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// 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);
}
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