315 Count of Smaller Numbers After Self

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題目:
https://leetcode.com/problems/count-of-smaller-numbers-after-self/

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// 一個更好的解法是Fenwick Tree (Binary Indexed Tree BIT)，這邊用歸併分治
class Solution {
public:
    vector<int> arr, help, ans, idx, help_idx;

    vector<int> countSmaller(vector<int>& nums) {
        int size = nums.size();

        arr.assign(size, 0);
        help.assign(size, 0);
        ans.assign(size, 0);
        idx.assign(size, 0);
        help_idx.assign(size, 0); 

        for(int i=0; i<size; ++i){
            arr[i] = nums[i];
            idx[i] = i;
        }
        merge(0, size-1);

        return ans;
    }

    void merge(int l, int r){
    if (l >= r) return; // l shouldn't be greater than r, and return when l == r
    
    int m = l + (r-l)/2; // prevent overflow
    merge(l, m);
    merge(m+1, r);
    solve(l, m, r);
}

void solve(int l, int m, int r){
    // count
    for(int i=l, j = m+1; i<=m; ++i){
        while(j <= r && arr[i] > arr[j]){
            j++;
        }
        ans[idx[i]] += (j-(m+1)); // total count = end - begin
    }

    // sort
    int i = l;
    int a = l;
    int b = m+1;
    while(a <= m && b <= r){
        if(arr[a] <= arr[b]){
            help_idx[i] = idx[a];
            help[i++] = arr[a++];
        }
        else{
            help_idx[i] = idx[b];
            help[i++] = arr[b++];
        }
    }
    while(a <= m){
        help_idx[i] = idx[a];
        help[i++] = arr[a++];
    }
    while(b <= r){
        help_idx[i] = idx[b];
        help[i++] = arr[b++];
    }
    for(int k=l; k <= r; ++k){
        arr[k] = help[k];
        idx[k] = help_idx[k];
    }
}
};