Medium
You are given an integer array nums of size n and a positive integer k.
An array capped by value x is obtained by replacing every element nums[i] with min(nums[i], x).
For each integer x from 1 to n, determine whether it is possible to choose a subsequence from the array capped by x such that the sum of the chosen elements is exactly k.
Return a 0-indexed boolean array answer of size n, where answer[i] is true if it is possible when using x = i + 1, and false otherwise.
Example 1:
Input: nums = [4,3,2,4], k = 5
Output: [false,false,true,true]
Explanation:
x = 1, the capped array is [1, 1, 1, 1]. Possible sums are 1, 2, 3, 4, so it is impossible to form a sum of 5.x = 2, the capped array is [2, 2, 2, 2]. Possible sums are 2, 4, 6, 8, so it is impossible to form a sum of 5.x = 3, the capped array is [3, 3, 2, 3]. A subsequence [2, 3] sums to 5, so it is possible.x = 4, the capped array is [4, 3, 2, 4]. A subsequence [3, 2] sums to 5, so it is possible.Example 2:
Input: nums = [1,2,3,4,5], k = 3
Output: [true,true,true,true,true]
Explanation:
For every value of x, it is always possible to select a subsequence from the capped array that sums exactly to 3.
Constraints:
1 <= n == nums.length <= 40001 <= nums[i] <= n1 <= k <= 4000public class Solution {
public boolean[] subsequenceSumAfterCapping(int[] nums, int k) {
int n = nums.length;
boolean[] answer = new boolean[n];
int[] freq = new int[n + 2];
for (int v : nums) {
if (v <= n) {
freq[v]++;
}
}
int[] cntGe = new int[n + 2];
cntGe[n] = freq[n];
for (int x = n - 1; x >= 1; x--) {
cntGe[x] = cntGe[x + 1] + freq[x];
}
boolean[] dp = new boolean[k + 1];
dp[0] = true;
for (int x = 1; x <= n; x++) {
int cnt = cntGe[x];
boolean ok = false;
int maxM = cnt;
int limit = k / x;
if (maxM > limit) {
maxM = limit;
}
for (int m = 0; m <= maxM; m++) {
int rem = k - m * x;
if (rem >= 0 && dp[rem]) {
ok = true;
break;
}
}
answer[x - 1] = ok;
int c = freq[x];
if (c == 0) {
continue;
}
int power = 1;
while (c > 0) {
int take = Math.min(power, c);
int weight = take * x;
for (int s = k; s >= weight; s--) {
if (!dp[s] && dp[s - weight]) {
dp[s] = true;
}
}
c -= take;
power <<= 1;
}
}
return answer;
}
}