// 3362. Zero Array Transformation III #include "leetcode.h" /* * you are given an integer array 'nums' of length 'n' and a 2d array 'queries' * where 'queries[i] = [li, ri]'. each 'queries[i]' represents the following * action on 'nums': decrement the value at each index in the range '[li, ri]' * in 'nums' by at most 1. the amount by which the value is decremented can be * chosen independently for each index. a zero array is an array with all its * elements equal to 0. return the maximum number of elements that can be * removed from 'queries' such that 'nums' can still be converted to a zero * array using the remaining queries. if it is not possible to convert 'nums' to * a zero array, return -1. */ int cmp(const void *a, const void *b) { int *aa = *(int **)a, *bb = *(int **)b; return aa[0] - bb[0]; } struct heap { int *data; int cap; int size; }; struct heap *heap_create(int cap) { struct heap *obj = (struct heap *)malloc(sizeof(struct heap)); obj->data = (int *)malloc(cap * sizeof(int)); obj->cap = cap; obj->size = 0; return obj; } void heap_free(struct heap *obj) { free(obj->data); free(obj); } void swap(int *a, int *b) { int tmp = *a; *a = *b; *b = tmp; } void heap_up(struct heap *obj, int idx) { while (idx > 0 && obj->data[(idx - 1) / 2] < obj->data[idx]) { swap(&obj->data[(idx - 1) / 2], &obj->data[idx]); idx = (idx - 1) / 2; } } void heap_down(struct heap *obj, int idx) { int max = idx, left = 2 * idx + 1, right = 2 * idx + 2; if (left < obj->size && obj->data[left] > obj->data[max]) max = left; if (right < obj->size && obj->data[right] > obj->data[max]) max = right; if (max != idx) { swap(&obj->data[idx], &obj->data[max]); heap_down(obj, max); } } void heap_push(struct heap *obj, int val) { if (obj->size == obj->cap) return; obj->data[obj->size++] = val; heap_up(obj, obj->size - 1); } int heap_pop(struct heap *obj) { if (!obj->size) return -1; int max = obj->data[0]; obj->data[0] = obj->data[--obj->size]; heap_down(obj, 0); return max; } int heap_top(struct heap *obj) { if (!obj->size) return -1; return obj->data[0]; } int maxRemoval(int *nums, int numsSize, int **queries, int queriesSize, int *queriesColSize) { qsort(queries, queriesSize, sizeof(int *), cmp); struct heap *heap = heap_create(queriesSize); int *end = (int *)calloc(numsSize + 1, sizeof(int)); int curr = 0, j = 0; for (int i = 0; i < numsSize; ++i) { curr -= end[i]; while (j < queriesSize && queries[j][0] <= i) heap_push(heap, queries[j++][1]); while (curr < nums[i]) { if (!heap->size || heap_top(heap) < i) { heap_free(heap); free(end); return -1; } int top = heap_top(heap); end[top + 1]++; heap_pop(heap); curr++; } } int ans = heap->size; heap_free(heap); free(end); return ans; } int main() { int n1[] = {2, 0, 2}, q1i[3][2] = {{0, 2}, {0, 2}, {1, 1}}; int n2[] = {1, 1, 1, 1}, q2i[4][2] = {{1, 3}, {0, 2}, {1, 3}, {1, 2}}; struct two_d_arr q1, q2; two_d_arr_init(&q1, ARRAY_SIZE(q1i), ARRAY_SIZE(q1i[0]), q1i); two_d_arr_init(&q2, ARRAY_SIZE(q2i), ARRAY_SIZE(q2i[0]), q2i); printf("%d\n", maxRemoval(n1, ARRAY_SIZE(n1), q1.arr, q1.row_size, q1.col_size)); // expect: 1 printf("%d\n", maxRemoval(n2, ARRAY_SIZE(n2), q2.arr, q2.row_size, q2.col_size)); // expect: 2 two_d_arr_free(&q1); two_d_arr_free(&q2); }