GFG Problem Of The Day

Today - 17 May 2024

Que - Find Pair Given Difference

The problem can be found at the following link: Question Link

My Approach

• The input array arr is sorted to facilitate the two-pointer technique.
• Use two pointers low and high. Initialize low to 0 and high to 1.
• Use a while loop to traverse the array with the condition low < n and high < n.
• If the absolute difference between the elements at high and low is equal to x, return 1 (indicating a pair is found).
• If the absolute difference is less than x, increment the high pointer to increase the difference.
• If the absolute difference is greater than x, increment the low pointer to decrease the difference.
• If no such pair is found by the end of the loop, return -1.

Time and Auxiliary Space Complexity

• Time Complexity: The time complexity of this approach is O(NlogN), where N is the number of elements in the array.
• Auxiliary Space Complexity: The auxiliary space complexity is O(1).

Code (C++)

class Solution {
  public:
    int findPair(int n, int x, vector<int> &arr)
    {
        sort(arr.begin(), arr.end());
        int low=0, high=1;
        while (low<n && high<n)
        {
            if (abs(arr[high]-arr[low])==x)
                return 1;
            else if (abs(arr[high]-arr[low])<x)
                high++;
            else low++;
        }
        return -1;
    }
};

Contribution and Support

I always encourage contributors to participate in the discussion forum for this repository.

If you have a better solution or any queries / discussions related to the Problem of the Day solution, please visit our discussion section. We welcome your input and aim to foster a collaborative learning environment.

If you find this solution helpful, consider supporting us by giving a ⭐ star to the getlost01/gfg-potd repository.