POTD Solutions | 02 Nov’ 23 | Minimum distance between two numbers

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Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Arrays but will also help you build up problem-solving skills.

POTD 02 November 2023

POTD 02 November: Minimum Distance between two numbers

You are given an array A[], of n elements. Find the minimum index based distance between two distinct elements of the array, x and y. Return -1, if either x or y does not exist in the array.

Example 1:
Input: N = 4, A[] = {1, 2, 3, 2}, x = 1, y = 2
Output: 1
Explanation: x = 1 and y = 2. There are two distances between x and y, which are 1 and 3 out of which the least is 1.

Example 2:
Input: N = 7, A[] = {86, 39, 90, 67, 84, 66, 62}, x = 42, y = 12
Output: –1
Explanation: x = 42 and y = 12. We return -1 as x and y don’t exist in the array.

Minimum Distance Between Two Numbers By Checking Consecutive (x, y) Pairs:

The basic approach is to check only consecutive pairs of x and y. For every element x or y, check the index of the previous occurrence of x or y and if the previous occurring element is not similar to current element update the minimum distance. But a question arises what if an x is preceded by another x and that is preceded by y, then how to get the minimum distance between pairs. By analyzing closely it can be seen that every x followed by y or vice versa can only be the closest pair (minimum distance) so ignore all other pairs.

Below is the implementation of the above approach:

C++

class Solution {
public:
    int minDist(int a[], int n, int x, int y)
    {
        // previous index and min distance
        int p = -1, min_dist = INT_MAX;

        for (int i = 0; i < n; i++) {
            if (a[i] == x || a[i] == y) {
                // we will check if p is not equal to -1 and
                // If the element at current index matches
                // with the element at index p , If yes then
                // update the minimum distance if needed
                if (p != -1 && a[i] != a[p])
                    min_dist = min(min_dist, i - p);

                // update the previous index
                p = i;
            }
        }
        // If distance is equal to int max
        if (min_dist == INT_MAX)
            return -1;

        return min_dist;
    }
};

Java

class Solution {
    int minDist(int a[], int n, int x, int y)
    {
        // previous index and min distance
        int i = 0, p = -1, min_dist = Integer.MAX_VALUE;

        for (i = 0; i < n; i++) {
            if (a[i] == x || a[i] == y) {
                // we will check if p is not equal to -1 and
                // If the element at current index matches
                // with the element at index p , If yes then
                // update the minimum distance if needed
                if (p != -1 && a[i] != a[p])
                    min_dist = Math.min(min_dist, i - p);

                // update the previous index
                p = i;
            }
        }
        // If distance is equal to int max
        if (min_dist == Integer.MAX_VALUE)
            return -1;
        return min_dist;
    }
}

Python3

class Solution:
    def minDist(self, arr, n, x, y):

        # previous index and min distance
        i = 0
        p = -1
        min_dist = 1e6

        for i in range(n):

            if(arr[i] == x or arr[i] == y):

                # we will check if p is not equal to -1 and
                # If the element at current index matches with
                # the element at index p , If yes then update
                # the minimum distance if needed
                if(p != -1 and arr[i] != arr[p]):
                    min_dist = min(min_dist, i-p)

                # update the previous index
                p = i

        # If distance is equal to int max
        if(min_dist == 1e6):
            return -1
        return min_dist

Javascript

class Solution {

    minDist(a, n, x, y)
    {
        // previous index and min distance
        var i=0,p=-1, min_dist=Number.MAX_VALUE;

        for(i=0 ; i<n ; i++)
        {
            if(a[i] ==x || a[i] == y)
            {
             // we will check if p is not equal to -1 and
            // If the element at current index matches with
            // the element at index p , If yes then update
            // the minimum distance if needed
            if(p != -1 && a[i] != a[p])
                min_dist = Math.min(min_dist,i-p);

            // update the previous index
                p=i;
            }
        }
        // If distance is equal to var max
        if(min_dist==Number.MAX_VALUE)
           return -1;
        return min_dist;
    }
}

Time Complexity: O(N), where N is the size of input array
Auxiliary Space: O(1)