MinimumAbsoluteSumDifference.cpp

// Source : https://leetcode.com/problems/minimum-absolute-sum-difference/
// Author : Hao Chen
// Date   : 2021-04-05

/***************************************************************************************************** 
 *
 * 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
 ******************************************************************************************************/

class Solution {
public:
    int minAbsoluteSumDiff(vector<int>& nums1, vector<int>& nums2) {
        int max=0, idx=0;
        long sum=0;
        int len = nums1.size();
        for (int i=0; i<len; i++) {
            int m = abs(nums1[i] - nums2[i]);
            if (max < m) {
                max = m; 
                idx = i;
            }
            sum += m ;
        }
        int min = max;
        for (int i=0; i<len; i++) {
            int m = abs(nums1[i] - nums2[idx]);
            if (m < min) {
                min = m;
            }
        }
        
        sum -= (max - min);
        return sum % 1000000007 ;
    }
};