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.

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.

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

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++

// 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); 
            } 
        } 
    } 
    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 <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

// 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

<?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); 
            } 
        } 
    } 
    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)
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++

// C++ implementation of above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
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 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. 

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, 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)); 
    } 
} 

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. 

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)  
    { 
        int i = 0; 
        int min_dist = int.MaxValue; 
        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 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 
// PHP program to Find the minimum  
// distance between two numbers 
  
function minDist($arr, $n, $x, $y) 
{ 
    $i = 0; 
    $min_dist = PHP_INT_MAX; 
    $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 or $arr[$i] == $y) 
        { 
            $prev = $i; 
            break; 
        } 
    } 
      
    // Traverse after the first occurence 
    for ( ; $i < $n; $i++) 
    { 
        if ($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; 
        } 
    } 
      
    return $min_dist; 
} 
  
/* Driver program to test above fnction */
    $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. 
?> 

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.

My Personal Notes

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

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python3

C#

PHP

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