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Medium
Greedy
Array
Sorting

2098. Subsequence of Size K With the Largest Even Sum 🔒

中文文档

Description

You are given an integer array nums and an integer k. Find the largest even sum of any subsequence of nums that has a length of k.

Return this sum, or -1 if such a sum does not exist.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: nums = [4,1,5,3,1], k = 3
Output: 12
Explanation:
The subsequence with the largest possible even sum is [4,5,3]. It has a sum of 4 + 5 + 3 = 12.

Example 2:

Input: nums = [4,6,2], k = 3
Output: 12
Explanation:
The subsequence with the largest possible even sum is [4,6,2]. It has a sum of 4 + 6 + 2 = 12.

Example 3:

Input: nums = [1,3,5], k = 1
Output: -1
Explanation:
No subsequence of nums with length 1 has an even sum.

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 105
  • 1 <= k <= nums.length

Solutions

Solution 1: Greedy + Sorting

We notice that the problem involves selecting a subsequence, so we can consider sorting the array first.

Next, we greedily select the largest $k$ numbers. If the sum of these numbers is even, we directly return this sum $ans$.

Otherwise, we have two greedy strategies:

  1. Among the largest $k$ numbers, find the smallest even number $mi1$, and then among the remaining $n - k$ numbers, find the largest odd number $mx1$. Replace $mi1$ with $mx1$. If such a replacement exists, then the sum after replacement $ans - mi1 + mx1$ is guaranteed to be even;
  2. Among the largest $k$ numbers, find the smallest odd number $mi2$, and then among the remaining $n - k$ numbers, find the largest even number $mx2$. Replace $mi2$ with $mx2$. If such a replacement exists, then the sum after replacement $ans - mi2 + mx2$ is guaranteed to be even.

We take the largest even sum as the answer. If no even sum exists, return $-1$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

Python3

class Solution:
    def largestEvenSum(self, nums: List[int], k: int) -> int:
        nums.sort()
        ans = sum(nums[-k:])
        if ans % 2 == 0:
            return ans
        n = len(nums)
        mx1 = mx2 = -inf
        for x in nums[: n - k]:
            if x & 1:
                mx1 = x
            else:
                mx2 = x
        mi1 = mi2 = inf
        for x in nums[-k:][::-1]:
            if x & 1:
                mi2 = x
            else:
                mi1 = x
        ans = max(ans - mi1 + mx1, ans - mi2 + mx2, -1)
        return -1 if ans < 0 else ans

Java

class Solution {
    public long largestEvenSum(int[] nums, int k) {
        Arrays.sort(nums);
        long ans = 0;
        int n = nums.length;
        for (int i = 0; i < k; ++i) {
            ans += nums[n - i - 1];
        }
        if (ans % 2 == 0) {
            return ans;
        }
        final int inf = 1 << 29;
        int mx1 = -inf, mx2 = -inf;
        for (int i = 0; i < n - k; ++i) {
            if (nums[i] % 2 == 1) {
                mx1 = nums[i];
            } else {
                mx2 = nums[i];
            }
        }
        int mi1 = inf, mi2 = inf;
        for (int i = n - 1; i >= n - k; --i) {
            if (nums[i] % 2 == 1) {
                mi2 = nums[i];
            } else {
                mi1 = nums[i];
            }
        }
        ans = Math.max(ans - mi1 + mx1, ans - mi2 + mx2);
        return ans < 0 ? -1 : ans;
    }
}

C++

class Solution {
public:
    long long largestEvenSum(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        long long ans = 0;
        int n = nums.size();
        for (int i = 0; i < k; ++i) {
            ans += nums[n - i - 1];
        }
        if (ans % 2 == 0) {
            return ans;
        }
        const int inf = 1 << 29;
        int mx1 = -inf, mx2 = -inf;
        for (int i = 0; i < n - k; ++i) {
            if (nums[i] % 2) {
                mx1 = nums[i];
            } else {
                mx2 = nums[i];
            }
        }
        int mi1 = inf, mi2 = inf;
        for (int i = n - 1; i >= n - k; --i) {
            if (nums[i] % 2) {
                mi2 = nums[i];
            } else {
                mi1 = nums[i];
            }
        }
        ans = max(ans - mi1 + mx1, ans - mi2 + mx2);
        return ans < 0 ? -1 : ans;
    }
};

Go

func largestEvenSum(nums []int, k int) int64 {
	sort.Ints(nums)
	ans := 0
	n := len(nums)
	for i := 0; i < k; i++ {
		ans += nums[n-1-i]
	}
	if ans%2 == 0 {
		return int64(ans)
	}
	const inf = 1 << 29
	mx1, mx2 := -inf, -inf
	for _, x := range nums[:n-k] {
		if x%2 == 1 {
			mx1 = x
		} else {
			mx2 = x
		}
	}
	mi1, mi2 := inf, inf
	for i := n - 1; i >= n-k; i-- {
		if nums[i]%2 == 1 {
			mi2 = nums[i]
		} else {
			mi1 = nums[i]
		}
	}
	ans = max(-1, max(ans-mi1+mx1, ans-mi2+mx2))
	if ans%2 < 0 {
		return -1
	}
	return int64(ans)
}

TypeScript

function largestEvenSum(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < k; ++i) {
        ans += nums[n - i - 1];
    }
    if (ans % 2 === 0) {
        return ans;
    }
    const inf = 1 << 29;
    let mx1 = -inf,
        mx2 = -inf;
    for (let i = 0; i < n - k; ++i) {
        if (nums[i] % 2 === 1) {
            mx1 = nums[i];
        } else {
            mx2 = nums[i];
        }
    }
    let mi1 = inf,
        mi2 = inf;
    for (let i = n - 1; i >= n - k; --i) {
        if (nums[i] % 2 === 1) {
            mi2 = nums[i];
        } else {
            mi1 = nums[i];
        }
    }
    ans = Math.max(ans - mi1 + mx1, ans - mi2 + mx2);
    return ans < 0 ? -1 : ans;
}