Find First Non-Repeating Element in Array Using Hash Map

Given an array of integers, find the first element that appears exactly once. If no such element exists, return -1.

Difficulty:

JavaScript Solution

function firstNonRepeating(arr) {
  const map = {};
  for (let num of arr) map[num] = (map[num] || 0) + 1;
  for (let num of arr) if (map[num] === 1) return num;
  return -1;
}

Explanation: Count frequency of each element using a hash map. Then iterate to find the first element with frequency 1.

⏱ O(n) | 💾 O(n)

Java Solution

class Solution {
  public int firstNonRepeating(int[] arr) {
    Map<Integer, Integer> map = new LinkedHashMap<>();
    for (int num : arr) map.put(num, map.getOrDefault(num, 0) + 1);
    for (int num : map.keySet()) if (map.get(num) == 1) return num;
    return -1;
  }
}

Explanation: LinkedHashMap preserves insertion order to find the first non-repeating element efficiently.

⏱ O(n) | 💾 O(n)

Python Solution

def first_non_repeating(arr):
    from collections import OrderedDict
    count = OrderedDict()
    for num in arr: count[num] = count.get(num,0)+1
    for num in count: 
        if count[num]==1: return num
    return -1

Explanation: Use OrderedDict to maintain insertion order while counting elements. The first key with count 1 is the answer.

⏱ O(n) | 💾 O(n)

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Find First Non-Repeating Element in Array Using Hash Map