Two Sum Less Than K

class Solution {
public:
    int pairSumLessThanK(vector<int>& arr, int limit) {
        int bestSum = -1;
        int frequency[1001] = {};
        for (int v : arr) {
            frequency[v]++;
        }
        int left = 1;
        int right = 1000;
        while (left <= right) {
            if (left + right >= limit || frequency[right] == 0) {
                right--;
            } else {
                if (frequency[left] > (left < right ? 0 : 1)) {
                    bestSum = max(bestSum, left + right);
                }
                left++;
            }
        }
        return bestSum;
    }
};

The solution for the "Two Sum Less Than K" problem in C++ involves finding the maximum sum of pairs in an array whose sum is less than a specified value K (here indicated as limit). The given approach uses a two-pointer technique combined with a frequency array to efficiently solve the problem.

• Begin by initializing the bestSum to -1, to handle cases where no valid pair sum exists.
• Create a frequency array of size 1001 (assuming the values range from 1 to 1000) to count occurrences of each integer in the input vector arr.
• Populate the frequency array by iterating through each value in arr.
• Use two pointers left and right, initially set to 1 and 1000 respectively, to explore potential pairs.
• Use a while loop to consider pairs from the smallest and largest possible values inward. Terminate if left is greater than right.
• If the sum of left and right is greater than or equal to limit or if there are no occurrences of right in arr, decrement right.
• Otherwise, if the frequency of left is sufficient (considering if left equals right, at least 2 occurrences are needed), update bestSum with the maximum of bestSum and the current sum of left and right. Then, increment left.
• Continue this process until all possible pairs have been considered.
• Return bestSum, which holds the maximum sum of any pair with a sum less than limit found in the array.

This method is efficient as it avoids the naive O(n^2) complexity associated with checking all possible pairs directly, leveraging the counting sort principle and two-pointer strategy to locate the optimal pair.

class Solution {
    public int maxSumUnderK(int[] elements, int threshold) {
        int maxSum = -1;
        int[] frequency = new int[1001];
        for (int element : elements) {
            frequency[element]++;
        }
        int low = 1;
        int high = 1000;
        while (low <= high) {
            if (low + high >= threshold || frequency[high] == 0) {
                high--;
            } else {
                if (frequency[low] > (low < high ? 0 : 1)) {
                    maxSum = Math.max(maxSum, low + high);
                }
                low++;
            }
        }
        return maxSum;
    }
}

The code snippet provided is a Java method named maxSumUnderK designed to solve the problem of finding two distinct elements in an array whose sum is the maximum possible under a given threshold, threshold.

Here's how it works:

• An auxiliary array, frequency, is used to track the occurrences of each element in the input array elements.
• The variable maxSum initializes at -1 and will store the maximum sum that is less than threshold.
• The approach uses a two-pointer technique with low starting at 1 (smallest possible element value) and high at 1000 (largest possible for this logic), iterating and adjusting to find the element pairs.
• The conditions inside the loop manage the low and high pointers:
  • If low + high is not less than threshold or frequency[high] is zero (i.e., high is not in elements), decrement high.
  • Otherwise, if frequency[low] is greater than a certain condition (accounts for distinct elements criteria in a slightly optimized manner), update maxSum if the new pair sum is greater.
  • Then, increment low.
• Finally, the method returns maxSum, which at the end of method execution would hold the highest sum of any pair of numbers from the array that is less than threshold.

Important elements affecting the solution approach and optimizations include:

• Use of a frequency array for constant time access to determine the presence of a specific integer in the input (effective for bounded range of integers).
• Two-pointer technique, a common strategy for array problems, optimizing the search for valid pairs in linear time relative to the range of values.
• The conditions and updates within the loop ensure that only valid pairs (distinct and sum under threshold) are considered.

This algorithm is efficient for problems bound within a known range (1 to 1000 here) and leverages space-time tradeoffs (auxiliary space in exchange for faster checks) typical in competitive programming or constrained environments.

class Solution:
    def sumPairsBelowK(self, numbers: List[int], limitK: int) -> int:
        bestSum = -1
        numCount = [0] * 1001
        for number in numbers:
            numCount[number] += 1
        low = 1
        high = 1000
        while low <= high:
            if low + high >= limitK or numCount[high] == 0:
                high -= 1
            else:
                if numCount[low] > (0 if low != high else 1):
                    bestSum = max(bestSum, low + high)
                low += 1
        return bestSum

The provided Python program tackles the problem of finding the maximum summation of a pair of numbers from a given list, where the sum is less than a specified limit, ( K ). The solution employs a two-pointer technique to efficiently explore possible pairs within an array of numbers.

Below is an outline of the logic utilized in the solution:

• Initial Setup:

  • A bestSum variable is initialized to -1 to keep track of the highest sum found that is less than ( K ).
  • A list numCount of size 1001 is initialized to zero. This list is used to count occurrences of each number in the input list numbers.

• Counting Elements:

  • Iterate through each number in the numbers list, updating the numCount list to reflect the frequency of each number.

• Finding Pairs:

  • Two pointers, low starting at 1 and high starting at 1000, are used to explore pairs.
  • The while loop continues as long as low is less than or equal to high.
  • The condition inside the loop checks if the sum of low and high is less than ( K ) and if the current high number is present in the list (checked using numCount[high]).
  • If the sum is greater than or equal to ( K ) or if high is not in the list, decrement high.
  • Otherwise, check if there are enough occurrences of low (taking care to handle the case where low equals high) to form a pair and potentially update bestSum.
  • If conditions are met, increment low.

• Return Result:

  • After exiting the loop, bestSum contains the maximum sum of pairs that is less than ( K ) or remains at -1 if no such pair exists.

This approach efficiently finds the desired sum without needing to evaluate every possible pair directly, leveraging the frequency count and two-pointer strategy for a more optimal solution.