CP_Problem_9.md

Write the JAVA code for the below problem statement: You are given two positive integer arrays nums1 and nums2, both of length n. The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed). You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference. Return the minimum absolute sum difference after replacing at most one element in the array nums1. Since the answer may be large, return it modulo 10^9 + 7. |x| is defined as: x if x >= 0, or -x if x < 0. Example 1: Input: nums1 = [1,7,5], nums2 = [2,3,5] Output: 3 Explanation: There are two possible optimal solutions: - Replace the second element with the first: [1,7,5] => [1,1,5], or - Replace the second element with the third: [1,7,5] => [1,5,5]. Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3. Example 2: Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10] Output: 0 Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an absolute sum difference of 0. Example 3: Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4] Output: 20 Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7]. This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20 Constraints: n == nums1.length n == nums2.length 1 <= n <= 10^5 1 <= nums1[i], nums2[i] <= 10^5 Use the below template: class Solution { public int minAbsoluteSumDiff(int[] nums1, int[] nums2) { } }

Here's the Java code for the given problem statement using the template provided:

arduino

class Solution {
    public int minAbsoluteSumDiff(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int[] sortedNums1 = nums1.clone();
        Arrays.sort(sortedNums1);
        long res = 0, maxDiff = 0;
        for (int i = 0; i < n; i++) {
            int diff = Math.abs(nums1[i] - nums2[i]);
            res += diff;
            int j = Arrays.binarySearch(sortedNums1, nums2[i]);
            if (j < 0) j = -j - 1;
            if (j < n) maxDiff = Math.max(maxDiff, diff - Math.abs(sortedNums1[j] - nums2[i]));
            if (j > 0) maxDiff = Math.max(maxDiff, diff - Math.abs(sortedNums1[j-1] - nums2[i]));
        }
        return (int)((res - maxDiff) % 1000000007);
    }
}

Explanation:

We create a clone of the nums1 array and sort it in ascending order. Then, we iterate through both arrays and calculate the absolute difference between their corresponding elements. We add up all these differences to get the initial result res.

Next, for each element of nums2, we find its closest match in the sorted nums1 array using the binary search algorithm. We then calculate the maximum difference that can be reduced by replacing the corresponding element in nums1 with this closest match. We keep track of the maximum of all such differences in the maxDiff variable.

Finally, we subtract maxDiff from res and return the result modulo 10^9 + 7.

Note: We use long data type to prevent overflow while calculating res - maxDiff.