You are given an array of n integers, nums, where there are at most 50 unique values in the array. You are also given an array of m customer order quantities, quantity, where quantity[i] is the amount of integers the ith customer ordered. Determine if it is possible to distribute nums such that:
The ith customer gets exactly quantity[i] integers,
The integers the ith customer gets are all equal, and
Every customer is satisfied.
Return true if it is possible to distribute nums according to the above conditions.
Example 1:
Input: nums = [1,2,3,4], quantity = [2]
Output: false
Explanation: The 0th customer cannot be given two different integers.
Example 2:
Input: nums = [1,2,3,3], quantity = [2]
Output: true
Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used.
Example 3:
Input: nums = [1,1,2,2], quantity = [2,2]
Output: true
Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2].
Example 4:
Input: nums = [1,1,2,3], quantity = [2,2]
Output: false
Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied.
Example 5:
Input: nums = [1,1,1,1,1], quantity = [2,3]
Output: true
Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1].
Constraints:
n == nums.length
1 <= n <= 105
1 <= nums[i] <= 1000
m == quantity.length
1 <= m <= 10
1 <= quantity[i] <= 105
There are at most 50 unique values in nums.
Solution:
class Solution {
public boolean canDistribute(int[] nums, int[] quantity) {
int[] count = new int[1001];
TreeSet<Integer> set = new TreeSet(Collections.reverseOrder());
for (int val : nums) {
count[val] ++;
set.add(val);
}
int[] c = new int[50];
int i = 0;
while (!set.isEmpty()) {
c[i] = set.first();
set.remove(c[i]);
i ++;
}
Arrays.sort(quantity);
return helper(c, count, quantity, quantity.length - 1);
}
private boolean helper(int[] c, int[] count, int[] quantity, int start) {
if (start < 0) return true;
for (int i = 0; i < 50; i ++) {
int item = c[i];
if (count[item] >= quantity[start]) {
count[item] -= quantity[start];
if (helper(c, count, quantity, start - 1)) {
return true;
}
count[item] += quantity[start];
}
}
return false;
}
}
dp[i][j] = for numbers in subsequence [0, i], whether it can satisfy j (bit mask of quantity)
class Solution {
public boolean canDistribute(int[] nums, int[] quantity) {
Map<Integer, Integer> count = new HashMap();
for (int val : nums) {
count.put(val, count.getOrDefault(val, 0) + 1);
}
List<Integer> items = new ArrayList(count.keySet());
int n = items.size(), m = quantity.length;
int[] requirements = new int[1 << m];
for (int i = 0; i < (1 << m); i ++) {
for (int j = 0; j < m; j ++) {
if ((i & (1 << j)) > 0) {
requirements[i] += quantity[j];
}
}
}
boolean[][] dp = new boolean[n + 1][1 << m];
dp[0][0] = true;
for (int i = 0; i < n; i ++) {
for (int j = 0; j < (1 << m); j ++) {
dp[i + 1][j] |= dp[i][j];
for (int s = j; s > 0; s = (s - 1) & j) {
if (requirements[s] <= count.get(items.get(i)) && dp[i][j ^ s]) {
dp[i + 1][j] = true;
break;
}
}
}
}
return dp[n][(1 << m) - 1];
}
}