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-rw-r--r-- | zero_array_transformation2.c | 71 | ||||
-rw-r--r-- | zero_array_transformation2.py | 41 |
2 files changed, 112 insertions, 0 deletions
diff --git a/zero_array_transformation2.c b/zero_array_transformation2.c new file mode 100644 index 0000000..4532469 --- /dev/null +++ b/zero_array_transformation2.c @@ -0,0 +1,71 @@ +// 3356. Zero Array Transformation II +#include "leetcode.h" + +/* + * you are given an integer array 'nums' of length 'n' and a 2d array 'queries' + * where 'queries[i] [l_i, r_i, val_i]'. each 'queries[i]' represents the + * following action on 'nums': decrement the value at each index in the range + * '[l_i, r_i]' in 'nums' by at most 'val_i'. the amount by which each 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 minimum possible + * non-negative value of 'k', such that after processing the first 'k' queries + * in sequence, 'nums' becomes a zero array. if no such 'k' exists, return -1. + */ + +bool can_cover(int *nums, int n, int **queries, int k) { + int *diff = (int *)calloc(n + 1, sizeof(int)); + for (int i = 0; i < k; i++) { + int l = queries[i][0]; + int r = queries[i][1]; + int val = queries[i][2]; + diff[l] += val; + if (r + 1 < n) + diff[r + 1] -= val; + } + int sum = 0; + for (int i = 0; i < n; i++) { + sum += diff[i]; + if (sum < nums[i]) { + free(diff); + return false; + } + } + free(diff); + return true; +} + +int minZeroArray(int *nums, int numsSize, int **queries, int queriesSize, + int *queriesColSize) { + int already_zero = 1; + for (int i = 0; i < numsSize; i++) + if (nums[i]) { + already_zero = 0; + break; + } + if (already_zero) + return 0; + int l = 1, r = queriesSize, ans = -1; + while (l <= r) { + int mid = l + (r - l) / 2; + if (can_cover(nums, numsSize, queries, mid)) { + ans = mid; + r = mid - 1; + } else + l = mid + 1; + } + return ans; +} + +int main() { + int n1[] = {2, 0, 2}, q1i[3][3] = {{0, 2, 1}, {0, 2, 1}, {1, 1, 3}}; + int n2[] = {4, 3, 2, 1}, q2i[2][3] = {{1, 3, 2}, {0, 2, 1}}; + 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", minZeroArray(n1, ARRAY_SIZE(n1), q1.arr, q1.row_size, + q1.col_size)); // expect: 2 + printf("%d\n", minZeroArray(n2, ARRAY_SIZE(n2), q2.arr, q2.row_size, + q2.col_size)); // expect: -1 + two_d_arr_free(&q1); + two_d_arr_free(&q2); +} diff --git a/zero_array_transformation2.py b/zero_array_transformation2.py new file mode 100644 index 0000000..1b462e5 --- /dev/null +++ b/zero_array_transformation2.py @@ -0,0 +1,41 @@ +# 3356. Zero Array Transformation II + +""" +you are given an integer array 'nums' of length 'n' and a 2d array 'queries' +where 'queries[i] [l_i, r_i, val_i]'. each 'queries[i]' represents the +following action on 'nums': decrement the value at each index in the range +'[l_i, r_i]' in 'nums' by at most 'val_i'. the amount by which each 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 minimum possible +non-negative value of 'k', such that after processing the first 'k' queries +in sequence, 'nums' becomes a zero array. if no such 'k' exists, return -1. +""" + + +class Solution(object): + def minZeroArray(self, nums, queries): + """ + :type nums: List[int] + :type queries: List[List[int]] + :rtype: int + """ + n, ans, k = len(nums), 0, 0 + cnt = [0 for _ in range(n + 1)] + for i in range(n): + while k + cnt[i] < nums[i]: + ans += 1 + if ans - 1 >= len(queries): + return -1 + l, r, val = queries[ans - 1] + if r < i: + continue + cnt[max(l, i)] += val + cnt[r + 1] -= val + k += cnt[i] + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.minZeroArray(nums=[2, 0, 2], queries=[[0, 2, 1], [0, 2, 1], [1, 1, 3]])) + print(obj.minZeroArray(nums=[4, 3, 2, 1], queries=[[1, 3, 2], [0, 2, 1]])) |