3907. Count Smaller Elements With Opposite Parity 🔒

Description

You are given an integer array nums of length n.

The score of an index i is defined as the number of indices j such that:

i < j < n
nums[j] < nums[i]
nums[i] and nums[j] have different parity (one is even and the other is odd).

Return an integer array answer of length n, where answer[i] is the score of index i.

Example 1:

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

Output: [2,1,2,0,0]

Explanation:

For i = 0, the elements nums[1] = 2 and nums[2] = 4 are smaller and have different parity.
For i = 1, the element nums[3] = 1 is smaller and has different parity.
For i = 2, the elements nums[3] = 1 and nums[4] = 3 are smaller and have different parity.
No valid elements exist for the remaining indices.

Thus, the answer = [2, 1, 2, 0, 0].

Example 2:

Input: nums = [4,4,1]

Output: [1,1,0]

Explanation:

For i = 0 and i = 1, the element nums[2] = 1 is smaller and has different parity. Thus, the answer = [1, 1, 0].

Example 3:

Input: nums = [7]

Output: [0]

Explanation:

No elements exist to the right of index 0, so its score is 0. Thus, the answer = [0].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Solutions

Solution 1: Ordered List or Binary Indexed Tree

We can use two ordered lists (or Binary Indexed Trees) to separately maintain even and odd elements. For each element, we query the number of smaller elements in the other list, and then add the current element to its corresponding list.

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

class BinaryIndexedTree {
    int n;
    int[] c;

    BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[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 > 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class Solution {
    public int[] countSmallerOppositeParity(int[] nums) {
        int n = nums.length;
        int[] sorted = nums.clone();
        Arrays.sort(sorted);
        int m = 0;
        for (int i = 0; i < n; i++) {
            if (i == 0 || sorted[i] != sorted[i - 1]) {
                sorted[m++] = sorted[i];
            }
        }
        BinaryIndexedTree[] bit = {new BinaryIndexedTree(m), new BinaryIndexedTree(m)};
        int[] ans = new int[n];
        for (int i = n - 1; i >= 0; --i) {
            int x = Arrays.binarySearch(sorted, 0, m, nums[i]) + 1;
            ans[i] = bit[nums[i] & 1 ^ 1].query(x - 1);
            bit[nums[i] & 1].update(x, 1);
        }
        return ans;
    }
}

struct BIT {
    int n;
    vector<int> c;
    BIT(int n)
        : n(n)
        , c(n + 1, 0) {}
    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> countSmallerOppositeParity(vector<int>& nums) {
        int n = nums.size();
        vector<int> sorted = nums;
        sort(sorted.begin(), sorted.end());
        sorted.erase(unique(sorted.begin(), sorted.end()), sorted.end());

        int m = sorted.size();
        BIT* bits[2] = {new BIT(m), new BIT(m)};
        vector<int> ans(n);

        for (int i = n - 1; i >= 0; --i) {
            int x = lower_bound(sorted.begin(), sorted.end(), nums[i]) - sorted.begin() + 1;
            ans[i] = bits[nums[i] & 1 ^ 1]->query(x - 1);
            bits[nums[i] & 1]->update(x, 1);
        }
        return ans;
    }
};

class Solution:
    def countSmallerOppositeParity(self, nums: list[int]) -> list[int]:
        n = len(nums)
        ans = [0] * n
        sl = [SortedList(), SortedList()]
        for i in range(n - 1, -1, -1):
            ans[i] = sl[nums[i] & 1 ^ 1].bisect_left(nums[i])
            sl[nums[i] & 1].add(nums[i])
        return ans

type BIT struct {
	n int
	c []int
}

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

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

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

func countSmallerOppositeParity(nums []int) []int {
	n := len(nums)
	sorted := make([]int, n)
	copy(sorted, nums)
	sort.Ints(sorted)

	m := 0
	if n > 0 {
		m = 1
		for i := 1; i < n; i++ {
			if sorted[i] != sorted[i-1] {
				sorted[m] = sorted[i]
				m++
			}
		}
		sorted = sorted[:m]
	}

	bits := []*BIT{newBIT(m), newBIT(m)}
	ans := make([]int, n)

	for i := n - 1; i >= 0; i-- {
		x := sort.SearchInts(sorted, nums[i]) + 1
		ans[i] = bits[nums[i]&1^1].query(x - 1)
		bits[nums[i]&1].update(x, 1)
	}
	return ans
}

class BIT {
    private c: Int32Array;
    constructor(private n: number) {
        this.c = new Int32Array(n + 1);
    }
    update(x: number, delta: number) {
        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 countSmallerOppositeParity(nums: number[]): number[] {
    const n = nums.length;
    const sorted = _.sortedUniq(_.sortBy(nums));
    const m = sorted.length;

    const bits = [new BIT(m), new BIT(m)];
    const ans = new Array(n);

    for (let i = n - 1; i >= 0; i--) {
        const rank = _.sortedIndex(sorted, nums[i]) + 1;
        ans[i] = bits[(nums[i] & 1) ^ 1].query(rank - 1);
        bits[nums[i] & 1].update(rank, 1);
    }
    return ans;
}

3907 - Count Smaller Elements With Opposite Parity

3907. Count Smaller Elements With Opposite Parity 🔒

Description

Solutions

Solution 1: Ordered List or Binary Indexed Tree

All Problems

All Solutions

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3907. Count Smaller Elements With Opposite Parity 🔒

Description

Solutions

Solution 1: Ordered List or Binary Indexed Tree

All Problems

All Solutions