LeetCode Challenge Day 109 — 961. N-Repeated Element in Size 2N Array

Hey folks 👋

This is Day 109 of my LeetCode streak 🚀
Today's problem is 961. N-Repeated Element in Size 2N Array — a deceptively simple problem that rewards logical observation over brute force.

📌 Problem Statement

You are given:

• An integer array nums of length 2n
• Exactly n + 1 unique elements
• One element is repeated exactly n times

Goal:
Return the element that is repeated n times.

💡 Intuition

Since the array has length 2n and one element appears n times, that element is everywhere in the array.

This means:

• The repeated element cannot be spaced far apart everywhere
• In any small window of consecutive elements, the repeated element must appear at least twice

So instead of counting frequencies, we can detect repetition by simply comparing nearby elements.

🔑 Approach

1. Traverse the array from left to right
2. For each index i, compare nums[i] with:
   • nums[i + 1]
   • nums[i + 2]
   • nums[i + 3]
3. If any match is found, return nums[i]

Why this works:

• With one element repeated n times in a 2n array, at least two occurrences must fall within any window of size 4
• This guarantees an early detection without extra space

⏱️ Complexity Analysis

• Time Complexity:
  O(n) — single traversal of the array

• Space Complexity:
  O(1) — no additional data structures used

🧑‍💻 Code (JavaScript)

/**
 * @param {number[]} nums
 * @return {number}
 */
var repeatedNTimes = function(nums) {
    for (let i = 0; i < nums.length - 1; i++) {
        if (
            nums[i] === nums[i + 1] ||
            nums[i] === nums[i + 2] ||
            nums[i] === nums[i + 3]
        ) {
            return nums[i];
        }
    }
};

🎯 Reflection

This problem is a great reminder that:

• Constraints often hide the real solution

• Not every problem needs a hash map

• Observation can reduce both time and space complexity

That wraps up Day 109 of my LeetCode challenge 🔥
Onward to Day 110 — showing up daily beats solving everything at once 🚀

Happy Coding 👨‍💻