LeetCode-in-Java
3444. Minimum Increments for Target Multiples in an Array
Hard
You are given two arrays, nums and target.
In a single operation, you may increment any element of nums by 1.
Return the minimum number of operations required so that each element in target has at least one multiple in nums.
Example 1:
Input: nums = [1,2,3], target = [4]
Output: 1
Explanation:
The minimum number of operations required to satisfy the condition is 1.
- Increment 3 to 4 with just one operation, making 4 a multiple of itself.
Example 2:
Input: nums = [8,4], target = [10,5]
Output: 2
Explanation:
The minimum number of operations required to satisfy the condition is 2.
- Increment 8 to 10 with 2 operations, making 10 a multiple of both 5 and 10.
Example 3:
Input: nums = [7,9,10], target = [7]
Output: 0
Explanation:
Target 7 already has a multiple in nums, so no additional operations are needed.
Constraints:
1 <= nums.length <= 5 * 1041 <= target.length <= 4target.length <= nums.length1 <= nums[i], target[i] <= 104
Solution
public class Solution {
public int minimumIncrements(int[] nums, int[] target) {
int m = target.length;
int fullMask = (1 << m) - 1;
long[] lcmArr = new long[1 << m];
for (int mask = 1; mask < (1 << m); mask++) {
long l = 1;
for (int j = 0; j < m; j++) {
if ((mask & (1 << j)) != 0) {
l = lcm(l, target[j]);
}
}
lcmArr[mask] = l;
}
long inf = Long.MAX_VALUE / 2;
long[] dp = new long[1 << m];
for (int i = 0; i < (1 << m); i++) {
dp[i] = inf;
}
dp[0] = 0;
for (int x : nums) {
long[] newdp = dp.clone();
for (int mask = 1; mask < (1 << m); mask++) {
long l = lcmArr[mask];
long r = x % l;
long cost = (r == 0 ? 0 : l - r);
for (int old = 0; old < (1 << m); old++) {
int newMask = old | mask;
newdp[newMask] = Math.min(newdp[newMask], dp[old] + cost);
}
}
dp = newdp;
}
return (int) dp[fullMask];
}
private long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
private long lcm(long a, long b) {
return a / gcd(a, b) * b;
}
}