Question

775. Global and Local Inversions

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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

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

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;
  }
}

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;
  }
};

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

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

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;
  }
}

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;
  }
};

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
}