> For the complete documentation index, see [llms.txt](https://gl01.gitbook.io/gfg-editorials/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://gl01.gitbook.io/gfg-editorials/2023/11-2023-nov-1/02-minimum-distance-between-two-numbers.md).

# 02. Minimum Distance Between Two Numbers

The problem can be found at the following link: [Question Link](https://www.geeksforgeeks.org/problems/minimum-distance-between-two-numbers/1)

![](https://badgen.net/badge/Level/Easy/green)

## My Approach

A simple and straightforward question, to find the minimum distance between two numbers, x and y, in an array,

* I traverse the array once. I maintain two variables, `lastX` and `lastY`, which represent the most recent positions of x and y in the array.
* I also maintain a variable `out` to store the minimum distance found so far. As I traverse the array, I update `lastX` and `lastY` whenever I encounter x or y. If both `lastX` and `lastY` have been updated, I calculate the distance and update `out` accordingly.
* Finally, I return `out`, which represents the minimum distance between x and y. If `out` remains equal to `INT_MAX`, it means one of the numbers was not found in the array, and I return -1.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(n)`, where `n` is the number of elements in the array.
* **Auxiliary Space Complexity**: `O(1)`, as we are using a constant amount of additional space.

## Code (C++)

```cpp
class Solution {
public:
    int minDist(int a[], int n, int x, int y) {
        int lastX, lastY, out;
        lastX = lastY = -1;
        out = INT_MAX;
        
        for(int i = 0; i < n; ++i) {
            if(a[i] == x)
                lastX = i;
            if(a[i] == y)
                lastY = i;
            
            if(lastX != -1 && lastY != -1)
                out = min(out, abs(lastX - lastY));
        }
        return out == INT_MAX ? -1 : out;
    }
};
```

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