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0307 - Range Sum Query - Mutable (Medium)

https://leetcode.com/problems/range-sum-query-mutable/

Problem Statement​

Given an integer array nums, handle multiple queries of the following types:

  1. Update the value of an element in nums.
  2. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

  • NumArray(int[] nums) Initializes the object with the integer array nums.
  • void update(int index, int val) Updates the value of nums[index] to be val.
  • int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

Example 1:

Input
["NumArray", "sumRange", "update", "sumRange"]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
Output
[null, 9, null, 8]

Explanation
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // return 1 + 3 + 5 = 9
numArray.update(1, 2);   // nums = [1, 2, 5]
numArray.sumRange(0, 2); // return 1 + 2 + 5 = 8

Constraints:

  • 1 <= nums.length <= 3 * 10^4
  • -100 <= nums[i] <= 100
  • 0 <= index < nums.length
  • -100 <= val <= 100
  • 0 <= left <= right < nums.length
  • At most 3 * 10^4 calls will be made to update and sumRange.

Approach 1: Segment Tree​

Written by @wingkwong
struct segtree {
    vector<long long> sums;
    int size;
    
    void init(int n) {
        size = 1;
        while (size < n) size *= 2;
        sums.assign(size * 2, 0LL);
    }
    
    void set(int i, int v, int x, int lx, int rx) {
        if (rx - lx == 1) {
            sums[x] = v;
            return;
        }
        int m = (lx + rx) / 2;
        if (i < m) {
            set(i, v, 2 * x + 1, lx, m);
        } else {
            set(i, v, 2 * x + 2, m, rx);
        }
        sums[x] = sums[2 * x + 1] + sums[2 * x + 2];
    }
    
    void set(int i, int v) {
        set(i, v, 0, 0, size);
    }
    
    long long sum(int l, int r, int x, int lx, int rx) {
        // no intersection
        if (lx >= r || l >= rx) return 0;
        // inside
        if (lx >= l && rx <= r) return sums[x];
        int m = (lx + rx) / 2;
        long long s1 = sum(l, r, 2 * x + 1, lx, m);
        long long s2 = sum(l, r, 2 * x + 2, m, rx);
        return s1 + s2;
    }
    
    
    long long sum(int l, int r) {
        return sum(l, r, 0, 0, size);
    }
};

class NumArray {
public:
    NumArray(vector<int>& nums) {
        n = nums.size();
        st.init(n);
        for (int i = 0; i < n; i++) {
            st.set(i, nums[i]);
        }
    }
    
    void update(int index, int val) {
        st.set(index, val);
    }
    
    int sumRange(int left, int right) {
        return st.sum(left, right + 1);
    }

private: 
    segtree st;
    int n;
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray* obj = new NumArray(nums);
 * obj->update(index,val);
 * int param_2 = obj->sumRange(left,right);
 */