From f2c22c74e61b8f8621401bccaa4f78fde13b8a8b Mon Sep 17 00:00:00 2001
From: Hao Chen <haoel@hotmail.com>
Date: Sun, 9 May 2021 13:22:03 +0800
Subject: [PATCH] New Problem Solution - "1855. Maximum Distance Between a Pair
 of Values"

---
 README.md                                     |  1 +
 .../MaximumDistanceBetweenAPairOfValues.cpp   | 87 +++++++++++++++++++
 2 files changed, 88 insertions(+)
 create mode 100644 algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp

diff --git a/README.md b/README.md
index 25c6cacb..9e77903a 100644
--- a/README.md
+++ b/README.md
@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1855|[Maximum Distance Between a Pair of Values](https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/) | [C++](./algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp)|Medium|
 |1854|[Maximum Population Year](https://leetcode.com/problems/maximum-population-year/) | [C++](./algorithms/cpp/maximumPopulationYear/MaximumPopulationYear.cpp)|Easy|
 |1851|[Minimum Interval to Include Each Query](https://leetcode.com/problems/minimum-interval-to-include-each-query/) | [C++](./algorithms/cpp/minimumIntervalToIncludeEachQuery/MinimumIntervalToIncludeEachQuery.cpp)|Hard|
 |1850|[Minimum Adjacent Swaps to Reach the Kth Smallest Number](https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/) | [C++](./algorithms/cpp/minimumAdjacentSwapsToReachTheKthSmallestNumber/MinimumAdjacentSwapsToReachTheKthSmallestNumber.cpp)|Medium|
diff --git a/algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp b/algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp
new file mode 100644
index 00000000..32891188
--- /dev/null
+++ b/algorithms/cpp/maximumDistanceBetweenAPairOfValues/MaximumDistanceBetweenAPairOfValues.cpp
@@ -0,0 +1,87 @@
+// Source : https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
+// Author : Hao Chen
+// Date   : 2021-05-09
+
+/***************************************************************************************************** 
+ *
+ * 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.
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    int maxDistance(vector<int>& nums1, vector<int>& nums2) {
+        return maxDistance2(nums1, nums2);
+        return maxDistance1(nums1, nums2);
+    }
+    
+    
+    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;
+    }
+
+    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;
+    }
+    
+    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;
+    }
+};