315. 计算右侧小于当前元素的个数

题目描述

给你一个整数数组 nums ，按要求返回一个新数组 counts 。数组 counts 有该性质： counts[i] 的值是 nums[i] 右侧小于 nums[i] 的元素的数量。

示例 1：

输入：nums = [5,2,6,1]
输出：[2,1,1,0] 
解释：
5 的右侧有 2 个更小的元素 (2 和 1)
2 的右侧仅有 1 个更小的元素 (1)
6 的右侧有 1 个更小的元素 (1)
1 的右侧有 0 个更小的元素

示例 2：

输入：nums = [-1]
输出：[0]

示例 3：

输入：nums = [-1,-1]
输出：[0,0]

提示：

1 <= nums.length <= 10⁵
-10⁴ <= nums[i] <= 10⁴

解法

方法一：树状数组

树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。它可以高效地实现如下两个操作：

单点更新 update(x, delta)：把序列 x 位置的数加上一个值 delta；
前缀和查询 query(x)：查询序列 [1,...x] 区间的区间和，即位置 x 的前缀和。

这两个操作的时间复杂度均为 \(O(\log n)\)。

树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。

比如给定数组 a[5] = {2, 5, 3, 4, 1}，求 b[i] = 位置 i 左边小于等于 a[i] 的数的个数。对于此例，b[5] = {0, 1, 1, 2, 0}。

解决方案是直接遍历数组，每个位置先求出 query(a[i])，然后再修改树状数组 update(a[i], 1) 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。

Python3JavaC++Go

class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

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


class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        alls = sorted(set(nums))
        m = {v: i for i, v in enumerate(alls, 1)}
        tree = BinaryIndexedTree(len(m))
        ans = []
        for v in nums[::-1]:
            x = m[v]
            tree.update(x, 1)
            ans.append(tree.query(x - 1))
        return ans[::-1]

class Solution {
    public List<Integer> countSmaller(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int v : nums) {
            s.add(v);
        }
        List<Integer> alls = new ArrayList<>(s);
        alls.sort(Comparator.comparingInt(a -> a));
        int n = alls.size();
        Map<Integer, Integer> m = new HashMap<>(n);
        for (int i = 0; i < n; ++i) {
            m.put(alls.get(i), i + 1);
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        LinkedList<Integer> ans = new LinkedList<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            int x = m.get(nums[i]);
            tree.update(x, 1);
            ans.addFirst(tree.query(x - 1));
        }
        return ans;
    }
}

class BinaryIndexedTree {
    private int n;
    private int[] c;

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

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

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

    public static int lowbit(int x) {
        return x & -x;
    }
}

class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

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

    void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

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

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        vector<int> alls(s.begin(), s.end());
        sort(alls.begin(), alls.end());
        unordered_map<int, int> m;
        int n = alls.size();
        for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
        BinaryIndexedTree* tree = new BinaryIndexedTree(n);
        vector<int> ans(nums.size());
        for (int i = nums.size() - 1; i >= 0; --i) {
            int x = m[nums[i]];
            tree->update(x, 1);
            ans[i] = tree->query(x - 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) lowbit(x int) int {
    return x & -x
}

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

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

func countSmaller(nums []int) []int {
    s := make(map[int]bool)
    for _, v := range nums {
        s[v] = true
    }
    var alls []int
    for v := range s {
        alls = append(alls, v)
    }
    sort.Ints(alls)
    m := make(map[int]int)
    for i, v := range alls {
        m[v] = i + 1
    }
    ans := make([]int, len(nums))
    tree := newBinaryIndexedTree(len(alls))
    for i := len(nums) - 1; i >= 0; i-- {
        x := m[nums[i]]
        tree.update(x, 1)
        ans[i] = tree.query(x - 1)
    }
    return ans
}

方法二：线段树

线段树将整个区间分割为多个不连续的子区间，子区间的数量不超过 log(width)。更新某个元素的值，只需要更新 log(width) 个区间，并且这些区间都包含在一个包含该元素的大区间内。

线段树的每个节点代表一个区间；
线段树具有唯一的根节点，代表的区间是整个统计范围，如 [1, N]；
线段树的每个叶子节点代表一个长度为 1 的元区间 [x, x]；
对于每个内部节点 [l, r]，它的左儿子是 [l, mid]，右儿子是 [mid + 1, r], 其中 mid = ⌊(l + r) / 2⌋ (即向下取整)。

Python3JavaC++Go

class Node:
    def __init__(self):
        self.l = 0
        self.r = 0
        self.v = 0


class SegmentTree:
    def __init__(self, n):
        self.tr = [Node() for _ in range(n << 2)]
        self.build(1, 1, n)

    def build(self, u, l, r):
        self.tr[u].l = l
        self.tr[u].r = r
        if l == r:
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)

    def modify(self, u, x, v):
        if self.tr[u].l == x and self.tr[u].r == x:
            self.tr[u].v += v
            return
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        if x <= mid:
            self.modify(u << 1, x, v)
        else:
            self.modify(u << 1 | 1, x, v)
        self.pushup(u)

    def query(self, u, l, r):
        if self.tr[u].l >= l and self.tr[u].r <= r:
            return self.tr[u].v
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        v = 0
        if l <= mid:
            v += self.query(u << 1, l, r)
        if r > mid:
            v += self.query(u << 1 | 1, l, r)
        return v

    def pushup(self, u):
        self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v


class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        s = sorted(set(nums))
        m = {v: i for i, v in enumerate(s, 1)}
        tree = SegmentTree(len(s))
        ans = []
        for v in nums[::-1]:
            x = m[v]
            ans.append(tree.query(1, 1, x - 1))
            tree.modify(1, x, 1)
        return ans[::-1]

class Solution {
    public List<Integer> countSmaller(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int v : nums) {
            s.add(v);
        }
        List<Integer> alls = new ArrayList<>(s);
        alls.sort(Comparator.comparingInt(a -> a));
        int n = alls.size();
        Map<Integer, Integer> m = new HashMap<>(n);
        for (int i = 0; i < n; ++i) {
            m.put(alls.get(i), i + 1);
        }
        SegmentTree tree = new SegmentTree(n);
        LinkedList<Integer> ans = new LinkedList<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            int x = m.get(nums[i]);
            tree.modify(1, x, 1);
            ans.addFirst(tree.query(1, 1, x - 1));
        }
        return ans;
    }
}

class Node {
    int l;
    int r;
    int v;
}

class SegmentTree {
    private Node[] tr;

    public SegmentTree(int n) {
        tr = new Node[4 * n];
        for (int i = 0; i < tr.length; ++i) {
            tr[i] = new Node();
        }
        build(1, 1, n);
    }

    public void build(int u, int l, int r) {
        tr[u].l = l;
        tr[u].r = r;
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }

    public void modify(int u, int x, int v) {
        if (tr[u].l == x && tr[u].r == x) {
            tr[u].v += v;
            return;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (x <= mid) {
            modify(u << 1, x, v);
        } else {
            modify(u << 1 | 1, x, v);
        }
        pushup(u);
    }

    public void pushup(int u) {
        tr[u].v = tr[u << 1].v + tr[u << 1 | 1].v;
    }

    public int query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return tr[u].v;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        int v = 0;
        if (l <= mid) {
            v += query(u << 1, l, r);
        }
        if (r > mid) {
            v += query(u << 1 | 1, l, r);
        }
        return v;
    }
}

class Node {
public:
    int l;
    int r;
    int v;
};

class SegmentTree {
public:
    vector<Node*> tr;

    SegmentTree(int n) {
        tr.resize(4 * n);
        for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
        build(1, 1, n);
    }

    void build(int u, int l, int r) {
        tr[u]->l = l;
        tr[u]->r = r;
        if (l == r) return;
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }

    void modify(int u, int x, int v) {
        if (tr[u]->l == x && tr[u]->r == x) {
            tr[u]->v += v;
            return;
        }
        int mid = (tr[u]->l + tr[u]->r) >> 1;
        if (x <= mid)
            modify(u << 1, x, v);
        else
            modify(u << 1 | 1, x, v);
        pushup(u);
    }

    void pushup(int u) {
        tr[u]->v = tr[u << 1]->v + tr[u << 1 | 1]->v;
    }

    int query(int u, int l, int r) {
        if (tr[u]->l >= l && tr[u]->r <= r) return tr[u]->v;
        int mid = (tr[u]->l + tr[u]->r) >> 1;
        int v = 0;
        if (l <= mid) v += query(u << 1, l, r);
        if (r > mid) v += query(u << 1 | 1, l, r);
        return v;
    }
};

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        vector<int> alls(s.begin(), s.end());
        sort(alls.begin(), alls.end());
        unordered_map<int, int> m;
        int n = alls.size();
        for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
        SegmentTree* tree = new SegmentTree(n);
        vector<int> ans(nums.size());
        for (int i = nums.size() - 1; i >= 0; --i) {
            int x = m[nums[i]];
            tree->modify(1, x, 1);
            ans[i] = tree->query(1, 1, x - 1);
        }
        return ans;
    }
};

type Pair struct {
    val   int
    index int
}

var (
    tmp   []Pair
    count []int
)

func countSmaller(nums []int) []int {
    tmp, count = make([]Pair, len(nums)), make([]int, len(nums))
    array := make([]Pair, len(nums))
    for i, v := range nums {
        array[i] = Pair{val: v, index: i}
    }
    sorted(array, 0, len(array)-1)
    return count
}

func sorted(arr []Pair, low, high int) {
    if low >= high {
        return
    }
    mid := low + (high-low)/2
    sorted(arr, low, mid)
    sorted(arr, mid+1, high)
    merge(arr, low, mid, high)
}

func merge(arr []Pair, low, mid, high int) {
    left, right := low, mid+1
    idx := low
    for left <= mid && right <= high {
        if arr[left].val <= arr[right].val {
            count[arr[left].index] += right - mid - 1
            tmp[idx], left = arr[left], left+1
        } else {
            tmp[idx], right = arr[right], right+1
        }
        idx++
    }
    for left <= mid {
        count[arr[left].index] += right - mid - 1
        tmp[idx] = arr[left]
        idx, left = idx+1, left+1
    }
    for right <= high {
        tmp[idx] = arr[right]
        idx, right = idx+1, right+1
    }
    // 排序
    for i := low; i <= high; i++ {
        arr[i] = tmp[i]
    }
}

方法三：归并排序

在归并排序的合并阶段，当左侧元素 \(\textit{left}[i] \leq\) 右侧元素 \(\textit{right}[j]\) 时，

说明右侧恰好有 \(j\) 个元素比 \(\textit{left}[i]\) 小，因此将 \(j\) 累加到左侧元素的计数中。

当右侧元素全部清空，其体现右侧所有元素均小于剩余左侧元素，累加右侧数组长度到每个剩余左侧元素的计数即可。

注意：C++做超大数组的归并排序时，有内存爆炸风险，需使用 缓存数组 避免Memory Limit Exceeded。

复杂度：

时间复杂度：\(O(n \log n)\)，归并排序的标准时间复杂度。
空间复杂度：\(O(n)\)，递归栈的标准空间复杂度。

Python3C++

class Solution:
    def countSmaller(self, nums: list[int]) -> list[int]:
        self.right_smaller_counts = [0] * len(nums)

        nums_indices = [(num, idx) for idx, num in enumerate(nums)]
        self.merge_sort(nums_indices)

        return self.right_smaller_counts

    def combine_arrays(
        self,
        left_nums_indices: list[tuple[int, int]],
        right_nums_indices: list[tuple[int, int]],
    ) -> list[tuple[int, int]]:
        merged_nums_indices: list[tuple[int, int]] = []
        left_idx, right_idx = 0, 0

        while left_idx < len(left_nums_indices) and right_idx < len(right_nums_indices):
            if left_nums_indices[left_idx][0] <= right_nums_indices[right_idx][0]:
                # Iterated left side element finalizes its right smaller count.
                left_num_idx = left_nums_indices[left_idx][1]
                self.right_smaller_counts[left_num_idx] += right_idx

                merged_nums_indices.append(left_nums_indices[left_idx])
                left_idx += 1
                continue

            merged_nums_indices.append(right_nums_indices[right_idx])
            right_idx += 1

        while left_idx < len(left_nums_indices):
            # Iterated left side element finalizes its right smaller count.
            left_num_idx = left_nums_indices[left_idx][1]
            self.right_smaller_counts[left_num_idx] += len(right_nums_indices)

            merged_nums_indices.append(left_nums_indices[left_idx])
            left_idx += 1

        while right_idx < len(right_nums_indices):
            merged_nums_indices.append(right_nums_indices[right_idx])
            right_idx += 1

        return merged_nums_indices

    def merge_sort(self, nums_indices: list[tuple[int, int]]) -> list[tuple[int, int]]:
        if len(nums_indices) == 1:
            return nums_indices  # Single element.

        split_idx = len(nums_indices) // 2

        left_nums_indices = self.merge_sort(nums_indices[:split_idx])
        right_nums_indices = self.merge_sort(nums_indices[split_idx:])

        return self.combine_arrays(left_nums_indices, right_nums_indices)

class Solution {
private:
    vector<int> rightSmallerCounts;
    vector<pair<int, int>> buffer;

    void combineArrays(
        vector<pair<int, int>>& numsIndices, int leftBound, int splitIdx, int rightBound) {
        // Left side array = numsIndices[leftBound: splitIdx].
        // Right side array = numsIndices[splitIdx: rightBound + 1].
        int leftIdx = leftBound, rightIdx = splitIdx;
        int bufferIdx = leftBound;

        while (leftIdx < splitIdx && rightIdx <= rightBound) {
            if (numsIndices[leftIdx].first <= numsIndices[rightIdx].first) {
                // Iterated left side element finalizes its right smaller count.
                int leftNumIdx = numsIndices[leftIdx].second;
                rightSmallerCounts[leftNumIdx] += rightIdx - splitIdx;

                buffer[bufferIdx++] = numsIndices[leftIdx++];
            }

            else
                buffer[bufferIdx++] = numsIndices[rightIdx++];
        }

        while (leftIdx < splitIdx) {
            // Iterated left side element finalizes its right smaller count.
            int leftNumIdx = numsIndices[leftIdx].second;
            rightSmallerCounts[leftNumIdx] += rightIdx - splitIdx;

            buffer[bufferIdx++] = numsIndices[leftIdx++];
        }

        while (rightIdx <= rightBound)
            buffer[bufferIdx++] = numsIndices[rightIdx++];

        for (int idx = leftBound; idx <= rightBound; idx++)
            numsIndices[idx] = buffer[idx]; // Put buffer data back to original array.
    }

    void mergeSort(vector<pair<int, int>>& numsIndices, int leftBound, int rightBound) {
        if (leftBound == rightBound) return; // Single element.

        // Plus 1: ensure splitIdx > leftBound.
        int splitIdx = (leftBound + rightBound + 1) / 2;

        mergeSort(numsIndices, leftBound, splitIdx - 1);
        mergeSort(numsIndices, splitIdx, rightBound);

        combineArrays(numsIndices, leftBound, splitIdx, rightBound);
    }

public:
    vector<int> countSmaller(vector<int>& nums) {
        buffer.resize(nums.size()); // Against memory explosions.

        vector<pair<int, int>> numsIndices(nums.size());
        for (int idx = 0; idx < nums.size(); idx++)
            numsIndices[idx] = {nums[idx], idx};

        rightSmallerCounts.assign(nums.size(), 0);
        mergeSort(numsIndices, 0, nums.size() - 1);
        return rightSmallerCounts;
    }
};