## Find the minimum distance between two numbers tutorials

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Find the minimum distance between two numbers

Problem:
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)
Our task is to find the distance between two given numbers , So we can find the distance between any two elements using two 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.

## 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[0]); ` `    ``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[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

 `// 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`

Time Complexity: O(n^2)

Method 2 (Tricky)
In our previous approach, we use two loops(without any jumps) which makes out time complexity to be O(n^2). We will try to reduce that to O(n). So our basic approach this time will be only checking consecutive pairs of x and y . We need not check all values of y for a single element (x) . What we can do is , for every element x or y , we will check the index of previous occurrence of x or y and if the previous occurring element is not similar to current element we will update the minimum distance . But a question arises what if a x is preceded by another x and that is preceded by a y ,then how we get the minimum distance between pairs . By analyzing closely we can see that every x followed by a y or vice versa can only be the closest pair (minimum distance ) so we can ignore all other pairs .

Thanks to wgpshashank for suggesting this approach.

## 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

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Output:

`Minimum distance between 3 and 6 is 1`

Time Complexity: O(n)

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

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