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New Problem Solution - "1855. Maximum Distance Between a Pair of Values"
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algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp
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// Source : https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/ | ||
// Author : Hao Chen | ||
// Date : 2021-05-09 | ||
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/***************************************************************************************************** | ||
* | ||
* You are given two non-increasing 0-indexed integer arrays nums1 and nums2. | ||
* | ||
* A pair of indices (i, j), where 0 <= i < nums1.length and 0 <= j < nums2.length, is valid if both i | ||
* <= j and nums1[i] <= nums2[j]. The distance of the pair is j - i. | ||
* | ||
* Return the maximum distance of any valid pair (i, j). If there are no valid pairs, return 0. | ||
* | ||
* An array arr is non-increasing if arr[i-1] >= arr[i] for every 1 <= i < arr.length. | ||
* | ||
* Example 1: | ||
* | ||
* Input: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5] | ||
* Output: 2 | ||
* Explanation: The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4). | ||
* The maximum distance is 2 with pair (2,4). | ||
* | ||
* Example 2: | ||
* | ||
* Input: nums1 = [2,2,2], nums2 = [10,10,1] | ||
* Output: 1 | ||
* Explanation: The valid pairs are (0,0), (0,1), and (1,1). | ||
* The maximum distance is 1 with pair (0,1). | ||
* | ||
* Example 3: | ||
* | ||
* Input: nums1 = [30,29,19,5], nums2 = [25,25,25,25,25] | ||
* Output: 2 | ||
* Explanation: The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4). | ||
* The maximum distance is 2 with pair (2,4). | ||
* | ||
* Example 4: | ||
* | ||
* Input: nums1 = [5,4], nums2 = [3,2] | ||
* Output: 0 | ||
* Explanation: There are no valid pairs, so return 0. | ||
* | ||
* Constraints: | ||
* | ||
* 1 <= nums1.length <= 10^5 | ||
* 1 <= nums2.length <= 10^5 | ||
* 1 <= nums1[i], nums2[j] <= 10^5 | ||
* Both nums1 and nums2 are non-increasing. | ||
******************************************************************************************************/ | ||
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class Solution { | ||
public: | ||
int maxDistance(vector<int>& nums1, vector<int>& nums2) { | ||
return maxDistance2(nums1, nums2); | ||
return maxDistance1(nums1, nums2); | ||
} | ||
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int binary_search(vector<int>& nums, int start, int target) { | ||
int end = nums.size() - 1; | ||
while (start <= end) { | ||
int mid = start + (end - start) /2; | ||
if(nums[mid] < target) end = mid - 1; | ||
else start = mid+1; | ||
} | ||
return end; | ||
} | ||
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int maxDistance1(vector<int>& nums1, vector<int>& nums2) { | ||
int mDist=0; | ||
int right = nums2.size() - 1; | ||
for(int i=0; i<nums1.size(); i++) { | ||
int j = binary_search(nums2, i, nums1[i]); | ||
mDist = max(mDist, j-i); | ||
} | ||
return mDist; | ||
} | ||
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int maxDistance2(vector<int>& nums1, vector<int>& nums2) { | ||
int i=0, j=0, dist = 0; | ||
while (i < nums1.size() && j < nums2.size() ){ | ||
if ( nums1[i] > nums2[j] ) i++; | ||
else dist = max(dist, j++ - i); | ||
} | ||
return dist; | ||
} | ||
}; |