Find Unique Number

Summary

This problem involves finding a unique number in an array where all numbers except one are repeated. The goal is to identify this unique number efficiently using algorithms. This problem requires attention to detail and understanding of basic algorithmic concepts.

Detailed Explanation

The given array contains multiple occurrences of each element, except for one, which appears only once. Our task is to find this unique number in the array. A simple approach would be to iterate through the array, keeping a count of each number's occurrence. However, this solution has a time complexity of O(n), where n is the size of the array.

A more efficient approach involves using the concept of XOR (exclusive OR) operations. The idea is that when we XOR all elements in the array, the unique element will remain as it doesn't have a pair to cancel out with. We can then XOR all elements again to get the unique number.

Here's the step-by-step breakdown:

Initialize a variable result to 0.
Iterate through the array and for each element:
- XOR the current element with result.
The final value of result will be the unique number in the array.

This algorithm has a time complexity of O(n) and a space complexity of O(1), as it only requires a single variable to store the result.

Optimized Solutions

Java

View GeeksforGeeks Problem

public int findUniqueNumber(int[] arr) {
    int result = 0;
    for (int num : arr) {
        result ^= num;
    }
    return result;
}

Code Editor (Testing phase)

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893307. Unique Number I