80 lines
2.5 KiB
C++
80 lines
2.5 KiB
C++
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// Source : https://leetcode.com/problems/minimum-absolute-sum-difference/
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// Author : Hao Chen
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// Date : 2021-04-05
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/*****************************************************************************************************
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*
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* You are given two positive integer arrays nums1 and nums2, both of length n.
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*
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* The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] -
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* nums2[i]| for each 0 <= i < n (0-indexed).
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*
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* You can replace at most one element of nums1 with any other element in nums1 to minimize the
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* absolute sum difference.
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*
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* Return the minimum absolute sum difference after replacing at most one element in the array nums1.
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* Since the answer may be large, return it modulo 10^9 + 7.
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*
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* |x| is defined as:
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*
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* x if x >= 0, or
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* -x if x < 0.
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*
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* Example 1:
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*
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* Input: nums1 = [1,7,5], nums2 = [2,3,5]
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* Output: 3
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* Explanation: There are two possible optimal solutions:
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* - Replace the second element with the first: [1,7,5] => [1,1,5], or
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* - Replace the second element with the third: [1,7,5] => [1,5,5].
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* Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3.
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*
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* Example 2:
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*
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* Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
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* Output: 0
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* Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an
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* absolute sum difference of 0.
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*
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* Example 3:
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*
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* Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
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* Output: 20
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* Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
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* This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20
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*
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* Constraints:
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*
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* n == nums1.length
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* n == nums2.length
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* 1 <= n <= 10^5
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* 1 <= nums1[i], nums2[i] <= 10^5
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******************************************************************************************************/
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class Solution {
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public:
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int minAbsoluteSumDiff(vector<int>& nums1, vector<int>& nums2) {
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int max=0, idx=0;
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long sum=0;
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int len = nums1.size();
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for (int i=0; i<len; i++) {
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int m = abs(nums1[i] - nums2[i]);
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if (max < m) {
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max = m;
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idx = i;
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}
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sum += m ;
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}
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int min = max;
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for (int i=0; i<len; i++) {
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int m = abs(nums1[i] - nums2[idx]);
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if (m < min) {
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min = m;
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}
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}
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sum -= (max - min);
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return sum % 1000000007 ;
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}
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};
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