Find the minimum distance between two numbers

2.4

Given an unsorted array arr[] and two numbers x and y, find the minimum distance between x and y in arr[]. The array might also contain duplicates. You may assume that both x and y are different and present in arr[].

Examples:
Input: arr[] = {1, 2}, x = 1, y = 2
Output: Minimum distance between 1 and 2 is 1.

Input: arr[] = {3, 4, 5}, x = 3, y = 5
Output: Minimum distance between 3 and 5 is 2.

Input: arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3}, x = 3, y = 6
Output: Minimum distance between 3 and 6 is 4.

Input: arr[] = {2, 5, 3, 5, 4, 4, 2, 3}, x = 3, y = 2
Output: Minimum distance between 3 and 2 is 1.

Method 1 (Simple)
Use two loops: The outer loop picks all the elements of arr[] one by one. The inner loop picks all the elements after the element picked by outer loop. If the elements picked by outer and inner loops have same values as x or y then if needed update the minimum distance calculated so far.

C

#include <stdio.h>
#include <stdlib.h> // for abs()
#include <limits.h> // for INT_MAX

int minDist(int arr[], int n, int x, int y)
{
   int i, j;
   int min_dist = INT_MAX;
   for (i = 0; i < n; i++)
   {
     for (j = i+1; j < n; j++)
     {
         if( (x == arr[i] && y == arr[j] ||
              y == arr[i] && x == arr[j]) && min_dist > abs(i-j))
         {
              min_dist = abs(i-j);
         }
     }
   }
   return min_dist;
}

/* Driver program to test above fnction */
int main() 
{
    int arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    int x = 3;
    int y = 6;

    printf("Minimum distance between %d and %d is %d\n", x, y, 
              minDist(arr, n, x, y));
    return 0;
}

Java

class MinimumDistance 
{
    int minDist(int arr[], int n, int x, int y) 
    {
        int i, j;
        int min_dist = Integer.MAX_VALUE;
        for (i = 0; i < n; i++) 
        {
            for (j = i + 1; j < n; j++) 
            {
                if ((x == arr[i] && y == arr[j]
                    || y == arr[i] && x == arr[j])
                    && min_dist > Math.abs(i - j)) 
                    min_dist = Math.abs(i - j);
            }
        }
        return min_dist;
    }

    public static void main(String[] args) 
    {
        MinimumDistance min = new MinimumDistance();
        int arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3};
        int n = arr.length;
        int x = 3;
        int y = 6;

        System.out.println("Minimum distance between " + x + " and " + y 
                + " is " + min.minDist(arr, n, x, y));
    }
}

Python3

# Python3 code to Find the minimum
# distance between two numbers

def minDist(arr, n, x, y):
    min_dist = 99999999
    for i in range(n):
        for j in range(i + 1, n):
            if (x == arr[i] and y == arr[j] or
            y == arr[i] and x == arr[j]) and min_dist > abs(i-j):
                min_dist = abs(i-j)
        return min_dist


# Driver code
arr = [3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3]
n = len(arr)
x = 3
y = 6
print("Minimum distance between ",x," and ",
     y,"is",minDist(arr, n, x, y))

# This code is contributed by "Abhishek Sharma 44"


Output:
Minimum distance between 3 and 6 is 4

Time Complexity: O(n^2)

Method 2 (Tricky)
1) Traverse array from left side and stop if either x or y is found. Store index of this first occurrence in a variable say prev
2) Now traverse arr[] after the index prev. If the element at current index i matches with either x or y then check if it is different from arr[prev]. If it is different then update the minimum distance if needed. If it is same then update prev i.e., make prev = i.

Thanks to wgpshashank for suggesting this approach.

C

#include <stdio.h>
#include <limits.h>  // For INT_MAX

int minDist(int arr[], int n, int x, int y)
{
   int i = 0;
   int min_dist = INT_MAX;
   int prev;

   // Find the first occurence of any of the two numbers (x or y)
   // and store the index of this occurence in prev
   for (i = 0; i < n; i++)
   {
     if (arr[i] == x || arr[i] == y)
     {
       prev = i;
       break;
     }
   }

   // Traverse after the first occurence
   for ( ; i < n; i++)
   {
      if (arr[i] == x || arr[i] == y)
      {
          // If the current element matches with any of the two then
          // check if current element and prev element are different
          // Also check if this value is smaller than minimm distance so far
          if ( arr[prev] != arr[i] && (i - prev) < min_dist )
          {
             min_dist = i - prev;
             prev = i;
          }
          else
             prev = i;
      }
   }

   return min_dist;
}

/* Driver program to test above fnction */
int main()
{
    int arr[] ={3, 5, 4, 2, 6, 3, 0, 0, 5, 4, 8, 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    int x = 3;
    int y = 6;

    printf("Minimum distance between %d and %d is %d\n", x, y,
              minDist(arr, n, x, y));
    return 0;
}

Java

class MinimumDistance
{
    int minDist(int arr[], int n, int x, int y) 
    {
        int i = 0;
        int min_dist = Integer.MAX_VALUE;
        int prev=0;

        // Find the first occurence of any of the two numbers (x or y)
        // and store the index of this occurence in prev
        for (i = 0; i < n; i++) 
        {
            if (arr[i] == x || arr[i] == y) 
            {
                prev = i;
                break;
            }
        }

        // Traverse after the first occurence
        for (; i < n; i++) 
        {
            if (arr[i] == x || arr[i] == y) 
            {
                // If the current element matches with any of the two then
                // check if current element and prev element are different
                // Also check if this value is smaller than minimum distance 
                // so far
                if (arr[prev] != arr[i] && (i - prev) < min_dist) 
                {
                    min_dist = i - prev;
                    prev = i;
                } 
                else
                    prev = i;
            }
        }

        return min_dist;
    }

    /* Driver program to test above functions */
    public static void main(String[] args) {
        MinimumDistance min = new MinimumDistance();
        int arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3};
        int n = arr.length;
        int x = 3;
        int y = 6;

        System.out.println("Minimum distance between " + x + " and " + y
                + " is " + min.minDist(arr, n, x, y));
    }
}


Output:
Minimum distance between 3 and 6 is 1

Time Complexity: O(n)

Asked in: Paytm

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