Question

1679. Max Number of K-Sum Pairs

中文文档

Description

You are given an integer array nums and an integer k.

In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.

Return _the maximum number of operations you can perform on the array_.

 

Example 1:

Input: nums = [1,2,3,4], k = 5
Output: 2
Explanation: Starting with nums = [1,2,3,4]:
- Remove numbers 1 and 4, then nums = [2,3]
- Remove numbers 2 and 3, then nums = []
There are no more pairs that sum up to 5, hence a total of 2 operations.

Example 2:

Input: nums = [3,1,3,4,3], k = 6
Output: 1
Explanation: Starting with nums = [3,1,3,4,3]:
- Remove the first two 3's, then nums = [1,4,3]
There are no more pairs that sum up to 6, hence a total of 1 operation.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= k <= 109

Solutions

Solution 1: Sorting

We sort $nums$. Then $l$ and $r$ point to the first and last elements of $nums$ respectively, and we compare the sum $s$ of the two integers with $k$.

• If $s = k$, it means that we have found two integers whose sum is $k$. We increment the answer and then move $l$ and $r$ towards the middle;
• If $s > k$, then we move the $r$ pointer to the left;
• If $s < k$, then we move the $l$ pointer to the right;
• We continue the loop until $l \geq r$.

After the loop ends, we return the answer.

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

class Solution:
  def maxOperations(self, nums: List[int], k: int) -> int:
    nums.sort()
    l, r, ans = 0, len(nums) - 1, 0
    while l < r:
      s = nums[l] + nums[r]
      if s == k:
        ans += 1
        l, r = l + 1, r - 1
      elif s > k:
        r -= 1
      else:
        l += 1
    return ans

class Solution {
  public int maxOperations(int[] nums, int k) {
    Arrays.sort(nums);
    int l = 0, r = nums.length - 1;
    int ans = 0;
    while (l < r) {
      int s = nums[l] + nums[r];
      if (s == k) {
        ++ans;
        ++l;
        --r;
      } else if (s > k) {
        --r;
      } else {
        ++l;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int maxOperations(vector<int>& nums, int k) {
    sort(nums.begin(), nums.end());
    int cnt = 0;
    int i = 0, j = nums.size() - 1;
    while (i < j) {
      if (nums[i] + nums[j] == k) {
        i++;
        j--;
        cnt++;
      } else if (nums[i] + nums[j] > k) {
        j--;
      } else {
        i++;
      }
    }
    return cnt;
  }
};

func maxOperations(nums []int, k int) int {
  sort.Ints(nums)
  l, r, ans := 0, len(nums)-1, 0
  for l < r {
    s := nums[l] + nums[r]
    if s == k {
      ans++
      l++
      r--
    } else if s > k {
      r--
    } else {
      l++
    }
  }
  return ans
}

function maxOperations(nums: number[], k: number): number {
  const cnt = new Map();
  let ans = 0;
  for (const x of nums) {
    if (cnt.get(k - x)) {
      cnt.set(k - x, cnt.get(k - x) - 1);
      ++ans;
    } else {
      cnt.set(x, (cnt.get(x) | 0) + 1);
    }
  }
  return ans;
}

impl Solution {
  pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
    let mut nums = nums.clone();
    nums.sort();
    let (mut l, mut r, mut ans) = (0, nums.len() - 1, 0);
    while l < r {
      match nums[l] + nums[r] {
        sum if sum == k => {
          ans += 1;
          l += 1;
          r -= 1;
        }
        sum if sum > k => {
          r -= 1;
        }
        _ => {
          l += 1;
        }
      }
    }
    ans
  }
}

Solution 2: Hash Table

We use a hash table $cnt$ to record the current remaining integers and their occurrence counts.

We iterate over $nums$. For the current integer $x$, we check if $k - x$ is in $cnt$. If it exists, it means that we have found two integers whose sum is $k$. We increment the answer and then decrement the occurrence count of $k - x$; otherwise, we increment the occurrence count of $x$.

After the iteration ends, we return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $nums$.

class Solution:
  def maxOperations(self, nums: List[int], k: int) -> int:
    cnt = Counter()
    ans = 0
    for x in nums:
      if cnt[k - x]:
        ans += 1
        cnt[k - x] -= 1
      else:
        cnt[x] += 1
    return ans

class Solution {
  public int maxOperations(int[] nums, int k) {
    Map<Integer, Integer> cnt = new HashMap<>();
    int ans = 0;
    for (int x : nums) {
      if (cnt.containsKey(k - x)) {
        ++ans;
        if (cnt.merge(k - x, -1, Integer::sum) == 0) {
          cnt.remove(k - x);
        }
      } else {
        cnt.merge(x, 1, Integer::sum);
      }
    }
    return ans;
  }
}

class Solution {
public:
  int maxOperations(vector<int>& nums, int k) {
    unordered_map<int, int> cnt;
    int ans = 0;
    for (int& x : nums) {
      if (cnt[k - x]) {
        --cnt[k - x];
        ++ans;
      } else {
        ++cnt[x];
      }
    }
    return ans;
  }
};

func maxOperations(nums []int, k int) (ans int) {
  cnt := map[int]int{}
  for _, x := range nums {
    if cnt[k-x] > 0 {
      cnt[k-x]--
      ans++
    } else {
      cnt[x]++
    }
  }
  return
}

impl Solution {
  pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
    let mut cnt = std::collections::HashMap::new();
    let mut ans = 0;
    for x in nums {
      let m = k - x;
      if let Some(v) = cnt.get_mut(&m) {
        ans += 1;
        *v -= 1;
        if *v == 0 {
          cnt.remove(&m);
        }
      } else if let Some(v) = cnt.get_mut(&x) {
        *v += 1;
      } else {
        cnt.insert(x, 1);
      }
    }
    ans
  }
}