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 <= 10^51 <= nums[i] <= 10^91 <= k <= 10^9
My Solution
// using Two Pointers
var maxOperations = function (nums, k) {
nums.sort((a, b) => a - b);
let count = 0;
let left = 0;
let right = nums.length - 1;
while(left < right) {
if (nums[left] + nums[right] === k) {
count++;
left++;
right--;
} else if (nums[left] + nums[right] < k) {
left++;
} else {
right--;
}
}
return count;
}
Solution from others
// using Hash Map
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var maxOperations = function (nums, k) {
let map = new Map(); // A hashmap to store the frequency of each number
let count = 0;
for (let num of nums) {
let complement = k - num; // Calculate the complement of the current number that would sum up to k
// If the complement is in the hashmap and has a frequency greater than 0,
// it means we've found a pair. So, increase the count and decrease the frequency of the complement.
if (map.get(complement) > 0) {
count++;
map.set(complement, map.get(complement) - 1);
} else {
// If the complement is not present, add or update the current number's frequency in the hashmap.
map.set(num, (map.get(num) || 0) + 1);
}
}
return count;
}