3072. Distribute Elements Into Two Arrays II

Description

You are given a 1-indexed array of integers nums of length n.

We define a function greaterCount such that greaterCount(arr, val) returns the number of elements in arr that are strictly greater than val.

You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the i^th operation:

If greaterCount(arr1, nums[i]) > greaterCount(arr2, nums[i]), append nums[i] to arr1.
If greaterCount(arr1, nums[i]) < greaterCount(arr2, nums[i]), append nums[i] to arr2.
If greaterCount(arr1, nums[i]) == greaterCount(arr2, nums[i]), append nums[i] to the array with a lesser number of elements.
If there is still a tie, append nums[i] to arr1.

The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6].

Return the integer array result.

Example 1:

Input: nums = [2,1,3,3]
Output: [2,3,1,3]
Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1].
In the 3^rd operation, the number of elements greater than 3 is zero in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1.
In the 4^th operation, the number of elements greater than 3 is zero in both arrays. As the length of arr2 is lesser, hence, append nums[4] to arr2.
After 4 operations, arr1 = [2,3] and arr2 = [1,3].
Hence, the array result formed by concatenation is [2,3,1,3].

Example 2:

Input: nums = [5,14,3,1,2]
Output: [5,3,1,2,14]
Explanation: After the first 2 operations, arr1 = [5] and arr2 = [14].
In the 3^rd operation, the number of elements greater than 3 is one in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1.
In the 4^th operation, the number of elements greater than 1 is greater in arr1 than arr2 (2 > 1). Hence, append nums[4] to arr1.
In the 5^th operation, the number of elements greater than 2 is greater in arr1 than arr2 (2 > 1). Hence, append nums[5] to arr1.
After 5 operations, arr1 = [5,3,1,2] and arr2 = [14].
Hence, the array result formed by concatenation is [5,3,1,2,14].

Example 3:

Input: nums = [3,3,3,3]
Output: [3,3,3,3]
Explanation: At the end of 4 operations, arr1 = [3,3] and arr2 = [3,3].
Hence, the array result formed by concatenation is [3,3,3,3].

Constraints:

3 <= n <= 10⁵
1 <= nums[i] <= 10⁹

Solutions

Solution 1: Discretization + Binary Indexed Tree

We can use two binary indexed trees tree1 and tree2 to maintain the number of elements in arr1 and arr2 that are less than or equal to a certain number. Each time, we query the number of elements that are less than or equal to the current number in the binary indexed tree, then the number of elements that are greater than the current number is the length of the current array minus the query result. Then we can decide which array to add the current number to based on this difference.

Since the range of numbers given in the problem is very large, we need to discretize these numbers. We can sort these numbers and remove duplicates, then use binary search to find the position of each number in the sorted array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$, where $n$ is the length of the array nums.

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

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

    public void update(int x, int delta) {
        for (; x <= n; x += x & -x) {
            c[x] += delta;
        }
    }

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

class Solution {
    public int[] resultArray(int[] nums) {
        int[] st = nums.clone();
        Arrays.sort(st);
        int n = st.length;
        BinaryIndexedTree tree1 = new BinaryIndexedTree(n + 1);
        BinaryIndexedTree tree2 = new BinaryIndexedTree(n + 1);
        tree1.update(Arrays.binarySearch(st, nums[0]) + 1, 1);
        tree2.update(Arrays.binarySearch(st, nums[1]) + 1, 1);
        int[] arr1 = new int[n];
        int[] arr2 = new int[n];
        arr1[0] = nums[0];
        arr2[0] = nums[1];
        int i = 1, j = 1;
        for (int k = 2; k < n; ++k) {
            int x = Arrays.binarySearch(st, nums[k]) + 1;
            int a = i - tree1.query(x);
            int b = j - tree2.query(x);
            if (a > b) {
                arr1[i++] = nums[k];
                tree1.update(x, 1);
            } else if (a < b) {
                arr2[j++] = nums[k];
                tree2.update(x, 1);
            } else if (i <= j) {
                arr1[i++] = nums[k];
                tree1.update(x, 1);
            } else {
                arr2[j++] = nums[k];
                tree2.update(x, 1);
            }
        }
        for (int k = 0; k < j; ++k) {
            arr1[i++] = arr2[k];
        }
        return arr1;
    }
}

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

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

    void update(int x, int delta) {
        for (; x <= n; x += x & -x) {
            c[x] += delta;
        }
    }

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

class Solution {
public:
    vector<int> resultArray(vector<int>& nums) {
        vector<int> st = nums;
        sort(st.begin(), st.end());
        int n = st.size();
        BinaryIndexedTree tree1(n + 1);
        BinaryIndexedTree tree2(n + 1);
        tree1.update(distance(st.begin(), lower_bound(st.begin(), st.end(), nums[0])) + 1, 1);
        tree2.update(distance(st.begin(), lower_bound(st.begin(), st.end(), nums[1])) + 1, 1);
        vector<int> arr1 = {nums[0]};
        vector<int> arr2 = {nums[1]};
        for (int k = 2; k < n; ++k) {
            int x = distance(st.begin(), lower_bound(st.begin(), st.end(), nums[k])) + 1;
            int a = arr1.size() - tree1.query(x);
            int b = arr2.size() - tree2.query(x);
            if (a > b) {
                arr1.push_back(nums[k]);
                tree1.update(x, 1);
            } else if (a < b) {
                arr2.push_back(nums[k]);
                tree2.update(x, 1);
            } else if (arr1.size() <= arr2.size()) {
                arr1.push_back(nums[k]);
                tree1.update(x, 1);
            } else {
                arr2.push_back(nums[k]);
                tree2.update(x, 1);
            }
        }
        arr1.insert(arr1.end(), arr2.begin(), arr2.end());
        return arr1;
    }
};

class BinaryIndexedTree:
    __slots__ = "n", "c"

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

    def update(self, x: int, delta: int) -> None:
        while x <= self.n:
            self.c[x] += delta
            x += x & -x

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


class Solution:
    def resultArray(self, nums: List[int]) -> List[int]:
        st = sorted(set(nums))
        m = len(st)
        tree1 = BinaryIndexedTree(m + 1)
        tree2 = BinaryIndexedTree(m + 1)
        tree1.update(bisect_left(st, nums[0]) + 1, 1)
        tree2.update(bisect_left(st, nums[1]) + 1, 1)
        arr1 = [nums[0]]
        arr2 = [nums[1]]
        for x in nums[2:]:
            i = bisect_left(st, x) + 1
            a = len(arr1) - tree1.query(i)
            b = len(arr2) - tree2.query(i)
            if a > b:
                arr1.append(x)
                tree1.update(i, 1)
            elif a < b:
                arr2.append(x)
                tree2.update(i, 1)
            elif len(arr1) <= len(arr2):
                arr1.append(x)
                tree1.update(i, 1)
            else:
                arr2.append(x)
                tree2.update(i, 1)
        return arr1 + arr2

type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) *BinaryIndexedTree {
	return &BinaryIndexedTree{n: n, c: make([]int, n+1)}
}

func (bit *BinaryIndexedTree) update(x, delta int) {
	for ; x <= bit.n; x += x & -x {
		bit.c[x] += delta
	}
}

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

func resultArray(nums []int) []int {
	st := make([]int, len(nums))
	copy(st, nums)
	sort.Ints(st)
	n := len(st)
	tree1 := NewBinaryIndexedTree(n + 1)
	tree2 := NewBinaryIndexedTree(n + 1)
	tree1.update(sort.SearchInts(st, nums[0])+1, 1)
	tree2.update(sort.SearchInts(st, nums[1])+1, 1)
	arr1 := []int{nums[0]}
	arr2 := []int{nums[1]}
	for _, x := range nums[2:] {
		i := sort.SearchInts(st, x) + 1
		a := len(arr1) - tree1.query(i)
		b := len(arr2) - tree2.query(i)
		if a > b {
			arr1 = append(arr1, x)
			tree1.update(i, 1)
		} else if a < b {
			arr2 = append(arr2, x)
			tree2.update(i, 1)
		} else if len(arr1) <= len(arr2) {
			arr1 = append(arr1, x)
			tree1.update(i, 1)
		} else {
			arr2 = append(arr2, x)
			tree2.update(i, 1)
		}
	}
	arr1 = append(arr1, arr2...)
	return arr1
}

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

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

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

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

function resultArray(nums: number[]): number[] {
    const st: number[] = nums.slice().sort((a, b) => a - b);
    const n: number = st.length;
    const search = (x: number): number => {
        let [l, r] = [0, n];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (st[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const tree1: BinaryIndexedTree = new BinaryIndexedTree(n + 1);
    const tree2: BinaryIndexedTree = new BinaryIndexedTree(n + 1);
    tree1.update(search(nums[0]) + 1, 1);
    tree2.update(search(nums[1]) + 1, 1);
    const arr1: number[] = [nums[0]];
    const arr2: number[] = [nums[1]];
    for (const x of nums.slice(2)) {
        const i: number = search(x) + 1;
        const a: number = arr1.length - tree1.query(i);
        const b: number = arr2.length - tree2.query(i);
        if (a > b) {
            arr1.push(x);
            tree1.update(i, 1);
        } else if (a < b) {
            arr2.push(x);
            tree2.update(i, 1);
        } else if (arr1.length <= arr2.length) {
            arr1.push(x);
            tree1.update(i, 1);
        } else {
            arr2.push(x);
            tree2.update(i, 1);
        }
    }
    return arr1.concat(arr2);
}

3072 - Distribute Elements Into Two Arrays II

3072. Distribute Elements Into Two Arrays II

Description

Solutions

Solution 1: Discretization + Binary Indexed Tree

All Problems

All Solutions

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3072. Distribute Elements Into Two Arrays II

Description

Solutions

Solution 1: Discretization + Binary Indexed Tree

All Problems

All Solutions