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)); } }

## Python3

import sys def minDist(arr, n, x, y): min_dist = sys.maxsize #Find the first occurence of any of the two numbers (x or y) # and store the index of this occurence in prev for i in range(n): if arr[i] == x or arr[i] == y: prev = i break # Traverse after the first occurence while i < n: if arr[i] == x or 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] and (i - prev) < min_dist : min_dist = i - prev prev = i else: prev = i i += 1 return min_dist # Driver program to test above fnction */ arr = [3, 5, 4, 2, 6, 3, 0, 0, 5, 4, 8, 3] n = len(arr) x = 3 y = 6 print ("Minimum distance between %d and %d is %d\n"%( x, y,minDist(arr, n, x, y))); # This code is contributed by Shreyanshi Arun.

Output:

Minimum distance between 3 and 6 is 1

Time Complexity: O(n)

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Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.