chore: add LeetCode daily solution

Author: TechnoBlogger14o3

Medium/2026-04-19-1855-Maximum-Distance-Between-a-Pair-of-Values/explanation.md

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+# Maximum Distance Between a Pair of Values (Medium)
+
+**Problem ID:** 1855  
+**Date:** 2026-04-19  
+**Link:** https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
+
+## Approach
+
+To solve the problem of finding the maximum distance between a pair of values from two non-increasing arrays, we can employ a two-pointer technique. Here’s a concise breakdown of the approach:
+
+### Main Idea:
+The goal is to find indices \(i\) and \(j\) such that \(0 \leq i < \text{nums1.length}\), \(0 \leq j < \text{nums2.length}\), \(i \leq j\), and \(\text{nums1}[i] \leq \text{nums2}[j]\). The distance is defined as \(j - i\), and we want to maximize this distance.
+
+### Approach:
+1. **Initialization**: Start with two pointers, one for each array. Let \(i\) begin at the start of `nums1` and \(j\) at the start of `nums2`.
+
+2. **Iterate with Two Pointers**:
+   - Use a while loop to iterate through the arrays. For each index \(i\) in `nums1`, increment \(j\) in `nums2` as long as the condition \(\text{nums1}[i] \leq \text{nums2}[j]\) holds true. This is possible because both arrays are non-increasing, meaning if \(\text{nums1}[i] \leq \text{nums2}[j]\), then \(\text{nums1}[i] \leq \text{nums2}[j+1]\) will also hold until \(j\) reaches the end of `nums2`.
+
+3. **Calculate Distance**: For every valid pair found (where \(\text{nums1}[i] \leq \text{nums2}[j]\)), calculate the distance \(j - i\) and update the maximum distance found.
+
+4. **Increment \(i\)**: After processing all valid \(j\) for a given \(i\), increment \(i\) to check for the next element in `nums1`.
+
+5. **Termination**: The process continues until all elements in `nums1` have been considered or \(j\) has reached the end of `nums2`.
+
+### Data Structures:
+- Two pointers (integers) for indexing `nums1` and `nums2`.
+
+### Complexity:
+- **Time Complexity**: \(O(n + m)\), where \(n\) is the length of `nums1` and \(m\) is the length of `nums2`. Each pointer traverses its respective array at most once.
+- **Space Complexity**: \(O(1)\), as we are using a constant amount of extra space.
+
+This efficient approach ensures we can handle the upper limits of the input constraints effectively, yielding the maximum distance between valid pairs in linear time.

Medium/2026-04-19-1855-Maximum-Distance-Between-a-Pair-of-Values/solution.java

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+class Solution {
+    public int maxDistance(int[] nums1, int[] nums2) {
+        int maxDist = 0;
+        int j = 0;
+        for (int i = 0; i < nums1.length; i++) {
+            while (j < nums2.length && nums1[i] <= nums2[j]) {
+                j++;
+            }
+            maxDist = Math.max(maxDist, j - 1 - i);
+        }
+        return maxDist;
+    }
+}

Medium/2026-04-19-1855-Maximum-Distance-Between-a-Pair-of-Values/solution.js

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+var maxDistance = function(nums1, nums2) {
+    let maxDist = 0;
+    let j = 0;
+    
+    for (let i = 0; i < nums1.length; i++) {
+        while (j < nums2.length && nums1[i] <= nums2[j]) {
+            j++;
+        }
+        maxDist = Math.max(maxDist, j - 1 - i);
+    }
+    
+    return maxDist;
+};

Medium/2026-04-19-1855-Maximum-Distance-Between-a-Pair-of-Values/solution.py

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+class Solution:
+    def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
+        max_distance = 0
+        j = 0
+        n1, n2 = len(nums1), len(nums2)
+
+        for i in range(n1):
+            while j < n2 and nums1[i] <= nums2[j]:
+                j += 1
+            max_distance = max(max_distance, j - 1 - i)
+
+        return max_distance

README.md

@@ -428,3 +428,4 @@ Through completing the Blind 75 and NeetCode 150, you will have mastered:
 - 2026-04-16 — [Closest Equal Element Queries](https://leetcode.com/problems/closest-equal-element-queries/) (Medium) → `Medium/2026-04-16-3488-Closest-Equal-Element-Queries`
 - 2026-04-17 — [Minimum Absolute Distance Between Mirror Pairs](https://leetcode.com/problems/minimum-absolute-distance-between-mirror-pairs/) (Medium) → `Medium/2026-04-17-3761-Minimum-Absolute-Distance-Between-Mirror-Pairs`
 - 2026-04-18 — [Mirror Distance of an Integer](https://leetcode.com/problems/mirror-distance-of-an-integer/) (Easy) → `Easy/2026-04-18-3783-Mirror-Distance-of-an-Integer`
+- 2026-04-19 — [Maximum Distance Between a Pair of Values](https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/) (Medium) → `Medium/2026-04-19-1855-Maximum-Distance-Between-a-Pair-of-Values`