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How to solve Leetcode 1679. Max Number of K-Sum Pairs

An example of using unordered_map

Published
2 min read
How to solve Leetcode 1679. Max Number of K-Sum Pairs
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Hi! My name is Nhut Nguyen. I am a software engineer and a writer in Copenhagen, Denmark.

Learn more about me at nhutnguyen.com

Problem statement

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 <= 10^5.

  • 1 <= nums[i] <= 10^9.

  • 1 <= k <= 10^9.

Solution: Count the appearances

You can use a map to count the appearances of the elements of nums.

Example 2

For nums = [3,1,3,4,3] and k = 6:

  • Initialize count = 0.

  • For i = 0: m[3] = 1; k - 3 = 3 but m[3] is only 1, not enough to have two numbers.

  • For i = 1: m[1] = 1; k - 1 = 5 and m[5] = 0.

  • For i = 2: m[3] = 2; k - 3 = 3 and m[3] = 2 just enough to have two numbers to perform the sum. count = 1. Erase those two values 3's from the map: m[3] = 0.

  • For i = 3: m[4] = 1; k - 4 = 2 and m[2] = 0.

  • For i = 4: m[3] = 1; k - 3 = 3 but m[3] is only 1, not enough to have two numbers.

  • Final count = 1.

Code

#include <vector>
#include <iostream>
#include <unordered_map>
using namespace std;
int maxOperations(vector<int>& nums, int k) {
    unordered_map<int,int> m;
    int count = 0;
    for (int a : nums) {
        m[a]++;
        if (m[k - a] > 0) {
            if (a != k - a || m[a] >= 2) {
                count++;
                m[a]--;
                m[k - a]--;
            }
        }
    }
    return count;
}
int main() {
    vector<int> nums{1,2,3,4};
    cout << maxOperations(nums, 5) << endl;
    nums = {3,1,3,4,3};
    cout << maxOperations(nums, 6) << endl;
}
Output:
2
1

Complexity

  • Runtime: O(N), where N = nums.length.

  • Extra space: O(N).

References

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