1679. Max Number of K-Sum Pairs

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 $nums$.

Python3JavaC++GoTypeScriptRust

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 $nums$.

Python3JavaC++GoRust

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
    }
}

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