# Find the minimum distance between two numbers

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.
Explanation: 1 is at index 0 and 2 is at
index 1, so the distance is 1

Input: arr[] = {3, 4, 5}, x = 3, y = 5
Output: Minimum distance between 3
and 5 is 2.
Explanation:3 is at index 0 and 5 is at
index 2, so the distance 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.
Explanation:3 is at index 0 and 6 is at
index 5, so the distance 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.
Explanation:3 is at index 7 and 2 is at
index 6, so the distance is 1
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1:

• Approach: The task is to find the distance between two given numbers, So find the distance between any two elements using nested loops. The outer loop for selecting the first element (x) and the inner loop for traversing the array in search for the other element (y) and taking the minimum distance between them.
• Algorithm:
1. Create a variable m = INT_MAX
2. Run a nested loop, the outer loop runs from start to end (loop counter i), the inner loop runs from i+1 to end (loop counter j).
3. If the ith element is x and jth element is y or vice versa, update m as m = min(m,j-i)
4. Print the value of m as minimum distance
• Implementation:

## C++

 `// C++ program to Find the minimum ` `// distance between two numbers ` `#include ` `using` `namespace` `std;  ` ` `  `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 code */` `int` `main()  ` `{ ` `    ``int` `arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``int` `x = 3; ` `    ``int` `y = 6; ` ` `  `    ``cout << ``"Minimum distance between "` `<< x <<  ` `                    ``" and "` `<< y << ``" is "` `<<  ` `                    ``minDist(arr, n, x, y) << endl; ` `} ` ` `  `// This code is contributed by Shivi_Aggarwal `

## C

 `// C program to Find the minimum ` `// distance between two numbers ` `#include ` `#include // for abs() ` `#include // 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 function */` `int` `main()  ` `{ ` `    ``int` `arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``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

 `// Java Program to Find the minimum ` `// distance between two numbers ` `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" `

## C#

 `// C# code to Find the minimum ` `// distance between two numbers ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `minDist(``int` `[]arr, ``int` `n, ` `                           ``int` `x, ``int` `y)  ` `    ``{ ` `        ``int` `i, j; ` `        ``int` `min_dist = ``int``.MaxValue; ` `        ``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; ` `    ``} ` `     `  `    ``// Driver function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = {3, 5, 4, 2, 6, ` `              ``5, 6, 6, 5, 4, 8, 3}; ` `        ``int` `n = arr.Length; ` `        ``int` `x = 3; ` `        ``int` `y = 6; ` ` `  `        ``Console.WriteLine(``"Minimum "` `               ``+ ``"distance between "` `         ``+ x +  ``" and "` `+ y + ``" is "`  `           ``+ minDist(arr, n, x, y)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` ``abs``(``\$i` `- ``\$j``)) ` `            ``{ ` `                ``\$min_dist` `= ``abs``(``\$i` `- ``\$j``); ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `\$min_dist``; ` `} ` ` `  `    ``// Driver Code ` `    ``\$arr` `= ``array``(3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3); ` `    ``\$n` `= ``count``(``\$arr``); ` `    ``\$x` `= 3; ` `    ``\$y` `= 6; ` ` `  `    ``echo` `"Minimum distance between "``,``\$x``, ``" and "``,``\$y``,``" is "``;  ` `    ``echo` `minDist(``\$arr``, ``\$n``, ``\$x``, ``\$y``); ` ` `  `// This code is contributed by anuj_67. ` `?> `

Output:

`Minimum distance between 3 and 6 is 4`
• Complexity Analysis:

• Time Complexity: O(n^2), Nested loop is used to traverse the array.
• Space Complexity: O(1), no extra space is required.

Method 2:

• Approach: So 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 a y, then how to get the minimum distance between pairs. By analyzing closely it can be seen that every x followed by a y or vice versa can only be the closest pair (minimum distance) so ignore all other pairs.
• Algorithm:

1. Create a variable prev=-1 and m= INT_MAX
2. Traverse through the array from start to end.
3. If the current element is x or y, prev is not equal to -1 and array[prev] is not equal to current element then update m = max(current_index – prev, m), i.e. find the distance between consecutive pairs and update m with it.
4. print the value of m
• Thanks to wgpshashank for suggesting this approach. Implementation.

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `minDist(``int` `arr[], ``int` `n, ``int` `x, ``int` `y) ` `{ ` `         `  `    ``//previous index and min distance ` `    ``int` `p = -1, min_dist = INT_MAX; ` `     `  `    ``for``(``int` `i=0 ; i

## C

 `#include ` `#include // For INT_MAX ` ` `  `//returns minimum of two numbers ` `int` `min(``int` `a ,``int` `b) ` `{ ` `    ``if``(a < b) ` `        ``return` `a; ` `    ``return` `b; ` `} ` ` `  `int` `minDist(``int` `arr[], ``int` `n, ``int` `x, ``int` `y) ` `{ ` `    ``//previous index and min distance ` `    ``int` `i=0,p=-1, min_dist=INT_MAX; ` `     `  `    ``for``(i=0 ; i

## Java

 `class` `MinimumDistance ` `{ ` `    ``int` `minDist(``int` `arr[], ``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

## Python3

 `import` `sys ` ` `  `def` `minDist(arr, n, x, y): ` `     `  `    ``#previous index and min distance ` `    ``i``=``0` `    ``p``=``-``1` `    ``min_dist ``=` `sys.maxsize; ` `     `  `    ``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 previos index  ` `            ``p``=``i ` `         `  `     `  `    ``#If distance is equal to int max  ` `    ``if``(min_dist ``=``=` `sys.maxsize): ` `       ``return` `-``1` `    ``return` `min_dist ` ` `  `  `  `# Driver program to test above function */ ` `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. `

## C#

 `// C# program to Find the minimum  ` `// distance between two numbers ` `using` `System; ` `class` `MinimumDistance { ` `     `  `    ``static` `int` `minDist(``int` `[]arr, ``int` `n, ` `                       ``int` `x, ``int` `y)  ` `    ``{ ` `    ``//previous index and min distance ` `    ``int` `i=0,p=-1, min_dist=``int``.MaxValue; ` `     `  `    ``for``(i=0 ; i

## PHP

 ` `

Output:

`Minimum distance between 3 and 6 is 1`
• Complexity Analysis:

• Time Complexity: O(n).
Only one traversal of the array is needed.
• Space Complexity: O(1).
As no extra space is required.

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

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