bonus32Logic.md

This code is an implementation of a function closest3Sum that returns the sum of three numbers in an array that is closest to a given target value. The function leverages a combination of sorting and the two-pointer technique to efficiently find the closest sum. Let’s break it down step by step.

Java Code:

static int closest3Sum(int[] arr, int target) {
    // Sort the array to apply the two-pointer technique
    Arrays.sort(arr);
    
    // Initialize variables to track the closest sum and the minimum difference
    int closestSum = Integer.MIN_VALUE;
    int minDiff = Integer.MAX_VALUE;

    // Outer loop to iterate over each element in the array (up to arr.length-2)
    for (int i = 0; i < arr.length - 2; i++) {
        // Set up two pointers: one at i+1 (left) and one at the last element (right)
        int left = i + 1;
        int right = arr.length - 1;

        // While left pointer is less than right pointer, check possible sums
        while (left < right) {
            // Calculate the current sum of the triplet
            int sum = arr[i] + arr[left] + arr[right];
            
            // Calculate the absolute difference between the current sum and the target
            int diff = Math.abs(sum - target);

            // If the current difference is smaller, or if it's the same but the sum is larger
            // (to prefer a larger sum when distances are equal), update closestSum and minDiff
            if (diff < minDiff || (diff == minDiff && sum > closestSum)) {
                closestSum = sum;
                minDiff = diff;
            }

            // If the current sum is less than the target, increment the left pointer to increase the sum
            if (sum < target) {
                left++;
            }
            // If the current sum is greater than the target, decrement the right pointer to decrease the sum
            else {
                right--;
            }
        }
    }

    // Return the closest sum found
    return closestSum;
}

Step-by-Step Explanation:

1. Sorting the Array:

   Arrays.sort(arr);

   • First, the array arr is sorted in ascending order. Sorting is necessary because it enables the two-pointer approach to efficiently narrow down possible sums.

2. Initial Variables:

   int closestSum = Integer.MIN_VALUE;
   int minDiff = Integer.MAX_VALUE;

   • closestSum is initialized to the smallest possible integer value (Integer.MIN_VALUE) to ensure that any valid sum will be larger initially.
   • minDiff is initialized to the largest possible integer value (Integer.MAX_VALUE) to ensure that any valid difference will be smaller initially.

3. Outer Loop:

   for (int i = 0; i < arr.length - 2; i++) {

   • This loop iterates over each element in the array (arr[i]) as the first element of the triplet. Since we are looking for triplets, the loop stops at arr.length - 2 to ensure there are at least two more elements available for the triplet.

4. Setting up Two Pointers:

   int left = i + 1;
   int right = arr.length - 1;

   • For each i, two pointers are initialized:
     • left is set to the element right after i (i + 1).
     • right is set to the last element of the array (arr.length - 1).
   • These two pointers help form the other two elements of the triplet (arr[left] and arr[right]).

5. Inner While Loop:

   while (left < right) {
       int sum = arr[i] + arr[left] + arr[right];

   • The while loop continues as long as left is less than right.
   • The sum of the triplet arr[i] + arr[left] + arr[right] is calculated.

6. Calculating the Difference:

   int diff = Math.abs(sum - target);

   • diff stores the absolute difference between the current sum and the target value. This helps track how close the current sum is to the target.

7. Updating the Closest Sum:

   if (diff < minDiff || (diff == minDiff && sum > closestSum)) {
       closestSum = sum;
       minDiff = diff;
   }

   • If the current diff is smaller than the previously recorded minDiff, it means we've found a closer sum to the target, so we update closestSum and minDiff.
   • If the diff is the same (i.e., the current sum is equally close to the target), we choose the larger sum, which is specified by the condition (diff == minDiff && sum > closestSum).

8. Adjusting the Pointers:

   if (sum < target) {
       left++;
   } else {
       right--;
   }

   • If the sum is less than the target, we want to increase the sum by moving the left pointer to the right (left++).
   • If the sum is greater than the target, we want to decrease the sum by moving the right pointer to the left (right--).

9. Return the Closest Sum:

   return closestSum;

   • After completing the loops, the function returns the closestSum, which is the sum of the triplet that is closest to the target.

Example Walkthrough:

Let’s take an example to understand how this works.

Example 1:

• Input:

  • arr = [-1, 2, 1, -4]
  • target = 1

• Step 1: Sort the array: [-4, -1, 1, 2]

• Step 2: Start the outer loop:

  • For i = 0 (arr[i] = -4):
    • left = 1, right = 3
    • Sum of triplet: -4 + (-1) + 2 = -3
    • diff = Math.abs(-3 - 1) = 4
    • Update closestSum = -3, minDiff = 4
    • Since the sum is less than the target, increment left.
  • For i = 1 (arr[i] = -1):
    • left = 2, right = 3
    • Sum of triplet: -1 + 1 + 2 = 2
    • diff = Math.abs(2 - 1) = 1
    • Update closestSum = 2, minDiff = 1
    • Since the sum is greater than the target, decrement right.
  • For i = 2 (arr[i] = 1):
    • The left pointer would be beyond the right pointer, so the loop ends.

• Step 3: Return closestSum = 2.

Example 2:

• Input:

  • arr = [1, 1, 1]
  • target = 3

• Output: The closest sum is 3.

Time Complexity:

• Sorting the array takes O(n log n).
• The two-pointer approach for each element takes O(n), where n is the length of the array.
• Therefore, the overall time complexity is O(n^2).

Space Complexity:

• The space complexity is O(1) because no additional data structures are used apart from the input array.

Conclusion:

This code efficiently finds the sum of three elements in an array that is closest to a given target by using sorting and the two-pointer technique, providing an optimal solution with a time complexity of O(n^2).