Dada una array desordenada arr[] y dos números x e y , encuentre la distancia mínima entre x e y en arr[] . La array también puede contener duplicados. Puede suponer que tanto x como y son diferentes y están presentes en arr[] .
Ejemplos:
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);
}
}
}
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
// 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
// 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);
}
}
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
# 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 - $j);
}
}
}
if($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.
?>
Javascript
<script>
// Javascript program to find the minimum
// distance between two numbers
function 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 code
var 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>
C++
// 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.
C
#include <stdio.h>
#include <limits.h> // 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<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;
}
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<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));
}
}
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 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 = 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<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
// 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.
?>
Javascript
<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>
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)
{
//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
// 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
Python3
# Python program to Find the minimum
# distance between two numbers
import sys
def minDist(arr, n, x, y) :
# idx1 and idx2 will store indices of
# x or y and min_dist will store the minimum difference
idx1=-1; idx2=-1; min_dist = sys.maxsize;
for i in range(n) :
# if current element is x then change idx1
if arr[i]==x :
idx1=i
# if current element is y then change idx2
elif 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 and 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 or idx2==-1 :
return -1
# return the minimum distance
else :
return min_dist
# Driver code
if __name__ == "__main__" :
arr = [ 3, 5, 4, 2, 6, 5, 6, 6, 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 aditya942003patil
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)
{
//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.MaxValue;
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()
{
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 aditya942003patil.
Javascript
<script>
function minDist(arr , n , x , y)
{
//idx1 and idx2 will store indices of
//x or y and min_dist will store the minimum difference
var idx1=-1,idx2=-1,min_dist = Number.MAX_VALUE;
for(var 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 program to test above functions */
var 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 contributed by aditya942003patil
</script>
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA