m = unique(arr, arr + (n + 1) * 3) - arr;
    BinaryIndexedTree tree(m);
    int ans = 0;
    for (int i = 0; i <= n; ++i) {
      int l = lower_bound(arr, arr + m, s[i] - upper) - arr + 1;
      int r = lower_bound(arr, arr + m, s[i] - lower) - arr + 1;
      ans += tree.query(r) - tree.query(l - 1);
      tree.update(lower_bound(arr, arr + m, s[i]) - arr + 1, 1);
    }
    return ans;
  }
};

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
}

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;
}