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-rw-r--r--zero_array_transformation3.c127
-rw-r--r--zero_array_transformation3.py41
2 files changed, 168 insertions, 0 deletions
diff --git a/zero_array_transformation3.c b/zero_array_transformation3.c
new file mode 100644
index 0000000..b966099
--- /dev/null
+++ b/zero_array_transformation3.c
@@ -0,0 +1,127 @@
+// 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);
+}
diff --git a/zero_array_transformation3.py b/zero_array_transformation3.py
new file mode 100644
index 0000000..61ceb98
--- /dev/null
+++ b/zero_array_transformation3.py
@@ -0,0 +1,41 @@
+# 3362. Zero Array Transformation III
+from sortedcontainers import SortedList
+
+"""
+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.
+"""
+
+
+class Solution(object):
+ def maxRemoval(self, nums, queries):
+ """
+ :type nums: List[int]
+ :type queries: List[List[int]]
+ :rtype: int
+ """
+ queries = sorted(queries)[::-1]
+ end, curr = SortedList(), SortedList()
+ for i in range(len(nums)):
+ while queries and queries[-1][0] <= i:
+ end.add(queries.pop()[1])
+ while curr and curr[0] < i:
+ curr.pop(0)
+ while nums[i] > len(curr):
+ if not end or end[-1] < i:
+ return -1
+ curr.add(end.pop())
+ return len(end)
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maxRemoval(nums=[2, 0, 2], queries=[[0, 2], [0, 2], [1, 1]]))
+ print(obj.maxRemoval(nums=[1, 1, 1, 1], queries=[[1, 3], [0, 2], [1, 3], [1, 2]]))