1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
|
// 1658. Minimum Operations to Reduce X to Zero
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
/*
* given an integer array 'nums' and an integer 'x'. in one operation, you can
* either remove the leftmost or the rightmost element from the array 'nums' and
* subtract its value from 'x'. note that this modifies the array for future
* operations. return the minimum number of operations to reduce 'x' to exactly
* 0. if it is possible, otherwise return -1
*/
int minOperations(int *nums, int nums_size, int x) {
int n = nums_size, mod = 1e5 + 7, i;
bool *table = calloc(mod, sizeof(bool));
int **hash = malloc(sizeof(int *) * mod);
long long pre_sum = nums[n - 1];
hash[pre_sum % mod] = calloc(2, sizeof(int));
hash[pre_sum % mod][0] = nums[n - 1];
hash[pre_sum % mod][1] = n - 1;
table[pre_sum % mod] = true;
for (i = n - 2; i >= 0; i--) {
pre_sum += nums[i];
if (pre_sum > x)
break;
if (!table[pre_sum % mod]) {
table[pre_sum % mod] = true;
hash[pre_sum % mod] = calloc(2, sizeof(int));
hash[pre_sum % mod][0] = pre_sum;
hash[pre_sum % mod][1] = i;
} else {
int pos = pre_sum + 1;
while (table[pos % mod])
pos++;
table[pos % mod] = true;
hash[pos % mod] = calloc(2, sizeof(int));
hash[pos % mod][0] = pre_sum;
hash[pos % mod][1] = i;
}
}
if (i == -1 && pre_sum == x)
return n;
int min = INT_MAX, sum = 0, k = 0;
while (sum <= x && k < n) {
int rem = x - sum;
if (!rem) {
min = fmin(min, k);
break;
}
if (table[rem % mod]) {
if (hash[rem % mod][0] == rem) {
if (k >= hash[rem % mod][1])
break;
if (n - hash[rem % mod][1] + k < min)
min = n - hash[rem % mod][1] + k;
} else {
int pos = rem + 1;
while (table[pos % mod]) {
if (hash[pos % mod][0] == rem) {
if (k >= hash[pos % mod][1])
break;
if (n - hash[pos % mod][1] + k < min)
min = n - hash[pos % mod][1] + k;
}
pos++;
}
}
}
sum += nums[k];
k++;
}
if (min == INT_MAX)
return -1;
else
return min;
}
int main() {
int n1[] = {1, 1, 4, 2, 3}, n2[] = {5, 6, 7, 8, 9},
n3[] = {3, 2, 20, 1, 1, 3};
printf("%d\n", minOperations(n1, 5, 5)); // expect: 2
printf("%d\n", minOperations(n2, 5, 4)); // expect: -1
printf("%d\n", minOperations(n3, 6, 10)); // expect: 5
}
|