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775. Global and Local Inversions

DifficultyMedium

Description

You are given an integer array nums of length n which represents a permutation of all the integers in the range [0, n - 1].

The number of global inversions is the number of the different pairs (i, j) where:

  • 0 <= i < j < n
  • nums[i] > nums[j]

The number of local inversions is the number of indices i where:

  • 0 <= i < n - 1
  • nums[i] > nums[i + 1]

Return true if the number of global inversions is equal to the number of local inversions.

 

Example 1:

Input: nums = [1,0,2]
Output: true
Explanation: There is 1 global inversion and 1 local inversion.

Example 2:

Input: nums = [1,2,0]
Output: false
Explanation: There are 2 global inversions and 1 local inversion.

 

Constraints:

  • n == nums.length
  • 1 <= n <= 105
  • 0 <= nums[i] < n
  • All the integers of nums are unique.
  • nums is a permutation of all the numbers in the range [0, n - 1].

Solutions

Solution 1

Thinking

Decide whether global inversions equal adjacent ones. \(n\le 10^5\). Equivalently, there is no inversion of distance greater than \(1\).

No \(i\le j-2\) with \(\textit{nums}[i]>\textit{nums}[j]\). Track the prefix max of \(\textit{nums}[0..i-2]\) and fail if it exceeds \(\textit{nums}[i]\).

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class Solution:
    def isIdealPermutation(self, nums: list[int]) -> bool:
        mx = 0
        for i in range(2, len(nums)):
            if (mx := max(mx, nums[i - 2])) > nums[i]:
                return False
        return True

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class Solution {
    public boolean isIdealPermutation(int[] nums) {
        int mx = 0;
        for (int i = 2; i < nums.length; ++i) {
            mx = Math.max(mx, nums[i - 2]);
            if (mx > nums[i]) {
                return false;
            }
        }
        return true;
    }
}

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class Solution {
public:
    bool isIdealPermutation(vector<int>& nums) {
        int mx = 0;
        for (int i = 2; i < nums.size(); ++i) {
            mx = max(mx, nums[i - 2]);
            if (mx > nums[i]) return false;
        }
        return true;
    }
};

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func isIdealPermutation(nums []int) bool {
    mx := 0
    for i := 2; i < len(nums); i++ {
        mx = max(mx, nums[i-2])
        if mx > nums[i] {
            return false
        }
    }
    return true
}

Solution 2

Thinking

Solution 1 uses the closed form. Counting instead: adjacent pairs for local inversions, a Fenwick tree for global ones.

Maintain their difference and fail when the global count pulls ahead. An extra \(\log n\).

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

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

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


class Solution:
    def isIdealPermutation(self, nums: list[int]) -> bool:
        n = len(nums)
        tree = BinaryIndexedTree(n)
        cnt = 0
        for i, v in enumerate(nums):
            cnt += i < n - 1 and v > nums[i + 1]
            cnt -= i - tree.query(v)
            if cnt < 0:
                return False
            tree.update(v + 1, 1)
        return True

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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 += x & -x;
        }
    }

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

class Solution {
    public boolean isIdealPermutation(int[] nums) {
        int n = nums.length;
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        int cnt = 0;
        for (int i = 0; i < n && cnt >= 0; ++i) {
            cnt += (i < n - 1 && nums[i] > nums[i + 1] ? 1 : 0);
            cnt -= (i - tree.query(nums[i]));
            tree.update(nums[i] + 1, 1);
        }
        return cnt == 0;
    }
}

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

    void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            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:
    bool isIdealPermutation(vector<int>& nums) {
        int n = nums.size();
        BinaryIndexedTree tree(n);
        long cnt = 0;
        for (int i = 0; i < n && ~cnt; ++i) {
            cnt += (i < n - 1 && nums[i] > nums[i + 1]);
            cnt -= (i - tree.query(nums[i]));
            tree.update(nums[i] + 1, 1);
        }
        return cnt == 0;
    }
};

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func isIdealPermutation(nums []int) bool {
    n := len(nums)
    tree := newBinaryIndexedTree(n)
    cnt := 0
    for i, v := range nums {
        if i < n-1 && v > nums[i+1] {
            cnt++
        }
        cnt -= (i - tree.query(v))
        if cnt < 0 {
            break
        }
        tree.update(v+1, 1)
    }
    return cnt == 0
}

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
}

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