Find the minimum distance between two numbers

20 Jun, 2022
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
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: Create a variable m = INT_MAXRun 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).If the ith element is x and jth element is y or vice versa, update m as m = min(m,j-i)Print the value of m as minimum distance
Create a variable m = INT_MAXRun 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).If the ith element is x and jth element is y or vice versa, update m as m = min(m,j-i)Print the value of m as minimum distance
Create a variable m = INT_MAX
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).
If the ith element is x and jth element is y or vice versa, update m as m = min(m,j-i)
Print the value of m as minimum distance
Implementation:
C++
C
Java
Python3
C#
PHP
Javascript
// C++ program to Find the minimum// distance between two numbers#include <bits/stdc++.h>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);            }        }    }    if (min_dist > n) {        return -1;    }    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 program to Find the minimum// distance between two numbers#include <limits.h> // for INT_MAX#include <stdio.h>#include <stdlib.h> // for abs() 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);            }        }    }    if (min_dist > n) {        return -1;    }    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 = 0;    int y = 6;     printf("Minimum distance between %d and %d is %d\n", x,           y, minDist(arr, n, x, y));    return 0;}
// Java Program to Find the minimum// distance between two numbersclass 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);            }        }        if (min_dist > n) {            return -1;        }        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 = 0;        int y = 6;         System.out.println("Minimum distance between " + x                           + " and " + y + " is "                           + min.minDist(arr, n, x, y));    }}
# 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 codearr = [3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3]n = len(arr)x = 3y = 6print("Minimum distance between ", x, " and ",      y, "is", minDist(arr, n, x, y)) # This code is contributed by "Abhishek Sharma 44"
// C# code to Find the minimum// distance between two numbersusing 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// PHP program to Find the minimum// distance between two numbers function minDist($arr, $n, $x, $y){    $i; $j;    $min_dist = PHP_INT_MAX;    for ($i = 0; $i < $n; $i++)    {        for ($j = $i + 1; $j < $n; $j++)        {            if( ($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 - $j);            }        }    }  if($min_dist>$n)      {        return -1;      }    return $min_dist;}     // Driver Code    $arr = array(3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3);    $n = count($arr);    $x = 0;    $y = 6;     echo "Minimum distance between ",$x, " and ",$y," is ";    echo minDist($arr, $n, $x, $y); // This code is contributed by anuj_67.?>
<script> // Javascript program to find the minimum// distance between two numbersfunction minDist(arr, n, x, y){    var i, j;    var min_dist = Number.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);        }    }    if(min_dist>n)      {        return -1;      }    return min_dist;} // Driver codevar arr = [ 3, 5, 4, 2, 6, 5,            6, 6, 5, 4, 8, 3 ];var n = arr.length;var x = 3;var y = 6; document.write("Minimum distance between " + x +               " and " + y + " is " +               minDist(arr, n, x, y)); // This code is contributed by gauravrajput1 </script>
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.
Time Complexity: O(n^2), Nested loop is used to traverse the array.
Space Complexity: O(1), no extra space is required.
Method 2: 
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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: Create a variable prev=-1 and m= INT_MAXTraverse through the array from start to end.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.print the value of m
Create a variable prev=-1 and m= INT_MAXTraverse through the array from start to end.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.print the value of m
Create a variable prev=-1 and m= INT_MAX
Traverse through the array from start to end.
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.
print the value of m
Thanks to wgpshashank for suggesting this approach.
Implementation.
C++
C
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>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<n ; i++)    {        if(arr[i]==x || 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 && 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==INT_MAX)        return -1;     return min_dist;} /* Driver code */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;     cout << "Minimum distance between " << x <<                        " and " << y << " is "<<                        minDist(arr, n, x, y) << endl;    return 0;} // This code is contributed by Mukul singh.
#include <stdio.h>#include <limits.h> // For INT_MAX //returns minimum of two numbersint 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<n ; i++)    {        if(arr[i] ==x || 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 && 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==INT_MAX)       return -1;    return min_dist;} /* Driver program to test above function */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;}
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<n ; i++)    {        if(arr[i] ==x || 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 && arr[i] != arr[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;}     /* Driver program to test above functions */    public static void main(String[] args) {        MinimumDistance min = new MinimumDistance();        int arr[] = {3, 5, 4, 2, 6, 3, 0, 0, 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));    }}
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 previous 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 = 3y = 6print ("Minimum distance between %d and %d is %d\n"%( x, y,minDist(arr, n, x, y))); # This code is contributed by Shreyanshi Arun.
// C# program to Find the minimum// distance between two numbersusing 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<n ; i++)    {        if(arr[i] ==x || 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 && arr[i] != arr[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==int.MaxValue)       return -1;     return min_dist;}    // Driver Code    public static void Main()    {                 int []arr = {3, 5, 4, 2, 6, 3,                     0, 0, 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 anuj_67.
<?php// PHP program to Find the minimum// distance between two numbers function minDist($arr, $n, $x, $y){         //previous index and min distance    $i=0;    $p=-1;    $min_dist=PHP_INT_MAX;         for($i=0 ; $i<$n ; $i++)    {        if($arr[$i] ==$x || $arr[$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 && $arr[$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==PHP_INT_MAX)       return -1;    return $min_dist;} /* Driver program to test above function */    $arr =array(3, 5, 4, 2, 6, 3, 0, 0, 5,                                    4, 8, 3);    $n = count($arr);    $x = 3;    $y = 6;     echo "Minimum distance between $x and ",         "$y is ", minDist($arr, $n, $x, $y); // This code is contributed by anuj_67.?>
<script>function minDist(arr , 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(arr[i] ==x || 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 && arr[i] != arr[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;}     /* Driver program to test above functions */     var arr = [3, 5, 4, 2, 6, 3, 0, 0, 5, 4, 8, 3];    var n = arr.length;    var x = 3;    var y = 6;     document.write("Minimum distance between " + x + " and " + y            + " is " + minDist(arr, n, x, y)); // This code contributed by shikhasingrajput </script>
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. 
Method 3:
Approach:  The problem says that we want a minimum distance between x and y. So the approach is traverse the array and while traversing in array if we got the number as x or y then we will store the difference between indices of previously found x or y and newly find x or y and like this for every time we will try to minimize the difference.
Algorithm: 
   Create variables idx1 = -1, idx2 = -1 and min_dist = INT_MAX;   Traverse the array from i = 0 to i = n-1 where n is the size of array.   While traversing if the current element is x then store index of current element in idx1 or if the current            element is y then store index of current element in idx2.   If idx1 and idx2 variable are not equal to -1 then store minimum of min_dist, difference of idx1 and idx2 into    ans.   At the end of traversal, if idx1 or idx2 are still -1(x or y not found in array) then return -1 or else return              min_dist.
   Create variables idx1 = -1, idx2 = -1 and min_dist = INT_MAX;
   Traverse the array from i = 0 to i = n-1 where n is the size of array.
   While traversing if the current element is x then store index of current element in idx1 or if the current            element is y then store index of current element in idx2.
   If idx1 and idx2 variable are not equal to -1 then store minimum of min_dist, difference of idx1 and idx2 into    ans.
   At the end of traversal, if idx1 or idx2 are still -1(x or y not found in array) then return -1 or else return              min_dist.
Implementation:
C++
Java
// C++ program to Find the minimum// distance between two numbers#include <bits/stdc++.h>using namespace std;  int minDist(int arr[], int n, int x, int y){    //idx1 and idx2 will store indices of    //x or y and min_dist will store the minimum difference    int idx1=-1,idx2=-1,min_dist = INT_MAX;    for(int i=0;i<n;i++)    {       //if current element is x then change idx1       if(arr[i]==x)       {          idx1=i;       }       //if current element is y then change idx2       else if(arr[i]==y)       {          idx2=i;       }       //if x and y both found in array       //then only find the difference and store it in min_dist       if(idx1!=-1 && idx2!=-1)       min_dist=min(min_dist,abs(idx1-idx2));    }         //if left or right did not found in array    //then return -1    if(idx1==-1||idx2==-1)    return -1;    //return the minimum distance    else    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;}
// Java program to Find the minimum// distance between two numbers public class GFG {     static int minDist(int arr[], int n, int x, int y)    {        // idx1 and idx2 will store indices of        // x or y and min_dist will store the minimum        // difference        int idx1 = -1, idx2 = -1,            min_dist = Integer.MAX_VALUE;        for (int i = 0; i < n; i++) {            // if current element is x then change idx1            if (arr[i] == x) {                idx1 = i;            }            // if current element is y then change idx2            else if (arr[i] == y) {                idx2 = i;            }            // if x and y both found in array            // then only find the difference and store it in            // min_dist            if (idx1 != -1 && idx2 != -1)                min_dist = Math.min(min_dist,                                    Math.abs(idx1 - idx2));        }         // if left or right did not found in array        // then return -1        if (idx1 == -1 || idx2 == -1)            return -1;        // return the minimum distance        else            return min_dist;    }     /* Driver code */    public static void main(String[] args)    {        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 "                           + minDist(arr, n, x, y));    }} // This code is contributed by Lovely Jain
Minimum distance between 3 and 6 is 4
Complexity Analysis:
Time Complexity: O(n). Only one traversal of the array is required.
Space Complexity: O(1). No extra space is required.
Find the minimum distance between two numbers | GeeksforGeeks - YouTubeGeeksforGeeks529K subscribersFind the minimum distance between two numbers | GeeksforGeeksWatch laterShareCopy linkInfoShoppingTap to unmuteIf playback doesn't begin shortly, try restarting your device.More videosMore videosYou're signed outVideos you watch may be added to the TV's watch history and influence TV recommendations. To avoid this, cancel and sign in to YouTube on your computer.CancelConfirmSwitch cameraShareInclude playlistAn error occurred while retrieving sharing information. Please try again later.Watch on0:000:000:00 / 15:56•Live•<div class="player-unavailable"><h1 class="message">An error occurred.</h1><div class="submessage"><a href="https://www.youtube.com/watch?v=hoceGcqQczM" target="_blank">Try watching this video on www.youtube.com</a>, or enable JavaScript if it is disabled in your browser.</div></div>
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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