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493. Reverse Pairs

Description

Given an integer array nums, return the number of reverse pairs in the array.

A reverse pair is a pair (i, j) where:

  • 0 <= i < j < nums.length and
  • nums[i] > 2 * nums[j].

 

Example 1:

Input: nums = [1,3,2,3,1]
Output: 2
Explanation: The reverse pairs are:
(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1
(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1

Example 2:

Input: nums = [2,4,3,5,1]
Output: 3
Explanation: The reverse pairs are:
(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1
(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1
(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • -231 <= nums[i] <= 231 - 1

Solutions

Solution 1

Thinking

A reverse pair is \(i<j\) and \(nums[i]>2\,nums[j]\). A double loop is \(O(n^2)\). During mergesort both halves are sorted, so cross pairs can be counted before merging.

Recurse on both halves, then for each right value advance a left pointer until \(nums[i]>2\,nums[j]\); the leftover left side all form pairs. Merge by raw value afterwards.

Counting and merging stay separate: the count compares against \(2\,nums[j]\), the merge compares the original values.

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class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def merge_sort(l, r):
            if l >= r:
                return 0
            mid = (l + r) >> 1
            ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
            t = []
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= 2 * nums[j]:
                    i += 1
                else:
                    ans += mid - i + 1
                    j += 1
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= nums[j]:
                    t.append(nums[i])
                    i += 1
                else:
                    t.append(nums[j])
                    j += 1
            t.extend(nums[i : mid + 1])
            t.extend(nums[j : r + 1])
            nums[l : r + 1] = t
            return ans

        return merge_sort(0, len(nums) - 1)

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class Solution {
    private int[] nums;
    private int[] t;

    public int reversePairs(int[] nums) {
        this.nums = nums;
        int n = nums.length;
        this.t = new int[n];
        return mergeSort(0, n - 1);
    }

    private int mergeSort(int l, int r) {
        if (l >= r) {
            return 0;
        }
        int mid = (l + r) >> 1;
        int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
        int i = l, j = mid + 1, k = 0;
        while (i <= mid && j <= r) {
            if (nums[i] <= nums[j] * 2L) {
                ++i;
            } else {
                ans += mid - i + 1;
                ++j;
            }
        }
        i = l;
        j = mid + 1;
        while (i <= mid && j <= r) {
            if (nums[i] <= nums[j]) {
                t[k++] = nums[i++];
            } else {
                t[k++] = nums[j++];
            }
        }
        while (i <= mid) {
            t[k++] = nums[i++];
        }
        while (j <= r) {
            t[k++] = nums[j++];
        }
        for (i = l; i <= r; ++i) {
            nums[i] = t[i - l];
        }
        return ans;
    }
}

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class Solution {
public:
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        int t[n];
        function<int(int, int)> mergeSort = [&](int l, int r) -> int {
            if (l >= r) {
                return 0;
            }
            int mid = (l + r) >> 1;
            int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
            int i = l, j = mid + 1, k = 0;
            while (i <= mid && j <= r) {
                if (nums[i] <= nums[j] * 2LL) {
                    ++i;
                } else {
                    ans += mid - i + 1;
                    ++j;
                }
            }
            i = l;
            j = mid + 1;
            while (i <= mid && j <= r) {
                if (nums[i] <= nums[j]) {
                    t[k++] = nums[i++];
                } else {
                    t[k++] = nums[j++];
                }
            }
            while (i <= mid) {
                t[k++] = nums[i++];
            }
            while (j <= r) {
                t[k++] = nums[j++];
            }
            for (i = l; i <= r; ++i) {
                nums[i] = t[i - l];
            }
            return ans;
        };
        return mergeSort(0, n - 1);
    }
};

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func reversePairs(nums []int) int {
    n := len(nums)
    t := make([]int, n)
    var mergeSort func(l, r int) int
    mergeSort = func(l, r int) int {
        if l >= r {
            return 0
        }
        mid := (l + r) >> 1
        ans := mergeSort(l, mid) + mergeSort(mid+1, r)
        i, j, k := l, mid+1, 0
        for i <= mid && j <= r {
            if nums[i] <= nums[j]*2 {
                i++
            } else {
                ans += mid - i + 1
                j++
            }
        }
        i, j = l, mid+1
        for i <= mid && j <= r {
            if nums[i] <= nums[j] {
                t[k] = nums[i]
                k, i = k+1, i+1
            } else {
                t[k] = nums[j]
                k, j = k+1, j+1
            }
        }
        for ; i <= mid; i, k = i+1, k+1 {
            t[k] = nums[i]
        }
        for ; j <= r; j, k = j+1, k+1 {
            t[k] = nums[j]
        }
        for i = l; i <= r; i++ {
            nums[i] = t[i-l]
        }
        return ans
    }
    return mergeSort(0, n-1)
}

Solution 2

Thinking

Solution 1 rewrites the array. After discretizing \(nums\) and \(2\,nums\), walk right to left, insert \(2\,nums[j]\) into a Fenwick tree, and query how many inserted values are less than \(nums[i]\).

The relative order of the original array is not needed beyond the right-to-left scan; the tree stores the \(2\,nums\) already seen.

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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 reversePairs(self, nums: List[int]) -> int:
        s = set()
        for num in nums:
            s.add(num)
            s.add(num * 2)
        alls = sorted(s)
        m = {v: i for i, v in enumerate(alls, 1)}
        ans = 0
        tree = BinaryIndexedTree(len(m))
        for num in nums[::-1]:
            ans += tree.query(m[num] - 1)
            tree.update(m[num * 2], 1)
        return ans

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class Solution {
    public int reversePairs(int[] nums) {
        TreeSet<Long> ts = new TreeSet<>();
        for (int num : nums) {
            ts.add((long) num);
            ts.add((long) num * 2);
        }
        Map<Long, Integer> m = new HashMap<>();
        int idx = 0;
        for (long num : ts) {
            m.put(num, ++idx);
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m.size());
        int ans = 0;
        for (int i = nums.length - 1; i >= 0; --i) {
            int x = m.get((long) nums[i]);
            ans += tree.query(x - 1);
            tree.update(m.get((long) nums[i] * 2), 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;
    }
}

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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:
    int reversePairs(vector<int>& nums) {
        set<long long> s;
        for (int num : nums) {
            s.insert(num);
            s.insert(num * 2ll);
        }
        unordered_map<long long, int> m;
        int idx = 0;
        for (long long num : s) m[num] = ++idx;
        BinaryIndexedTree* tree = new BinaryIndexedTree(m.size());
        int ans = 0;
        for (int i = nums.size() - 1; i >= 0; --i) {
            ans += tree->query(m[nums[i]] - 1);
            tree->update(m[nums[i] * 2ll], 1);
        }
        return ans;
    }
};

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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 reversePairs(nums []int) int {
    s := make(map[int]bool)
    for _, num := range nums {
        s[num] = true
        s[num*2] = true
    }
    var alls []int
    for num := range s {
        alls = append(alls, num)
    }
    sort.Ints(alls)
    m := make(map[int]int)
    for i, num := range alls {
        m[num] = i + 1
    }
    tree := newBinaryIndexedTree(len(m))
    ans := 0
    for i := len(nums) - 1; i >= 0; i-- {
        ans += tree.query(m[nums[i]] - 1)
        tree.update(m[nums[i]*2], 1)
    }
    return ans
}

Solution 3

Thinking

Replace the Fenwick tree with a segment tree: same discretization, insert \(2\,nums\) from the right, query the count below \(nums[i]\). Same algebra, with explicit interval nodes.

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

    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


class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        s = set()
        for num in nums:
            s.add(num)
            s.add(num * 2)
        alls = sorted(s)
        m = {v: i for i, v in enumerate(alls, 1)}
        tree = SegmentTree(len(m))
        ans = 0
        for v in nums[::-1]:
            x = m[v]
            ans += tree.query(1, 1, x - 1)
            tree.modify(1, m[v * 2], 1)
        return ans

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class Solution {
    public int reversePairs(int[] nums) {
        TreeSet<Long> ts = new TreeSet<>();
        for (int num : nums) {
            ts.add((long) num);
            ts.add((long) num * 2);
        }
        Map<Long, Integer> m = new HashMap<>();
        int idx = 0;
        for (long num : ts) {
            m.put(num, ++idx);
        }
        SegmentTree tree = new SegmentTree(m.size());
        int ans = 0;
        for (int i = nums.length - 1; i >= 0; --i) {
            int x = m.get((long) nums[i]);
            ans += tree.query(1, 1, x - 1);
            tree.modify(1, m.get((long) nums[i] * 2), 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;
    }
}

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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:
    int reversePairs(vector<int>& nums) {
        set<long long> s;
        for (int num : nums) {
            s.insert(num);
            s.insert(num * 2ll);
        }
        unordered_map<long long, int> m;
        int idx = 0;
        for (long long num : s) m[num] = ++idx;
        SegmentTree* tree = new SegmentTree(m.size());
        int ans = 0;
        for (int i = nums.size() - 1; i >= 0; --i) {
            ans += tree->query(1, 1, m[nums[i]] - 1);
            tree->modify(1, m[nums[i] * 2ll], 1);
        }
        return ans;
    }
};

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