LeetCode-in-Java

3685. Subsequence Sum After Capping Elements

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:

• For 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.
• For 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.
• For x = 3, the capped array is [3, 3, 2, 3]. A subsequence [2, 3] sums to 5, so it is possible.
• For 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 <= 4000
• 1 <= nums[i] <= n
• 1 <= k <= 4000

Solution

public 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;
    }
}

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