3187. Peaks in Array

Description

A peak in an array arr is an element that is greater than its previous and next element in arr.

You are given an integer array nums and a 2D integer array queries.

You have to process queries of two types:

queries[i] = [1, l_i, r_i], determine the count of peak elements in the subarray nums[l_i..r_i].
queries[i] = [2, index_i, val_i], change nums[index_i] to val_i.

Return an array answer containing the results of the queries of the first type in order.

Notes:

The first and the last element of an array or a subarray cannot be a peak.

Example 1:

Input: nums = [3,1,4,2,5], queries = [[2,3,4],[1,0,4]]

Output: [0]

Explanation:

First query: We change nums[3] to 4 and nums becomes [3,1,4,4,5].

Second query: The number of peaks in the [3,1,4,4,5] is 0.

Example 2:

Input: nums = [4,1,4,2,1,5], queries = [[2,2,4],[1,0,2],[1,0,4]]

Output: [0,1]

Explanation:

First query: nums[2] should become 4, but it is already set to 4.

Second query: The number of peaks in the [4,1,4] is 0.

Third query: The second 4 is a peak in the [4,1,4,2,1].

Constraints:

3 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
1 <= queries.length <= 10⁵
queries[i][0] == 1 or queries[i][0] == 2
For all i that:
- queries[i][0] == 1: 0 <= queries[i][1] <= queries[i][2] <= nums.length - 1
- queries[i][0] == 2: 0 <= queries[i][1] <= nums.length - 1, 1 <= queries[i][2] <= 10⁵

Solutions

Solution 1: Binary Indexed Tree

According to the problem description, for $0 < i < n - 1$, if it satisfies $nums[i - 1] < nums[i]$ and $nums[i] > nums[i + 1]$, we can consider $nums[i]$ as $1$, otherwise as $0$. Thus, for operation $1$, i.e., querying the number of peak elements in the subarray $nums[l..r]$, it is equivalent to querying the number of $1$s in the interval $[l + 1, r - 1]$. We can use a binary indexed tree to maintain the number of $1$s in the interval $[1, n - 1]$.

For operation $1$, i.e., updating $nums[idx]$ to $val$, it will only affect the values at positions $idx - 1$, $idx$, and $idx + 1$, so we only need to update these three positions. Specifically, we can first remove the peak elements at these three positions, then update the value of $nums[idx]$, and finally add back the peak elements at these three positions.

The time complexity is $O((n + q) \times \log n)$, and the space complexity is $O(n)$. Here, $n$ and $q$ are the lengths of the array nums and the query array queries, respectively.

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 {
    private BinaryIndexedTree tree;
    private int[] nums;

    public List<Integer> countOfPeaks(int[] nums, int[][] queries) {
        int n = nums.length;
        this.nums = nums;
        tree = new BinaryIndexedTree(n - 1);
        for (int i = 1; i < n - 1; ++i) {
            update(i, 1);
        }
        List<Integer> ans = new ArrayList<>();
        for (var q : queries) {
            if (q[0] == 1) {
                int l = q[1] + 1, r = q[2] - 1;
                ans.add(l > r ? 0 : tree.query(r) - tree.query(l - 1));
            } else {
                int idx = q[1], val = q[2];
                for (int i = idx - 1; i <= idx + 1; ++i) {
                    update(i, -1);
                }
                nums[idx] = val;
                for (int i = idx - 1; i <= idx + 1; ++i) {
                    update(i, 1);
                }
            }
        }
        return ans;
    }

    private void update(int i, int val) {
        if (i <= 0 || i >= nums.length - 1) {
            return;
        }
        if (nums[i - 1] < nums[i] && nums[i] > nums[i + 1]) {
            tree.update(i, val);
        }
    }
}

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> countOfPeaks(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        BinaryIndexedTree tree(n - 1);
        auto update = [&](int i, int val) {
            if (i <= 0 || i >= n - 1) {
                return;
            }
            if (nums[i - 1] < nums[i] && nums[i] > nums[i + 1]) {
                tree.update(i, val);
            }
        };
        for (int i = 1; i < n - 1; ++i) {
            update(i, 1);
        }
        vector<int> ans;
        for (auto& q : queries) {
            if (q[0] == 1) {
                int l = q[1] + 1, r = q[2] - 1;
                ans.push_back(l > r ? 0 : tree.query(r) - tree.query(l - 1));
            } else {
                int idx = q[1], val = q[2];
                for (int i = idx - 1; i <= idx + 1; ++i) {
                    update(i, -1);
                }
                nums[idx] = val;
                for (int i = idx - 1; i <= idx + 1; ++i) {
                    update(i, 1);
                }
            }
        }
        return ans;
    }
};

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 countOfPeaks(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        def update(i: int, val: int):
            if i <= 0 or i >= n - 1:
                return
            if nums[i - 1] < nums[i] and nums[i] > nums[i + 1]:
                tree.update(i, val)

        n = len(nums)
        tree = BinaryIndexedTree(n - 1)
        for i in range(1, n - 1):
            update(i, 1)
        ans = []
        for q in queries:
            if q[0] == 1:
                l, r = q[1] + 1, q[2] - 1
                ans.append(0 if l > r else tree.query(r) - tree.query(l - 1))
            else:
                idx, val = q[1:]
                for i in range(idx - 1, idx + 2):
                    update(i, -1)
                nums[idx] = val
                for i in range(idx - 1, idx + 2):
                    update(i, 1)
        return ans

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 countOfPeaks(nums []int, queries [][]int) (ans []int) {
	n := len(nums)
	tree := NewBinaryIndexedTree(n - 1)
	update := func(i, val int) {
		if i <= 0 || i >= n-1 {
			return
		}
		if nums[i-1] < nums[i] && nums[i] > nums[i+1] {
			tree.update(i, val)
		}
	}
	for i := 1; i < n-1; i++ {
		update(i, 1)
	}
	for _, q := range queries {
		if q[0] == 1 {
			l, r := q[1]+1, q[2]-1
			t := 0
			if l <= r {
				t = tree.query(r) - tree.query(l-1)
			}
			ans = append(ans, t)
		} else {
			idx, val := q[1], q[2]
			for i := idx - 1; i <= idx+1; i++ {
				update(i, -1)
			}
			nums[idx] = val
			for i := idx - 1; i <= idx+1; i++ {
				update(i, 1)
			}
		}
	}
	return
}

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 countOfPeaks(nums: number[], queries: number[][]): number[] {
    const n = nums.length;
    const tree = new BinaryIndexedTree(n - 1);
    const update = (i: number, val: number): void => {
        if (i <= 0 || i >= n - 1) {
            return;
        }
        if (nums[i - 1] < nums[i] && nums[i] > nums[i + 1]) {
            tree.update(i, val);
        }
    };
    for (let i = 1; i < n - 1; ++i) {
        update(i, 1);
    }
    const ans: number[] = [];
    for (const q of queries) {
        if (q[0] === 1) {
            const [l, r] = [q[1] + 1, q[2] - 1];
            ans.push(l > r ? 0 : tree.query(r) - tree.query(l - 1));
        } else {
            const [idx, val] = [q[1], q[2]];
            for (let i = idx - 1; i <= idx + 1; ++i) {
                update(i, -1);
            }
            nums[idx] = val;
            for (let i = idx - 1; i <= idx + 1; ++i) {
                update(i, 1);
            }
        }
    }
    return ans;
}

3187 - Peaks in Array

3187. Peaks in Array

Description

Solutions

Solution 1: Binary Indexed Tree

All Problems

All Solutions

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3187. Peaks in Array

Description

Solutions

Solution 1: Binary Indexed Tree

All Problems

All Solutions