New Problem Solution -"1818. Minimum Absolute Sum Difference"

Author: haoel

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1818|[Minimum Absolute Sum Difference](https://leetcode.com/problems/minimum-absolute-sum-difference/) | [C++](./algorithms/cpp/minimumAbsoluteSumDifference/MinimumAbsoluteSumDifference.cpp)|Medium|
 |1817|[Finding the Users Active Minutes](https://leetcode.com/problems/finding-the-users-active-minutes/) | [C++](./algorithms/cpp/findingTheUsersActiveMinutes/FindingTheUsersActiveMinutes.cpp)|Medium|
 |1816|[Truncate Sentence](https://leetcode.com/problems/truncate-sentence/) | [C++](./algorithms/cpp/truncateSentence/TruncateSentence.cpp)|Easy|
 |1808|[Maximize Number of Nice Divisors](https://leetcode.com/problems/maximize-number-of-nice-divisors/) | [C++](./algorithms/cpp/maximizeNumberOfNiceDivisors/MaximizeNumberOfNiceDivisors.cpp)|Hard|

algorithms/cpp/minimumAbsoluteSumDifference/MinimumAbsoluteSumDifference.cpp

@@ -0,0 +1,79 @@
+// 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 ;
+    }
+};