chore: add LeetCode daily solution

Author: TechnoBlogger14o3

Easy/2026-01-02-961-N-Repeated-Element-in-Size-2N-Array/explanation.md

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+# N-Repeated Element in Size 2N Array (Easy)
+
+**Problem ID:** 961  
+**Date:** 2026-01-02  
+**Link:** https://leetcode.com/problems/n-repeated-element-in-size-2n-array/
+
+## Approach
+
+To solve the problem of finding the N-repeated element in a size 2N array, we can utilize a straightforward approach leveraging a set or a dictionary to track occurrences of elements.
+
+### Approach:
+
+1. **Understanding the Problem**: The array `nums` has a length of `2N` and contains `N + 1` unique elements, with exactly one element appearing `N` times. Our goal is to identify this repeated element.
+
+2. **Data Structure**: We can use a dictionary (or a hash map) to count occurrences of each element as we iterate through the array. This allows us to efficiently check how many times each element appears.
+
+3. **Iterating through the Array**: As we traverse the `nums` array:
+   - For each element, we check if it already exists in our dictionary.
+   - If it does, we increment its count.
+   - If it doesn't, we add it to the dictionary with a count of 1.
+
+4. **Finding the N-Repeated Element**: After populating the dictionary, we can simply iterate through its entries to find the element whose count equals `N`.
+
+### Complexity:
+- **Time Complexity**: O(N), where N is the number of unique elements (which is at most 5000). This is because we make a single pass through the array to populate the dictionary and potentially another pass through the dictionary to find the repeated element.
+- **Space Complexity**: O(N) in the worst case, as we may store up to `N + 1` unique elements in our dictionary.
+
+This approach is efficient and straightforward, leveraging the properties of hash maps for quick lookups and insertions, making it well-suited for this problem.

Easy/2026-01-02-961-N-Repeated-Element-in-Size-2N-Array/solution.java

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+class Solution {
+    public int repeatedNTimes(int[] nums) {
+        Set<Integer> seen = new HashSet<>();
+        for (int num : nums) {
+            if (seen.contains(num)) {
+                return num;
+            }
+            seen.add(num);
+        }
+        return -1; // This line will never be reached due to problem constraints
+    }
+}

Easy/2026-01-02-961-N-Repeated-Element-in-Size-2N-Array/solution.js

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+var repeatedNTimes = function(nums) {
+    const count = {};
+    for (const num of nums) {
+        if (count[num]) {
+            return num;
+        }
+        count[num] = 1;
+    }
+};

Easy/2026-01-02-961-N-Repeated-Element-in-Size-2N-Array/solution.py

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+class Solution:
+    def repeatedNTimes(self, nums: List[int]) -> int:
+        count = {}
+        n = len(nums) // 2
+        for num in nums:
+            if num in count:
+                count[num] += 1
+                if count[num] == n:
+                    return num
+            else:
+                count[num] = 1

README.md

@@ -323,3 +323,4 @@ Through completing the Blind 75 and NeetCode 150, you will have mastered:
 - 2025-12-30 — [Magic Squares In Grid](https://leetcode.com/problems/magic-squares-in-grid/) (Medium) → `Medium/2025-12-30-840-Magic-Squares-In-Grid`
 - 2025-12-31 — [Last Day Where You Can Still Cross](https://leetcode.com/problems/last-day-where-you-can-still-cross/) (Hard) → `Hard/2025-12-31-1970-Last-Day-Where-You-Can-Still-Cross`
 - 2026-01-01 — [Plus One](https://leetcode.com/problems/plus-one/) (Easy) → `Easy/2026-01-01-66-Plus-One`
+- 2026-01-02 — [N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/) (Easy) → `Easy/2026-01-02-961-N-Repeated-Element-in-Size-2N-Array`