树状数组理解

数状数组理解：

1，为了解决什么问题

频繁查询数组前缀和+数组元素动态更新时

2，原理

本质上利用数字的二进制思想来存储前缀和

举个例子：

原数组arry1...n

数状数组bit_tree1...n

bit_tree1 = arry1

bit_tree2 = arry1 + arry2

....

bit_treei本质上存储sum = arryi - lowbit(i) + 1 +... arryi

3，这样的好处是什么呢？

正常数组比如arry1 += 1时，后面所有的idx都得做+1操作；用数组数组只需要更新几个节点即可，降低了时间复杂度～

来做题消化一下~

https://leetcode.cn/problems/minimum-inversion-count-in-subarrays-of-fixed-length

思路：

1，滑动窗口来计算当前当前逆序值~

2，使用树状数组来降低时间复杂度~

typedef struct {
    int* tree;
    int size;
} FenwickTree;

FenwickTree* createFenwickTree(int size) {
    FenwickTree* ft = (FenwickTree*)malloc(sizeof(FenwickTree));
    ft->size = size + 2; 
    ft->tree = (int*)calloc(ft->size, sizeof(int));
    return ft;
}

void update(FenwickTree* ft, int idx, int delta) {
    idx++; 
    while (idx < ft->size) {
        ft->tree[idx] += delta;
        idx += idx & -idx;
    }
}

int query(FenwickTree* ft, int idx) {
    idx++; 
    int sum = 0;
    while (idx > 0) {
        sum += ft->tree[idx];
        idx -= idx & -idx;
    }
    return sum;
}

int cmp(const void* a, const void* b) 
{ return *(int*)a - *(int*)b; }

void process_data_qsort_Deduplicate(int* nums, int n, int* processed_index, int* qsorted_data) {
    int* sorted = (int*)malloc(n * sizeof(int));
    memcpy(sorted, nums, n * sizeof(int));
    
    qsort(sorted, n, sizeof(int), cmp);
    
    int m = 0;
    for (int i = 0; i < n; i++) {
        if (i == 0 || sorted[i] != sorted[i - 1]) {
            qsorted_data[m++] = sorted[i];
        }
    }
    
    for (int i = 0; i < n; i++) {
        int left = 0, right = m - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (qsorted_data[mid] == nums[i]) {
                processed_index[i] = mid;
                break;
            } else if (qsorted_data[mid] < nums[i]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
    }
    
    free(sorted);
}
long long minInversionCount(int* nums, int numsSize, int k) {
    if (k == 1) return 0; // 单个元素没有逆序对
    int  n = numsSize;
    int* processed_index = (int*)malloc(n * sizeof(int));
    int* qsorted_data = (int*)malloc(n * sizeof(int));
    process_data_qsort_Deduplicate(nums, n, processed_index, qsorted_data);
    int* timberavos = (int*)malloc(n * sizeof(int));
    memcpy(timberavos, processed_index, n * sizeof(int));
    FenwickTree* ft = createFenwickTree(n);
    long long current_Inversion = 0;
    for (int i = 0; i < k; i++) {
        int greater_count = query(ft, n - 1) - query(ft, timberavos[i]);
        current_Inversion += greater_count;
        update(ft, timberavos[i], 1);
    }
    long long min_Inversion = current_Inversion;
    for (int i = k; i < n; i++) {
        int left_val = timberavos[i - k];
        update(ft, timberavos[i-k], -1);
        int left_greater = query(ft, timberavos[i-k]-1);
        current_Inversion -= left_greater;
        // printf("current_Inversion00:%d %d\n", i, current_Inversion);
        int right_val = timberavos[i];
        update(ft, timberavos[i], 1);
        int right_greater = query(ft, n - 1) - query(ft, timberavos[i]);
        current_Inversion += right_greater;
        // printf("current_Inversion11:%d %d\n", i, current_Inversion);
        if (current_Inversion < min_Inversion) {
            min_Inversion = current_Inversion;
        }
    }
    
    free(processed_index);
    free(qsorted_data);
    free(timberavos);
    free(ft->tree);
    free(ft);
    
    return min_Inversion;
}