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2659. Make Array Empty

Description

You are given an integer array nums containing distinct numbers, and you can perform the following operations until the array is empty:

  • If the first element has the smallest value, remove it
  • Otherwise, put the first element at the end of the array.

Return an integer denoting the number of operations it takes to make nums empty.

 

Example 1:

Input: nums = [3,4,-1]
Output: 5
Operation Array
1 [4, -1, 3]
2 [-1, 3, 4]
3 [3, 4]
4 [4]
5 []

Example 2:

Input: nums = [1,2,4,3]
Output: 5
Operation Array
1 [2, 4, 3]
2 [4, 3]
3 [3, 4]
4 [4]
5 []

Example 3:

Input: nums = [1,2,3]
Output: 3
Operation Array
1 [2, 3]
2 [3]
3 []

 

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • All values in nums are distinct.

Solutions

  • 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 long countOperationsToEmptyArray(int[] nums) {
        int n = nums.length;
        Map<Integer, Integer> pos = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            pos.put(nums[i], i);
        }
        Arrays.sort(nums);
        long ans = pos.get(nums[0]) + 1;
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int k = 0; k < n - 1; ++k) {
            int i = pos.get(nums[k]), j = pos.get(nums[k + 1]);
            long d = j - i - (tree.query(j + 1) - tree.query(i + 1));
            ans += d + (n - k) * (i > j ? 1 : 0);
            tree.update(i + 1, 1);
        }
        return ans;
    }
}

  • 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:
    long long countOperationsToEmptyArray(vector<int>& nums) {
        unordered_map<int, int> pos;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            pos[nums[i]] = i;
        }
        sort(nums.begin(), nums.end());
        BinaryIndexedTree tree(n);
        long long ans = pos[nums[0]] + 1;
        for (int k = 0; k < n - 1; ++k) {
            int i = pos[nums[k]], j = pos[nums[k + 1]];
            long long d = j - i - (tree.query(j + 1) - tree.query(i + 1));
            ans += d + (n - k) * int(i > j);
            tree.update(i + 1, 1);
        }
        return ans;
    }
};

  • from sortedcontainers import SortedList


class Solution:
    def countOperationsToEmptyArray(self, nums: List[int]) -> int:
        pos = {x: i for i, x in enumerate(nums)}
        nums.sort()
        sl = SortedList()
        ans = pos[nums[0]] + 1
        n = len(nums)
        for k, (a, b) in enumerate(pairwise(nums)):
            i, j = pos[a], pos[b]
            d = j - i - sl.bisect(j) + sl.bisect(i)
            ans += d + (n - k) * int(i > j)
            sl.add(i)
        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 countOperationsToEmptyArray(nums []int) int64 {
	n := len(nums)
	pos := map[int]int{}
	for i, x := range nums {
		pos[x] = i
	}
	sort.Ints(nums)
	tree := newBinaryIndexedTree(n)
	ans := pos[nums[0]] + 1
	for k := 0; k < n-1; k++ {
		i, j := pos[nums[k]], pos[nums[k+1]]
		d := j - i - (tree.query(j+1) - tree.query(i+1))
		if i > j {
			d += n - k
		}
		ans += d
		tree.update(i+1, 1)
	}
	return int64(ans)
}

  • class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    public update(x: number, v: number): void {
        while (x <= this.n) {
            this.c[x] += v;
            x += x & -x;
        }
    }

    public query(x: number): number {
        let s = 0;
        while (x > 0) {
            s += this.c[x];
            x -= x & -x;
        }
        return s;
    }
}

function countOperationsToEmptyArray(nums: number[]): number {
    const pos: Map<number, number> = new Map();
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        pos.set(nums[i], i);
    }
    nums.sort((a, b) => a - b);
    const tree = new BinaryIndexedTree(n);
    let ans = pos.get(nums[0])! + 1;
    for (let k = 0; k < n - 1; ++k) {
        const i = pos.get(nums[k])!;
        const j = pos.get(nums[k + 1])!;
        let d = j - i - (tree.query(j + 1) - tree.query(i + 1));
        if (i > j) {
            d += n - k;
        }
        ans += d;
        tree.update(i + 1, 1);
    }
    return ans;
}

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