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Formatted question description: https://leetcode.ca/all/2569.html

2569. Handling Sum Queries After Update

Description

You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:

  1. For a query of type 1, queries[i] = [1, l, r]. Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r. Both l and r are 0-indexed.
  2. For a query of type 2, queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.
  3. For a query of type 3, queries[i] = [3, 0, 0]. Find the sum of the elements in nums2.

Return an array containing all the answers to the third type queries.

 

Example 1:

Input: nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]
Output: [3]
Explanation: After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.

Example 2:

Input: nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]
Output: [5]
Explanation: After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.

 

Constraints:

  • 1 <= nums1.length,nums2.length <= 105
  • nums1.length = nums2.length
  • 1 <= queries.length <= 105
  • queries[i].length = 3
  • 0 <= l <= r <= nums1.length - 1
  • 0 <= p <= 106
  • 0 <= nums1[i] <= 1
  • 0 <= nums2[i] <= 109

Solutions

Solution 1: Segment Tree

According to the problem description:

  • Operation $1$ is to reverse all numbers in the index range $[l,..r]$ of array nums1, that is, change $0$ to $1$ and $1$ to $0$.
  • Operation $3$ is to sum all numbers in array nums2.
  • Operation $2$ is to add the sum of all numbers in array nums2 with $p$ times the sum of all numbers in array nums1, that is, $sum(nums2) = sum(nums2) + p * sum(nums1)$.

Therefore, we actually only need to maintain the segment sum of array nums1, which can be implemented through a segment tree.

We define each node of the segment tree as Node, each node contains the following attributes:

  • l: The left endpoint of the node, the index starts from $1$.
  • r: The right endpoint of the node, the index starts from $1$.
  • s: The segment sum of the node.
  • lazy: The lazy tag of the node.

The segment tree mainly has the following operations:

  • build(u, l, r): Build the segment tree.
  • pushdown(u): Propagate the lazy tag.
  • pushup(u): Update the information of the parent node with the information of the child nodes.
  • modify(u, l, r): Modify the segment sum. In this problem, it is to reverse each number in the segment, so the segment sum $s = r - l + 1 - s$.
  • query(u, l, r): Query the segment sum.

First, calculate the sum of all numbers in array nums2, denoted as $s$.

When executing operation $1$, we only need to call modify(1, l + 1, r + 1).

When executing operation $2$, we update $s = s + p \times query(1, 1, n)$.

When executing operation $3$, we just need to add $s$ to the answer array.

The time complexity is $O(n + m \times \log n)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of arrays nums1 and queries respectively.

  • class Node {
        int l, r;
        int s, lazy;
    }
    
    class SegmentTree {
        private Node[] tr;
        private int[] nums;
    
        public SegmentTree(int[] nums) {
            int n = nums.length;
            this.nums = nums;
            tr = new Node[n << 2];
            for (int i = 0; i < tr.length; ++i) {
                tr[i] = new Node();
            }
            build(1, 1, n);
        }
    
        private void build(int u, int l, int r) {
            tr[u].l = l;
            tr[u].r = r;
            if (l == r) {
                tr[u].s = nums[l - 1];
                return;
            }
            int mid = (l + r) >> 1;
            build(u << 1, l, mid);
            build(u << 1 | 1, mid + 1, r);
            pushup(u);
        }
    
        public void modify(int u, int l, int r) {
            if (tr[u].l >= l && tr[u].r <= r) {
                tr[u].lazy ^= 1;
                tr[u].s = tr[u].r - tr[u].l + 1 - tr[u].s;
                return;
            }
            pushdown(u);
            int mid = (tr[u].l + tr[u].r) >> 1;
            if (l <= mid) {
                modify(u << 1, l, r);
            }
            if (r > mid) {
                modify(u << 1 | 1, l, r);
            }
            pushup(u);
        }
    
        public int query(int u, int l, int r) {
            if (tr[u].l >= l && tr[u].r <= r) {
                return tr[u].s;
            }
            pushdown(u);
            int mid = (tr[u].l + tr[u].r) >> 1;
            int res = 0;
            if (l <= mid) {
                res += query(u << 1, l, r);
            }
            if (r > mid) {
                res += query(u << 1 | 1, l, r);
            }
            return res;
        }
    
        private void pushup(int u) {
            tr[u].s = tr[u << 1].s + tr[u << 1 | 1].s;
        }
    
        private void pushdown(int u) {
            if (tr[u].lazy == 1) {
                int mid = (tr[u].l + tr[u].r) >> 1;
                tr[u << 1].s = mid - tr[u].l + 1 - tr[u << 1].s;
                tr[u << 1].lazy ^= 1;
                tr[u << 1 | 1].s = tr[u].r - mid - tr[u << 1 | 1].s;
                tr[u << 1 | 1].lazy ^= 1;
                tr[u].lazy ^= 1;
            }
        }
    }
    
    class Solution {
        public long[] handleQuery(int[] nums1, int[] nums2, int[][] queries) {
            SegmentTree tree = new SegmentTree(nums1);
            long s = 0;
            for (int x : nums2) {
                s += x;
            }
            int m = 0;
            for (var q : queries) {
                if (q[0] == 3) {
                    ++m;
                }
            }
            long[] ans = new long[m];
            int i = 0;
            for (var q : queries) {
                if (q[0] == 1) {
                    tree.modify(1, q[1] + 1, q[2] + 1);
                } else if (q[0] == 2) {
                    s += 1L * q[1] * tree.query(1, 1, nums2.length);
                } else {
                    ans[i++] = s;
                }
            }
            return ans;
        }
    }
    
  • class Node {
    public:
        int l = 0, r = 0;
        int s = 0, lazy = 0;
    };
    
    class SegmentTree {
    public:
        SegmentTree(vector<int>& nums) {
            this->nums = nums;
            int n = nums.size();
            tr.resize(n << 2);
            for (int i = 0; i < tr.size(); ++i) {
                tr[i] = new Node();
            }
            build(1, 1, n);
        }
    
        void modify(int u, int l, int r) {
            if (tr[u]->l >= l && tr[u]->r <= r) {
                tr[u]->lazy ^= 1;
                tr[u]->s = tr[u]->r - tr[u]->l + 1 - tr[u]->s;
                return;
            }
            pushdown(u);
            int mid = (tr[u]->l + tr[u]->r) >> 1;
            if (l <= mid) {
                modify(u << 1, l, r);
            }
            if (r > mid) {
                modify(u << 1 | 1, l, r);
            }
            pushup(u);
        }
    
        int query(int u, int l, int r) {
            if (tr[u]->l >= l && tr[u]->r <= r) {
                return tr[u]->s;
            }
            pushdown(u);
            int mid = (tr[u]->l + tr[u]->r) >> 1;
            int res = 0;
            if (l <= mid) {
                res += query(u << 1, l, r);
            }
            if (r > mid) {
                res += query(u << 1 | 1, l, r);
            }
            return res;
        }
    
    private:
        vector<Node*> tr;
        vector<int> nums;
    
        void build(int u, int l, int r) {
            tr[u]->l = l;
            tr[u]->r = r;
            if (l == r) {
                tr[u]->s = nums[l - 1];
                return;
            }
            int mid = (l + r) >> 1;
            build(u << 1, l, mid);
            build(u << 1 | 1, mid + 1, r);
            pushup(u);
        }
    
        void pushup(int u) {
            tr[u]->s = tr[u << 1]->s + tr[u << 1 | 1]->s;
        }
    
        void pushdown(int u) {
            if (tr[u]->lazy) {
                int mid = (tr[u]->l + tr[u]->r) >> 1;
                tr[u << 1]->s = mid - tr[u]->l + 1 - tr[u << 1]->s;
                tr[u << 1]->lazy ^= 1;
                tr[u << 1 | 1]->s = tr[u]->r - mid - tr[u << 1 | 1]->s;
                tr[u << 1 | 1]->lazy ^= 1;
                tr[u]->lazy ^= 1;
            }
        }
    };
    
    class Solution {
    public:
        vector<long long> handleQuery(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {
            SegmentTree* tree = new SegmentTree(nums1);
            long long s = 0;
            for (int& x : nums2) {
                s += x;
            }
            vector<long long> ans;
            for (auto& q : queries) {
                if (q[0] == 1) {
                    tree->modify(1, q[1] + 1, q[2] + 1);
                } else if (q[0] == 2) {
                    s += 1LL * q[1] * tree->query(1, 1, nums1.size());
                } else {
                    ans.push_back(s);
                }
            }
            return ans;
        }
    };
    
  • class Node:
        def __init__(self):
            self.l = self.r = 0
            self.s = self.lazy = 0
    
    
    class SegmentTree:
        def __init__(self, nums):
            self.nums = nums
            n = len(nums)
            self.tr = [Node() for _ in range(n << 2)]
            self.build(1, 1, n)
    
        def build(self, u, l, r):
            self.tr[u].l, self.tr[u].r = l, r
            if l == r:
                self.tr[u].s = self.nums[l - 1]
                return
            mid = (l + r) >> 1
            self.build(u << 1, l, mid)
            self.build(u << 1 | 1, mid + 1, r)
            self.pushup(u)
    
        def modify(self, u, l, r):
            if self.tr[u].l >= l and self.tr[u].r <= r:
                self.tr[u].lazy ^= 1
                self.tr[u].s = self.tr[u].r - self.tr[u].l + 1 - self.tr[u].s
                return
            self.pushdown(u)
            mid = (self.tr[u].l + self.tr[u].r) >> 1
            if l <= mid:
                self.modify(u << 1, l, r)
            if r > mid:
                self.modify(u << 1 | 1, l, r)
            self.pushup(u)
    
        def query(self, u, l, r):
            if self.tr[u].l >= l and self.tr[u].r <= r:
                return self.tr[u].s
            self.pushdown(u)
            mid = (self.tr[u].l + self.tr[u].r) >> 1
            res = 0
            if l <= mid:
                res += self.query(u << 1, l, r)
            if r > mid:
                res += self.query(u << 1 | 1, l, r)
            return res
    
        def pushup(self, u):
            self.tr[u].s = self.tr[u << 1].s + self.tr[u << 1 | 1].s
    
        def pushdown(self, u):
            if self.tr[u].lazy:
                mid = (self.tr[u].l + self.tr[u].r) >> 1
                self.tr[u << 1].s = mid - self.tr[u].l + 1 - self.tr[u << 1].s
                self.tr[u << 1].lazy ^= 1
                self.tr[u << 1 | 1].s = self.tr[u].r - mid - self.tr[u << 1 | 1].s
                self.tr[u << 1 | 1].lazy ^= 1
                self.tr[u].lazy ^= 1
    
    
    class Solution:
        def handleQuery(
            self, nums1: List[int], nums2: List[int], queries: List[List[int]]
        ) -> List[int]:
            tree = SegmentTree(nums1)
            s = sum(nums2)
            ans = []
            for op, a, b in queries:
                if op == 1:
                    tree.modify(1, a + 1, b + 1)
                elif op == 2:
                    s += a * tree.query(1, 1, len(nums1))
                else:
                    ans.append(s)
            return ans
    
    
  • type node struct {
    	l, r, s, lazy int
    }
    
    type segmentTree struct {
    	nums []int
    	tr   []*node
    }
    
    func newSegmentTree(nums []int) *segmentTree {
    	n := len(nums)
    	tr := make([]*node, n<<2)
    	for i := range tr {
    		tr[i] = &node{}
    	}
    	t := &segmentTree{nums, tr}
    	t.build(1, 1, n)
    	return t
    }
    
    func (t *segmentTree) build(u, l, r int) {
    	t.tr[u].l, t.tr[u].r = l, r
    	if l == r {
    		t.tr[u].s = t.nums[l-1]
    		return
    	}
    	mid := (l + r) >> 1
    	t.build(u<<1, l, mid)
    	t.build(u<<1|1, mid+1, r)
    	t.pushup(u)
    }
    
    func (t *segmentTree) modify(u, l, r int) {
    	if t.tr[u].l >= l && t.tr[u].r <= r {
    		t.tr[u].lazy ^= 1
    		t.tr[u].s = t.tr[u].r - t.tr[u].l + 1 - t.tr[u].s
    		return
    	}
    	t.pushdown(u)
    	mid := (t.tr[u].l + t.tr[u].r) >> 1
    	if l <= mid {
    		t.modify(u<<1, l, r)
    	}
    	if r > mid {
    		t.modify(u<<1|1, l, r)
    	}
    	t.pushup(u)
    }
    
    func (t *segmentTree) query(u, l, r int) int {
    	if t.tr[u].l >= l && t.tr[u].r <= r {
    		return t.tr[u].s
    	}
    	t.pushdown(u)
    	mid := (t.tr[u].l + t.tr[u].r) >> 1
    	res := 0
    	if l <= mid {
    		res += t.query(u<<1, l, r)
    	}
    	if r > mid {
    		res += t.query(u<<1|1, l, r)
    	}
    	return res
    }
    
    func (t *segmentTree) pushup(u int) {
    	t.tr[u].s = t.tr[u<<1].s + t.tr[u<<1|1].s
    }
    
    func (t *segmentTree) pushdown(u int) {
    	if t.tr[u].lazy == 1 {
    		mid := (t.tr[u].l + t.tr[u].r) >> 1
    		t.tr[u<<1].s = mid - t.tr[u].l + 1 - t.tr[u<<1].s
    		t.tr[u<<1].lazy ^= 1
    		t.tr[u<<1|1].s = t.tr[u].r - mid - t.tr[u<<1|1].s
    		t.tr[u<<1|1].lazy ^= 1
    		t.tr[u].lazy ^= 1
    	}
    }
    
    func handleQuery(nums1 []int, nums2 []int, queries [][]int) (ans []int64) {
    	tree := newSegmentTree(nums1)
    	var s int64
    	for _, x := range nums2 {
    		s += int64(x)
    	}
    	for _, q := range queries {
    		if q[0] == 1 {
    			tree.modify(1, q[1]+1, q[2]+1)
    		} else if q[0] == 2 {
    			s += int64(q[1] * tree.query(1, 1, len(nums1)))
    		} else {
    			ans = append(ans, s)
    		}
    	}
    	return
    }
    

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